classical mechanics

QUESTION:
Here is the scenario. I have a A volt motor that supplies B watt to work. I connect the motor to a circular disc of R radius that is of M mass and has a density of 1 unit. I want to use the motor to move a conveyor belt of mass N. I will place an object of mass K on the conveyor belt. Is it possible to find the velocity of which the object will be moving at given the voltage? Or perhaps can we find the minimum about of voltage that is needed to move the object. Assume that the efficiency of all the motor as well as the whole thing is 100% and that frictional force and air resistance can be neglected. You can add othervariables in either.

ANSWER:
This is an engineering problem, not physics. However, it sounds to me like there is no answer if you have no frictional losses because the motor will just keep adding energy to the system and there is no loss.

FOLLOWUP QUESTION:
However, it should be able to work since the motor is attached to a conveyor belt, yes it will keep going but it will kep moving the conveyor belt only.

ANSWER:
If the motor is adding energy (B watts would mean B Joules of energy per second are being added) and there are no losses, where is that energy going? It will constantly accelerate the conveyer belt.

FOLLOWUP QUESTION:
Is it possible to find the acceleration of the conveyor belt?

ANSWER:
Yes! Refer to the picture above. Suppose the mass and the conveyor belt move with speed V; then the angular velocity of the disc is ω=V/R. The moment of inertia of the disk is IMR2. The kinetic energy of the belt plus mass is KE1=½(K+N)V2 and of the disc is KE22MV2, so the total kinetic energy is KE=½(½M+K+N)V2. The rate of change of the kinetic energy is equal to the power B, B=(½M+K+N)V(dV/dt)=(½M+K+N)Va where a is the acceleration. Solving, a=B/[(½M+K+N)V]≡C/V. Note that when V=0, a=∞; this just means that the motor cannot deliver power unless it is actually spinning with some rate. But, as soon as it gets moving it will have a very large acceleration which decreases as the speed gets bigger and bigger. So, if you are really interested in such a system, you will have to determine energy losses and probably also have a motor with an adjustable power output. (Also, note that the voltage has nothing to do with it, all you need is the power output.)


QUESTION:
If you have two weights, where one is 100 grams and the other is 200 grams and you use the same spring to create a pendulum with them (one at a time) why isn't the amplitude the double for the heavier weight? (This is the result of an exercise we did, my personal curiosity, and my teacher's unwillingness to explain because he says it's too complicated for our level.)

ANSWER:
I think you must surely mean a mass on a spring oscillating vertically, not a pendulum. That is what I will assume. You have probably not studied energy yet which is why your teacher did not want to get into it. Energy methods are, by far, the easiest way to answer your question which is what I will do. Two things you need to know: potential energy of a spring which is stretched by an amount y is ½ky2 and gravitational potential energy of something a distance y above where y=0 is mgy. Imagine taking a mass m and attaching it to a spring with spring constant k which is unstretched and holding it there for a moment. Where it is right now I will define to be y=0 and so, since the spring is unstretched, the total energy of the system is ½k∙02+mg∙0=0; the energy never changes. As it falls it speeds up for a while and then slows down for a while (acquiring and then losing what is called kinetic energy, energy by virtue of motion), finally stopping and going back up. If it has fallen some distance A (for amplitude) before turning around, the energy is now 0=½kA2+mg(-A) and so we find two solutions (it is a quadratic equation), A=0 (we already knew it was at rest there) and A=√(2mg/k). So, you see, the amplitude is proportional to the square root of the mass, not the mass. (Incidentally, most folks would call the amplitude of this oscillation to be ½A. I thought it would be clearer this way.)


QUESTION:
Is it possible to derive the formula for Kinetic Energy without using work? Or are they linked by definition?

ANSWER:
Well, I can tell you that you never have to utter the phrase "work is defined as…" The work-energy theorem is merely the integral form of Newton's second law. For simplicity, I will do this simple derivation in one dimension. F=ma=m(dv/dt)=m(dv/dt)(dx/dx)=m(dv/dx)(dx/dt)=mvdv/dx. Rearrange: Fdx=mvdv. Integrate:
Fdx=mvdvmv22mv12 where the integral on the left is from x1 to x2 and is usually called the work W. Generalizing to three dimensions, F∙drmv22mv12 where the integral on the left is from r1 to r2.


QUESTION:
I came across your site while looking for the answer of a physics problem I would like to program in a Smartphone app. A toy car is dragged from position A to B giving it an initial velocity (v=d/t). At point B the car is released where it travels in a straight line until the frictional force of the ground stops the toy car completely (position C). I would like to find a formula that relates the distance from B to C with the initial velocity provided from A to B.

ANSWER:
Of course, you have to know the frictional force f and the mass m of the car in general. Assuming that the force is constant all along the path B to C, the distance s can be written smv2/f. In most cases f is proportional to m so that you do not need to know the mass. If you are on level ground, f=μmg where μ is called the coefficient of friction and g=9.8 m/s2 is the acceleration due to gravity (mg is the weight of the car). The coefficient just is a parameter which is small if there is little friction. So, finally you have s=v2/(2μg). If the car is on a slope making an angle θ with the horizontal, s=v2/(2μg∙cosθ).


QUESTION:
Okay I have wondered this question for many years, and more over have wondered how to ask it! So I think a scenario is best; when an object is propelled forward or in any direction very rapidly, let's say for this case a bullet out of a gun, does it advance through every speed in between it's current speed, zero, and it's maximum velocity or does it simply "jump" from zero to it's maximum velocity? And also the same for when it hits its target, let's say a thick steel plate does it go from its current velocity straight to zero, or is there some sort of slowing down in which it goes through every speed in between?

ANSWER:
Fundamental to classical physics is Newton's second law. That which changes the speed v of something is a force F, a push or a pull. Further, the bigger the rate of change of speed (acceleration a), the bigger the force—double the force and you double the acceleration. This is often written as F=ma where m is the mass of the object. Suppose that you shoot a bullet from a gun. The bullet starts with a speed v=0 and ends with speed v and this happens in some time t and so the acceleration can be written as a=v/t. If, as you suggest, the bullet "jumps" instantaneously to v, then t=0. But for t=0 the acceleration would be infinitely large which would imply that you had to push on it with an infinite force. The same reasoning can be applied to stopping the bullet. You know that the force propelling or stopping a bullet in the real world is finite. It is a pretty good rule-of-thumb for everyday occurances that there are no infinities or discontinuities (instantaneous changes) in the universe.


QUESTION:
Please help a work related dispute. Can the weight of a patient in a wheelchair be calculated using f=Ma? Even if it's a guesstimate? I thought, if m= F/a Where a = V2-V1 / t Where V2 is the velocity of the chair (and porter?) and V1 is coming to a stop. Would F be the mass of the porter multiplied by the common acceleration of chair and porter? I'm guessing if I clock the speed of the porter I can calculate a.

ANSWER:
I guess that "porter" means the guy pushing the chair. You have it all wrong, I am afraid. But, it is worth talking about for a bit. There are three masses involved here, Mpatient, Mchair, and Mporter. There is an acceleration a which we can agree could be roughly measured by measuring times and distances. Suppose we first look at all three and call the sum of their masses M. Then M=Fall/a where Fall is the force which is causing the collective mass to stop. If you neglect the friction which would eventually stop the wheelchair with no porter, Fall is the frictional force between the porter's feet and the floor; you do not know that force. Suppose you focus your attention on the patient. Mpatient=Fpatient/a where Fpatient
is the force responsible for stopping the patient. This would be the frictional force the seat of the chair exerts on the patient's butt; you do not know that force. Suppose you focus your attention on the porter. Mporter=Fporter/a where Fporter is the force responsible for stopping the porter. If we call the force with which the chair pulls on the porter Fc-p, then Fporter=Fall-Fc-p; you do not know either of these forces. I could go on and focus on the chair alone next, but you can see that you do not know any of the forces which are responsible for stopping any or all of the masses, so you cannot infer the mass of any of them. Just knowing the acceleration, you cannot infer the mass.


QUESTION:
I am writing a couple of SF novels that take place in significant part on a beanstalk station, i.e. a space elevator station located in geosynchronous orbit, and my question concerns the design. In my stories, the station is about a half mile wide with a shape similar to a hockey puck. The center third of the puck is attached to the elevator ribbon (both up to the counterweight and down to the planet of course) and fixed, therefore essentially in freefall. My question concerns the outer section, which in my story spins in order to provide some g-force: is this a viable model? Some of my concerns are that the spin will create a gyroscopic effect as the station orbits around the Earth that will twist it off the ribbon or perhaps exert other forces on the hub that I'm not accounting for. One possible solution I thought of would be to split the outer section into two rings rotating in opposite directions to cancel out those effects. Assuming some clever engineer works out the minor details of moving between the hub and the rotating section(s) is this design plausible? I like the science in my science fiction to be accurate and reasonable.

ANSWER:
I believe you are right to worry about the rotation causing problems. If you just forget about the elevator and have a rotating "puck station" in orbit, its axis will always point in the same direction in space (conservation of angular momentum); so, if it were in a geosychronous orbit, from earth it would appear to do a 3600 flip every day. I think that having a counter-rotating ring would be a very good idea, the ring having the same angular momentum (moment of inertia times angular velocity) as the main station. If the station has a radius of about a quarter of a mile, about 400 m, the angular velocity ω needed to have an earth-like artificial gravity (g=9.8 m/s2) at the edge would be ω=√(g/R)=0.16 radians/s=0.025 revolutions/s≈1.5 rpm.


QUESTION:
I need a very brief explanation for Kindergarteners. (I just want to use the right words; they can learn the details of what they mean later.) In an amusement park ride in which long swings are spun around a central pole, why do the swings rise up as the pole spins faster? (I assume it's the same reason that a skirt rises up when the person wearing it twirls.) Is it something to do with centrifugal force? Centripetal force? Nothing I saw on the web about these forces seems to explain why the objects rise, only why they move to the inside or outside of the orbit.

ANSWER:
For Kindergarteners, use of centrifugal force is probably best. I guess the merry-go-rounds we had on playgrounds when I was a kid have been deemed unsafe, but get your students to appreciate that the faster they are spun on something akin to this, the harder it is to hang on. This is because the centrifugal force which tries to push them off gets bigger as the merry-go-round spins faster. Then you can just take a pendulum and demonstrate that the harder you pull, the higher it rises. The little diagram to the right shows the effect of the the centrifugal force F which will lift the pendulum bob higher as it pulls harder.


QUESTION:
I have a question that been thats been unanswered from quite some time now. I even tried asking that to a person in NASA houston and he couldn't even understand. Hope you can help. We know that space dust and debris keep on falling on earth. And I have read that its many tons a day. Considering mass of earth is increasing every day for millions of years now, how does it affect its revolution, speed and axis. Or does it even affect that.

ANSWER:
One estimate is that the earth gains 40,000 metric tons (4x107 kg) per year. So, in a million years that would be a gain of 4x1013 kg. The mass of the earth is 6x1024 kg. The moment of inertia will be proportional to the mass times the square of the radius, so, assuming the change in radius is negligible, the fractional change in moment of inertia over a million years would be approximately 4x1013/6x1024≈10-9=10-7 %. Since angular momentum (moment of inertia times angular velocity) is conserved, this would mean that the angular velocity would decrease by about 10-7 %, an increase in the length of a day of about (24 hr)(3600 s/hr)x10-9≈10-3 seconds in a million years!


QUESTION:
I was explaining conservation of energy to my daughter when she was spinning on our office chair, and her rotational velocity increased as her moment of inertia decreased when she pulled her arms and legs toward the rotational axis. Is there an analogous example for linear motion, where the linear velocity increases as the mass decreases? I can't think of a real-world example of a body whose mass decreases (or increases).

ANSWER:
Actually, you were demonstrating conservation of angular momentum, the product of the angular velocity and moment of inertia remains constant for an isolated system. Energy is not conserved because rotational energy is proportional to the product of the square of the angular velocity and moment of inertia. But, you explained it correctly for your example. So, your question is if it is possible for conservation linear momentum, the product of mass times velocity, to result in a changed velocity due to a change of mass. You do not usually see as many examples of this as for angular momentum because rotating things change shape frequently whereas moving objects usually do not have significant change of mass. The classic example is a conveyer belt onto which mass is being dropped from a hopper. If you had a very long frictionless conveyer belt with mass on it, it would have a certain linear momentum. Now, if you start dropping mass on it, it will slow down. Similarly, if you let mass drop off the end of the conveyer belt without replacing it, it will speed up.


QUESTION:
How long does it take to stop a 7000 lbs vehicle at 45 mph? Time and distance please.

ANSWER:
You have not given me enough information, in particular what are the wheels and the surface made of; surely you realize that a truck stopping on ice will go much farther than a truck stopping on a dry road. I will work it in general and then calculate it for a typical example. The quickest stop you can affect is to apply the brakes hard enough that the wheels are just about to start skidding; that is what anti-skid braking systems do. Therefore the force F which is stopping you (on a level road) is the static friction between the road and the wheels, F=μsW where μs is the coefficient of static friction between the wheels and the road, W=Mg is the weight of the vehicle, M is the mass, and g=32 ft/s2 is the acceleration due to gravity. But, Newton's second law tells us that also F=Ma where a is the acceleration of the vehicle. Therefore, the acceleration is independent of the mass of the vehicle and the acceleration is a=μsg. Now that you have the acceleration you can write the equations of motion for position x and velocity v as functions of time t: x=v0tat2 and v=v0-at where v0=45 mph=66 ft/s. Now, you need to specify what μs is. For example, μs0.9 for rubber on dry asphalt, so a≈0.9x32=28.8 ft/s2 and I find t≈66/28.8=2.3 s and x=66x2.3-½x28.8x2.32≈76 ft.

FOLLOWUP QUESTION:
I'm trying to figure out if a traffic light at an intersection is timed too short for a heavier vehicle to stop in time. The speed limit is 45mph, the vehicle weighs ~7000 lbs (7200 empty w/driver) , there is a downward slope of which I'm trying to find out. Yes to dry pavement. What would you need to figure surface area of rubber/tire? And I need to time the light still. With all of the above information would you be able to calculate that?

ANSWER:
As I showed above, the weight is irrelevant. Of course, this is an approximation as all friction calculations are, but a quite good one for this situation. It is also important that my calculation is the shortest time and minimum distance, what you would get by flooring the brake pedal with anti-skid braking operating. If you do not have anti-skid braking and you lock your wheels, it will take longer and go farther. Also, there is the possibility that μs could be different from 0.9 depending on local conditions (temperature, surface condition, etc.). It is important to include the slope in the calculation. If the slope is down as you say, the acceleration (I am assuming you are not interested in the details) is a=32(μscosθ-sinθ) where θ is the angle of the slope; for example, if θ=200 and μs=0.9, a=32(0.9x0.94-0.34)=16.2 ft/s2, quite a bit smaller than the value of 28.8 ft/s2
for a level road. Once you calculate the acceleration, the expressions you can use for time and distance are t=66/a and x=2178/a, respectively. So, for a 200 slope, t=4.1 s and x=134 ft. Again, this is the lower limit. I think an engineer would build in a factor of 2 safety factor. You do not need to know what the surface area of the contact between the rubber and the road is.


QUESTION::
Nothing can accelerate itself by applying force on itself. Besides, isolated forces do not exist and they exist in an action reaction pair. These are the essence of Newton's third law. When an engine exerts force on a bike how can it accelerate the bike taking bike as a system in this argument? After all, a part is exerting force on another part , right?

ANSWER:
The engine exerts a torque on a wheel trying to make it spin. If the bike were on ice, the wheel would spin, there would be no acceleration and therefore there must have been zero net force on the bike+engine as you surmise. However, if there is friction between the wheel and the ground, the ground exerts a force on the wheel which is forward; this is the force which drives the bike forward. Note that the wheel exerts a force backward on the ground (Newton's third law) but the ground does not move because it is, effectively, infinitely massive.


QUESTION:
When a bullet hits a door and gets embeded in it, no external force acts on the system of door and bullet but why is linear momentum not conserved and the angular momentum conserved? Can you give some examples where angular momentum is conserved but not linear? Can it be possible to have both linear and angular momentum conserved?

ANSWER:
I presume you are alluding to the classic introductory physics problem of a bullet hitting a door mounted on frictionless hinges. Angular momentum is conserved if there are no external torques, and since hinges cannot exert a torque, it is. Linear momentum is conserved if there are no external forces but the hinges exert a force on the door during the collision time and so it is not. If the hinges were not there, the door and the bullet would move forward and rotate about their center of mass conserving both linear and angular momentum.


QUESTION:
What is the change in velocity of the earth's rotation if a person (myself) who weighs 60 kg were to stand on something about a foot tall. Ps this is not a homework question, I'm just a curious teen who's never taken physics. Also it's the middle of the summer for me in New Orleans.

ANSWER:
The glib answer to this question would be, for all intents and purposes, the change in rotation would be zero. It is a good opportunity  to talk about the physics involved and to estimate how small small is here. The moment of inertia of the earth is about Ie=8x1037 kg∙m2. Your moment of inertia if you are on the earth's surface is about Iy=60x(6.4x106)2=2.5x1015 kg∙m2. The moment of the earth plus you is I=Ie+IyIe=8x1037. If you increase your distance by the amount 0.3 m, about 1 ft, your moment of inertia increases to IyI=60x(6.4x106+0.3)2=Iy(1+4.7x10-8)2≈2.5x1015(1+2x4.7x10-8)=2.5x1015I and so ΔI=9.4x10-8 kg∙m2. So you and the earth start with I=8x1037 and end with II=
8x1037+9.4x10-8. The operative physical principle here is conservation of angular momentum, the product of moment of inertia and angular frequency ω=2π/T where T is the period, 24 hours: Iω=(II)(ω+Δω)=Iω+IΔω+ΔIω+ΔIΔω. Neglecting ΔIΔω, Δω/ω=-ΔI/I=-1.2x10-45; note that since Δω/ω<0, the frequency decreases, the rotation slows down. Now, it is pretty easy to show that  ΔT/T≈-Δω/ω=1.2x10-45 or the day gets longer by 24x1.2x10-45=2.8x10-44 hours! Your contribution to the earth's moment of inertia is so tiny that anything you do to change your own moment of inertia will have no measurable effect on the rotation of the earth.


QUESTION:
If an astronaut caught a ball in space, the ball would cause the astronaut to move backwards with the force the ball was moving at correct? then, as the astronaut is still moving backwards and throws the ball back where it came from, would the astronaut move even faster with the force of throwing the ball? or would their speed remain the same?

ANSWER:
When she catches the ball, she exerts a force on it to stop it (relative to her); the ball exerts an equal and opposite force on her causing her (and the caught ball) to move in the direction the ball was originally moving. When she throws it back, she must exert a force on it opposite the direction she is moving; the ball exerts an equal and opposite force on her causing her to move even faster in the direction she was moving. These are examples of Newton's third law.


QUESTION:
What is mass?

ANSWER:
There two kinds of mass. Inertial mass is the property an object has which resists acceleration when a force is applied; the harder it is to accelerate something, the more inertial mass it has. Gravitational mass is the property an object has which allows it to feel and create gravitational forces; for example, the more gravitational mass an object has the greater the force it will feel due to the earth's gravity—the more it will weigh. It turns out that the two masses are actually identical; this fact is one of the cornerstones of the theory of general relativity.

QUESTION:
My question is for a legal nature. I was recently in a head on collision. The crime scene analysis could not determine the speed of my vehicle but police reports indicated the other vehicle was doing 45+- mph. We were on a dirt road on impact. The vehicle going 45+- continued to travel 20 ft past impact pushing my vehicle back 40 ft. His vehicle is a 1 1/2 ton dodge ram and my vehicle was a 1 3/4 ton dodge caravan. What speed was my vehicle? This will be instrumental in a lawsuit currently being filed. Your help would be greatly appreciated.

ANSWER:
I am afraid that the speed cannot be determined from this information for several reasons:

  • I do not know how much energy was lost in the collision.

  • I do not know whether the wheels were locked (brakes applied) or not.

  • I do not know the frictional forces which ultimately brought the vehicles to rest.

  • I believe the weight of the Dodge Ram is wrong. The Dodge Ram 1500 is referred to as 1½ tons because this is the maximum recommended load. An unladen Ram 1500 weighs more than 6000 lb; I found that the weight of the Caravan is about 1.75 tons, though.


QUESTION:
Friction opposes the relative motion between two surfaces. when a car travels on a circular path , how can the friction act sideways to provide necessary centripetal force. The friction should act backward relative to the motion of the car. the car doesn't tend to go sideways outward, then, how does friction act sideways?

ANSWER:
There are two important classes of friction, kinetic friction which occurs when two surfaces are sliding on each other and static friction when they are not sliding. Kinetic friction is the one which usually (but not always) acts opposite the direction of motion; an example of kinetic friction acting in the direction of motion is a car which is accelerating from rest and spinning its wheels—the friction force on the spinning wheels by the road is forward. Static friction can point in any direction, depending on the situation. If a box is sitting at rest on an incline, the frictional force points up the incline to keep it from sliding down. If a car is moving but not skidding, the appropriate friction to think about is the static friction between the wheels and the road. Think about a very icy road; to drive around a curve at high speed is impossible because there is no static friction and the car simply continues going straight regardless of whether you turn the steering wheel or not.

QUESTION:
Gravitational potential energy is the term that means the work done by the gravitational force to take an object to the gravitational field. Here the displacement is towards the force.So,it (Gravitational potential energy) should be positive.But it is negative.Why?

ANSWER:
You need to be a little more careful in how you define potential energy. And, what is actually defined is the potential energy difference between two points in space. The definition is ΔU=U(r')-U(r)=-
rr'Fdr, where F is the force of gravity on m due to the presence of M. Now, if I choose increasing r to be in the upward direction, Fdr=-(MmG/r2)dr. So, -rr'Fdr.=-rr'[-(MmG/r2)]dr=-MmG[(1/r')-(1/r)]=U(r')-U(r). This is completely general. It is customary to choose U(∞)=0, so if r'=∞, U(r)=-MmG/r. (You have to be very careful of all these minus signs!) So, you see that the potential energy is determined by where you choose it to be zero and the choice of coordinate system; if we had chosen r to increase in the downward direction and U=0 infinitely far away (at r'=-∞), U would have been everywhere positive.

QUESTION:
ok, two stones. Both spherical and same mass and density evenly spread in each stone. Set each about 1/2 the distance to the moon. One leading the earth's orbit and one following the earth's orbit. Both not moving relative to the earth, yet the same speed as the earth as it moves around the sun. No tangent or orbital speed, the stones are starting in freefall. Which gets to earth's surface first, neglecting air drag.

ANSWER:
Since you stipulate that the stones are not orbiting, the stones are at rest with respect to the sun and the earth is not. Therefore, the stone on the leading side of the orbiting earth will win the race because the earth is moving toward it and away from the other when the stones start dropping.

BETTER ANSWER:
I see that I misread this question. I guess you meant there is no orbital speed around the earth. To make this manageable at all I will neglect the influence of the moon and assume a spherically-symmetric mass distribution of the earth. In the figure above, I show your two stones and the forces (blue arrows) on them. The down-pointing forces are the from the sun (keeping them in orbit) and the horizontal forces are the weights making them want to fall toward the earth. The distances and forces are not drawn to scale; when the stones are about 30 earth radii away from the earth center (the moon is about 60 earth radii away), the weight forces are about 2 times larger than the sun forces. The big blue arrow shows the direction everyone is orbiting the sun. So, the leading stone slows down its orbital speed and so will slightly fall toward the sun as it falls toward the earth; the trailing stone increases its orbital speed and so will slightly fall away from the sun as it falls toward earth. These deflections are shown (probably quite exaggerated)  by the red arrows. Given the symmetry of the situation, I would expect the two to be at the same distance from the center of the earth at any given time—which is the crux of your question, I think. To actually do this more quantitatively, though, would be very hard because as the stones were deflected the weight force would change direction now having a vertical component in the figure.


QUESTION:
I have a 10' long trailer I am using to haul a 4,000# symetrical object that is 6 ft in length. I was told to move it forward of the axle on the trailer to put some weight on the tongue for safety and better hauling. The maximum wt my pick up truck can hold on its hitch is 650#. I have tried to find equations for this on-line as it would be very useful for me to know how to adjust other loads as well. There are plenty of equations on-line that deal with finding the CG, but I can't find one that discusses how the weight on the tongue changes as the load is moved fore or aft of the axle. Trailer/towing experts just wing it. Do you know of an equation that would help me?

ANSWER:
I will assume that the unladen trailer will have approximately zero force on the hitch (which would mean that its center of gravity (COG) is at the axle). You need to know where the COG of the load is; in your specific case, you know that it is at the geometrical center (you said it is symmetrical), 3' from either end. For loads not symmetrical, you need to find it. I will call the distance between the axle and the hitch L. Suppose that the COG is a distance x from the hitch. Then, the sum of the torques about the axle must be zero, so Hx=W(L-x) where H is the force (up) by the hitch and W is the weight (down) of the load. Solving this, x=[W/(H+W)])L. For your case, with H being 650 lb (although I cannot see why the maximum would be the optimal) x=(4000/4650)L=0.86L. If your trailer has more than one axle or its COG is not over the axel, I would need more information like the geometry and weight of the trailer.

QUESTION:
Here's a question that I have been pondering. If a truck is driving on the freeway and a car pulls in closely behind the moving truck to take advantage of the draft created by the truck is there an energy cost to the truck or is having a car in it's wake energy neutral? My gut feeling is that there would be a slight energy cost to the truck due to it's turbulence wake being interfered with.

ANSWER:
This, I discovered, is not a trivial question. For a lengthy discussion, see The Naked Scientists. Here is my take on it. There is no question that the trailing car consumes less gas. The reason for this is not so much that the truck is pulling the car but that the car experiences a much lower air drag when drafting; the drag is approximately proportional to the square of the velocity and the truck's wake is moving forward with the truck. What seems to be controversial is the crux of your question
—is there a cost to the truck? Some argue that the composite truck-car system has less total air drag, others that there is a net cost to the truck which need not (and almost certainly will not) equal the gain by the car. There is certainly no conservation principle here because the new system has different forces on it than the separate systems. My feeling it that there is at least a small cost to the truck and I base this on an observation from nature. Why do geese fly in a V? There is less overall air drag than if the flock all flew individually. But periodically, the leader drops back and another goose takes a turn at the front; must be because the leader has to do more work.

QUESTION:
A friend asked me this and we disagreed with the answer. If we put 25 kg of weight on top of 25 kg person, how much force would he feel?

ANSWER:
Technically, a kilogram is not a weight but a mass. But, since so many countries use it as a weight, I will do that for this problem. The person feels the downward force of her own weight, 25 kg; the downward force of the object pushing down on her, 25 kg; and the upward force of the floor pushing up on her, 50 kg. The net force is zero because she is in equilibrium.

QUESTION:
If an ultra high energy cosmic ray with energy of 1020 eV were to strike an astronaut will that kill an astronaut?

ANSWER:
This is only about 6 J of energy. That is the energy needed to lift 1 kg about 60 cm. And, it would probably not leave all its energy in the astronaut. Certainly would not kill her.

QUESTION:
If an ultra high energy cosmic ray with energy of 1020 eV were to strike an astronaut will that kill an astronaut?

ANSWER:
This is only about 6 J of energy. That is the energy needed to lift 1 kg about 60 cm. And, it would probably not leave all its energy in the astronaut. Certainly would not kill her.

QUESTION:
I have a question releated to weight/mass placement on a bar. My friend and I are weight lifters. We got into a discussion about the center of gravity on the bar. Here is the question. If we are using a 45 pound plate on each side and also have a 5 and 10 on each side. Each taking up the same space and the end of the bar is the same distance from the last weight and will not change. Does it change anything if the weights are not in the same order, from one side to the other? My friend says the side with the 45 pound plate close to the end is slightly heavier becuase the ratio has changed. I say nothing has changed becuase the weights on the bar are still taking up the same space. I believe it would only change if the distance to the end of the bar is changed, which it is not. I hope I explained this well enough.

ANSWER:
Assuming that the bar itself is uniform (has its center of gravity (COG) at its geometrical center), the COG of the total barbell depends on the location of the weights. Relative to the center of the bar, the position of the center of gravity may be written as COG=(45x1+10x2+5x3-45x4-10x5-5x6)/120 where the xis are the distances of weights from the center. Suppose that the weights are placed symmetrically (x1=x4, x2=x5, x3=x6); then COG=0, the center of the bar. Now, suppose we interchange two of the weights, exchange the 45 lb with the 10 lb on one side: COG=(45x2+10x1+5x3-45x4-10x5-5x6)/120=(45x1+10x2-45x2-10x1)/120=(35/120)(x1-x2); since x1x2, COG≠0, the barbell is no longer balanced. If that explanation is too mathematical for you, try a more qualitative argument. Each weight W a distance D from the center exerts a torque about the center and the magnitude of that torque is WD. The net torque due to all weights must be zero if the bar is to balance at its center. This means that the sum of all the WDs on one side must be precisely equal to those on the other if the barbell is to be balanced about its center. If you change the Ds on only one side, the bar will not be balanced at its center. (This qualitative argument is just the mathematical argument in words.) What certainly does not change is the total weight.

QUESTION:
After watching the Bond classic YOU ONLY LIVE TWICE, I read that the scene where a craft in space overtakes a capsule ahead in the same orbit in order to "swallow" it, but would be impossible because it would have to be in a separate orbit. Then when it catches up, turn vertical and move upward. Why is it not possible for an object to accelerate in the same orbit as a slower object?

ANSWER:
If the two satellites were in the same orbit, they would maintain the same separation. If they were in different but crossing orbits, you could have them come together if properly synchronized; if you were to observe this, say from the perspective of the "chased" satellite, it would appear that the other satellite was coming at you from slightly above or below. Finally, if the "chasing" satellite had rockets which he could point in any direction with any thrust, he could move exactly on the same path as the "chased" satellite but with a different speed.

FOLLOWUP QUESTION:
I don't grasp the physical law that would prevent the "chasing" capsule to catch up in the same orbit (as we see in the movie) if its thrusters accelerate it.

ANSWER:
It is easiest if we just think about circular orbits; near-earth orbits are nearly circular and I will consider only orbits whose altitude is very small compared to the radius of the earth. With each orbit there is one special speed v for an orbiting satellite where the centripetal force equals the weight, mv2/R=mg or v=√(gR) where g is the acceleration due to gravity, R is the radius of the orbit (approximately Rearth), and m is the mass of the satellite. If you are going faster or slower than that you will not be in that circular orbit but some elliptical orbit which happens to cross the circular orbit. But, let us just suppose that you are going a little faster than v, say v+u where u<<v; you got there by briefly firing your rockets out the rear tangent to the orbit. If you do nothing else, you leave that orbit. However, the force necessary F to keep you in that orbit would be m(v+u)2/R=F. But part of F is the weight, so you can write F=mg+f where f is what your rockets have to do. Therefore f=m(v+u)2/R-mg=m(v2+2vu+u2)/R-mg. Now, mv2/R=mg from above and you can neglect u2 because it will be very small compared to 2uv, so f≈2muv√(g/R); you would have to point your rockets away from the center of the earth so that this force would be down but you would keep on that circular orbit going faster than other satellites in that same orbit.

QUESTION:
For a statement to be a law it must be based on observations and experiments. Newton, certainly didn't perform experiments to verify his universal law of gravitation. Was it correct then to state it as a law?

ANSWER:
Newton may not have done the experiments, but his law was the result of experiments done by others. Most important were Kepler's three laws which were empirical summaries of a large body of data on the motions of the planets. His law of gravitation, F=-MmG/r2, provided a complete explanation of Kepler's laws. However, since the mass of the sun was not known, only the product MG could be determined from the data. A good measurement of G was not done until more than 70 years after Newton's death. Because gravity is such a weak force, this is a very difficult measurement to make on a laboratory scale.

QUESTION:
This question is in regards to flowing water and buoyancy. Lets say I have two connected reservoirs at different heights, therefore creating a pressure difference and fluid flow between them. There is a pump that refills the higher reservoir so the flow is constant, and there is a section of tubing that is vertical. If I were to put a buoyant object like a balloon in the vertical section of tubing with flow, could I keep it from floating to the top with enough flow? Basically can fluid flow in the opposite direction of the buoyancy force keep it from floating. I feel like it can but I am having trouble understanding why (seems like the only factor is density/displacement, maybe fluid flow increases drag?), just curious because my friend and I got into a random debate about it.

ANSWER:
Usually when we think of buoyant forces we are thinking about fluid statics, all fluid at rest. Your balloon in the tube will experience a buoyant force up and a force down from its weight just as it would in a nonmoving fluid. If the water is moving down, the balloon will also feel a downward force due to the drag it experiences. What this drag force is will depend on the size of the balloon, the size of the tube, and the speed of the water. It would be very complicated to calculate, but I am sure there would be a correct speed for any geometrical situation where the balloon would remain stationary. If that is enough, you can stop reading here. If not, here is an example below:

Basically, this is just a terminal velocity problem. Suppose that we imagine just releasing a spherical balloon with radius R, volume V=4πR3/3, cross sectional area A=πR2, and mass m under water. The net force upward would be F=ρgV-mg=4ρgπR3/3-mg where ρ=1000 kg/m3 is the density of water and g=9.8 m/s2 is the acceleration due to gravity. The drag force can be approximated as f=πv2R2ρCd/2 where the drag coefficient for a sphere is Cd=0.47 and v is the speed of the balloon. So the net force is Fnet=4ρgπR3/3-mg-πv2R2ρCd/2 and this is zero when v=vt, the terminal velocity, vt=√[(8gR/(3Cd))-2mg/(πR2ρCd))]; the second term in the square root is much smaller than the first because the mass of the balloon is very small (about 0.04 kg if R=0.1 m and the air is at atmospheric pressure) compared to the buoyant force. This is how fast a balloon would rise in still water. So, that would be the speed the water would have to be moving down for the balloon to stay in place. I did a rough calculation for R=0.1 m and found vt≈2.4 m/s. These estimates are all for the size of the pipe much greater than the size of the balloon. Things get much more complicated if that is not the case, but you would still be able to use the water flow to keep the balloon in place.


QUESTION:
Using real-world estimates for the coefficient of friction between his feet and the ground, how fast could the Flash run a quarter-mile? Assume that the limiting factor for his acceleration is the force parallel to the ground that his feet can apply.

ANSWER:
Suppose he is running on a dry asphalt road with rubber-sole shoes. Then the coefficient of static friction is approximately μ≈0.8. The maximum force of friction on level ground would be fmaxμN=μmg≈8m where m is his mass. So, his acceleration would be a=fmax/m=8 m/s2. A quarter mile is about 400 m, so assuming uniform acceleration the appropriate kinematic equation would be 400=½at2=4t2, so t=10 s.

QUESTION:
I have a doubt about static friction and number of wheels. As for elementary physics principles
1) static friction depends is mass times the coefficient of static friction
2) static friction does not depend on surface static friction is independent by the number of wheels.
... but it is hard to accept to me! Let's suppose to design a cart to be pushed by a worker. The total weight (cart + content) is about 1000 kg. The question is: as for the static friction it is better to use 4 or 6 wheels?

ANSWER:
If the cart is to be "pushed by a worker" it is not static but rather kinetic friction which is in play unless all the wheels are locked. And this is not friction due to the contact between the wheels and the ground but friction due to the axles rubbing on the wheels. But, let's talk about friction anyway because you seem to have a serious misconception. First of all, the friction is proportional to the normal force which presses the wheel to the road, not the mass. If there were one wheel, the maximum static frictional force you could get before the cart started slipping (call that fmax) would be the weight W times the coefficient of static friction μs (on level ground), fmax=μsW. If you had two wheels, each wheel would hold up half the weight so the maximum static frictional force you could get from each wheel would be μsW/2; but the total force is still μsW. Things are more complicated on a slope, but the conclusion is still that you do not gain an advantage regarding traction by having more wheels. The reason big trucks, for example, have many wheels is so that each wheel does not need to support so much weight, not to get more traction.

QUESTION:
I understand that acceleration due to gravity decreases with distance, specifically by the inverse square law. That being said, what is the maximum distance for which one can use 9.81 m/s2 as g for Earth?

ANSWER:
That depends entirely on how accurate you want to be, there is technically no place other than the surface of the earth where this is the acceleration. Furthermore, the number 9.81 is simply an average value; it varies over the surface of the earth due to local density variations, rotation of the earth, influences of the moon's gravity, altitude variation, etc. You need to ask something like "at what altitude h from the surface is the value of g changed by X%?" Then
X/100=((1/R)2-(1/(R+h)2))/(1/R)2 where R is the radius of the earth. Provided that h is small compared to R, you can solve this equation approximately as h≈XR/200. For example, g will be reduced by 2% when h≈R/100. Another example: the International Space Station is at an altitude of about 230 miles, about 6% of the earth's radius. Then XISS200x0.0612% smaller than 9.81 m/s2.

QUESTION:
If I am driving my car with a bowling ball in the trunk, does it take the same energy to accelerate the vehicle to a given speed at a given time if the ball is free to roll around as it would if it were fixed to the vehicle? I assume that the net energy use would be the same in both situations (same total vehicle mass), but the acceleration rates would be different - ie: the fixed ball would result in a constant acceleration to speed, while the rolling ball would result in a non-constant acceleration. If this is true, could I harness the energy of the ball's movement relative to the vehicle (using some sort of linear generator) without causing parasitic energy loss to the vehicle?

ANSWER:
As long as the ball and the car end up going the same speed, the total energy to get them there is the same (neglecting frictional and air drag forces). If you devise some way to take enegy away from the ball, that energy ultimately must come from the engine.

QUESTION:
Perhaps you can help solve a disagreement we have at work. The question being "Does a person's initial velocity during a jump equal their final velocity once the land?" My contention is "no" in that the jumper could theoretically produce any velocity on the way up, but downward would be limited to terminal velocity. Who's right?

ANSWER:
Technically, you are correct. If air drag is present, energy is lost which results in the landing speed being less than the launch speed. In practice, however, for a person jumping into the air the height acquired is not high enough for this to be a measurable effect; that is, this is an example where we can say, as we often do in an elementary physics course, that air drag is negligible. A typical terminal velocity for a human is about 120 mph≈54 m/s. If you jumped with this speed you go over 100 m high, obviously not in the cards. I did a rough estimate assuming the maximum height you could jump would be about 2 m; if the person drops from 2 m his speed at the ground would be about 6.32 m/s without drag, 6.30 m/s with drag, a 0.3% difference. For comparison, dropping from 100 m the speeds would be roughly 44.1 m/s and 37.1 m/s for no drag and drag, respectively. It is good to be precisely correct as you are, but it is also good to be able to make reasonable estimates in real-world situations.

QUESTION:
When a ball is thrown vertically upwards ignoring air resistance, and another ball is also thrown upwards with air resistance, the time taken is less for the ball with air resistance to reach max height. Why is this "because average acceleration/force is greater"? Wouldn't there be less acceleration/force because the air resistance cancels some out?

ANSWER:
The reason is that the ball with air resistance does not go as high. The force on the ball without resistance is the weight of the ball pointing in the downward direction; but the downward force is greater for the ball with air resistance because the drag force is also pointing down. Therefore the ball with resistance slows down faster so it stops more quickly. Think of an extreme example: if you throw the ball upwards in honey which has very great resistance, it stops almost immediately.

QUESTION:
Whenever we roll a ball or spin a quarter it will slow down and eventually stop, since energy cannot just dissapear where does it go?

ANSWER:
The kinetic energy is being taken away from the ball or coin by friction. That energy shows up as thermal energy, the ball/coin-table-air all get a little bit warmer. Also, since you can hear the ball rolling and the coin spinning, some of the energy must be lost to sound.

QUESTION:
Something ridiculous I thought of, if the Moon suddenly stopped moving and began to fall toward the Earth, how long would it take to impact? I'm stumped as to how to calculate this, as the force on the Moon gradually increases as it falls, and the Moon also pulls the Earth toward it, and the radius of each object would have to be included.

ANSWER:
I guess I am going to have to put questions like this one on the FAQ page. You should read the details of these earlier questions since I do not want to go over all the details again. It is tedious and uninstructive to try to do this kind of problem precisely. I, being a great advocate of "back of the envelope" estimates, use Kepler's laws to solve this kind of problem; I have found that a very excellent approximation to fall time can be found this way. I note that the mass of the moon is only about 1% of the mass of the earth, the period of the moon is about 28 days, and the moon's orbit is very nearly circular. The trick here is to use Kepler's third law and recognize that a vertical fall is equivalent to the very special orbit of a straight line which is an ellipse of semimajor axis half the length of the line. Kepler's third law tells us that (T2/T1)2=(R2/R1)3 where Ti is the period of orbit i and Ri is the semimajor axis of orbit i. Now, T1=28 and R2=R1/2 and so T2=T1/√8=9.9 days. But this is the time for this very eccentric orbit to complete a complete orbit, go back out to where it was dropped from; so, the time we want is half that time, 4.9 days.

But this is not what you really wanted since I have treated the earth and the moon as point masses. What you really want is when the two point masses are separated by a distance of the sum of the earth and moon radii, 6.4x106+1.7x104≈6.4x106 m. To see how much error this causes, I can use the equation for the velocity v at the position r=6.4x106 m if dropped from r=Rmoon-orbit=3.85x108 m which I derived in one of the earlier answers: v=√[2GM(1/6.4x106-1/3.85x108)]=1.1x104 m/s. It would continue speeding up if the collision did not happen, but even if it went with constant speed the time required would be about t=R/v=6.4x106/1.1x104=580 s=9.6 min. This is extremely small compared to the 4.9 day total time, so, to at least two significant figures, 4.9 days is the answer to your question.

An important part of doing physics, or any science, is knowing when to eliminate things which are of negligible importance!


QUESTION:
I'm a Science Olympiad coach trying to optimize the performance of our "Scrambler", a car which must be accelerated by only a falling mass. Most competitors simply tie a weight to a string and route that string over a set of pulleys (using no mechanical advantage to convert the vertical falling acceleration horizontal. …Read a whole lot more!

ANSWER:
Sorry, but if you read site groundrules you will see that "concise, well-focused questions" are required.

FOLLOWUP QUESTION:
I was hoping you'd like the challenge of a motion/force problem that must span across several formula -- PE, KE, PEspring, velocity solved by acceleration and distance only, etc. Something to sink your teeth into...

ANSWER:
It is really not that interesting to work the whole thing out, but on second thought it is interesting to talk qualitatively and generally about the questioner's proposal; so I will do that. I will summarize the situation since I am sure none of you loyal readers will want to read the whole original question. By using a falling mass M attached to a car of mass m, it is wished to maximize the speed v of the car for M having fallen through some some distance H. The car moves only horizontally. The simplest thing to do is to have the two simply attached by a string over a pulley. Then, using energy conservation, 0=½(M+m)v2-MgH or v=√(2MgH/(M+m)). What the questioner proposes is to hold the car at rest and insert a rubber band in the string so that the falling weight stretches the rubber band which has been carefully chosen to be just right that, when M has fallen H, it has just come to rest and is held there. Now, presumably, the rubber band has a potential energy of MgH. If the car is now released, the rubber band will presumably contract back to its original length giving its potential energy to the car, MgHmv2 or v=√(2MgH/m), a considerable improvement. My suggestion would be to use a spring rather than a rubber band since a rubber band has much more damping (energy loss due to internal friction) and hysteresis (will not return to its original length). Since the rules fix M and H, one obviously wants m to be as small as possible.


QUESTION:
Does acceleration have momentum? In other words if you fire a rifle, does the highest velocity of the bullet occur as it exits the barrel or does the acceleration increase after it leaves the barrel?

ANSWER:
In terms of physics nomenclature, your first question has no meaning. But your second question seems to clarify what you mean: if something has an acceleration does it keep accelerating even if there are no forces on it? The answer is an unequivocal no. The only thing which causes acceleration is force and when the bullet exits the barrel of the gun the force which was accelerating it disappears. If there were no new forces on it, it would continue with the same velocity it had when it exited. There are, however, two important forces on the bullet when it is outside the gun—gravity and air drag. Gravity causes it to accelerate toward the ground and air drag causes it to slow down.


QUESTION:
We have a metal ruler 1 yard in length. If held in exact balance center then "pinged" it causes vibrations. nothing new there BUT at exact equal distances to either side there is a point at which it appears that the vibrations stop (a calm spot) for about 1 inch then the vibrations start again and continue to the end. We've tested and the vibrations don't stop, they are just a much smaller wave length. What is is this "calm spot" phenomenon? What causes it? Does it happen with earthquakes too? Really geeked out about this! Way cool!

ANSWER:
You are exciting standing waves when you "ping" the stick. These are waves which bounce back and forth from the ends of the stick and, for special wavelengths, are just right to to resonate like a guitar string or an organ pipe. The various wavelengths for which resonance occurs are called the modes of oscillation. For a stick clamped at the middle, the lowest mode, called the fundamental, has approximately 1/4 of a wavelength on either side of the center as shown by the upper part of the figure above. A point with zero amplitude, the center for the fundamental, is called a node and points with maximum amplitude, the ends for the fundamental, are called antinodes. What you are seeing is the next mode, called the first overtone, which has approximately 3/4 of a wavelength on either side; this mode has three nodes and four antinodes.
To the right are animations for a stick clamped at the end but they are exactly what your stick is doing on one half. Here the nodes are near the darkest blue. Earthquakes are traveling waves and therefore do not have nodes.


QUESTION:
I've been reading about rotational space habitats for a while now and haven't found an answer to this question by googling. So how would a space habitat be rotated? By cogs? By propulsion systems?

ANSWER:
If you want to read more about the details of such habitats, see my earlier answers (1 and 2). In answer to your question, you would need thrusters to get the ring spinning and then to occasionally correct minor changes, but once it was spinning, it would continue to spin just the same forever if there were no external torques on it as would be the case in empty space.


QUESTION:
The moon's gravity is one sixth that of the earth. Thus if you kicked a box with a force of 60 N across a frictionless floor on earth, the box would travel the same distance in 1 second as when the same box was placed on a frictionless floor on the moon and kicked with a force of 10 N. Am I wrong?

ANSWER:
You are wrong on both your conclusion and on your "…kicked…with a force of…" premise. First the premise: you need to go the FAQ and read the link from the question about how much force does it take to make something move with some speed. Just knowing the force you cannot know the resulting speed; you need either how long the force was applied or over what distance it was applied. Now, your question implies that you think it will be easier to get the box on the moon moving with the same speed as a box on earth with that speed. But, in fact, the box has the same mass on both the earth and moon and you are not lifting it against gravity, so it is equally easy to move a box horizontally on earth or the moon. It is six time harder to lift a box on earth as on the moon. If there were friction, however, it would be harder to move the box on earth than on the moon because the frictional force on the moon would be six times smaller.


QUESTION:
If you have two permanent cylindrical magnets (the kind with a hole in the center) and you stack them with poles opposite on a pencil, the top magnet will "float" above the bottom magnet. Energy is being expended to keep the magnet "up" the pencil. Where is the energy coming from? The bottom magnet will be pushing down with an equal but opposite force, but that does not cancel the energy needed to float the top magnet as far as I can see.

ANSWER:
I am afraid you do not understand energy. The lower magnet exerts a force on the upper magnet. The force holds it there in equilibrium, it does not require energy to hold it there. It is no different from saying that if one of the magnets were hanging from a string, where does the energy to hold it there come from? Or, if one of the magnets were sitting on a table top, where does the energy to hold it there come from?


QUESTION:
I read of "gravity assist" swingy-bys of Jupiter to speed a spacecraft up to reach the outer planets. As the spacecraft approaches Jupiter, it speeds up. But it retreats from Jupiter on a symmetric path (a hyperbola I think) and Jupiter will therefore slow the spacecraft down by the same amount on the outbound path. It appears there should be no net increase in speed, just a bending of the spacecraft path. But bending a spacecraft path also takes energy. So Jupiter is providing the energy in some manner though it is unclear to me how.

ANSWER:
The trick is that the planet, with much greater mass than the spacecraft, is moving in its orbit and the boost comes from using the speed of the planet to speed up the spacecraft. The figure shows the idealized one-dimensional interaction with planet; because the mass of the spacecraft is much less than the mass of the planet, the spacecraft picks up twice the speed of the planet. For those who have studied elementary physics, this should look vaguely familiar: a perfectly elastic collision between a BB at rest and a bowling ball with speed U results in the BB going with speed 2U and the bowling ball still going U. (Of course, this is only approximately true if the mass of the BB is much smaller than that of the bowling ball; the bowling ball will actually lose a very tiny amount of its original speed.)


QUESTION:
My son is in 5th grade and he is supposed to be doing a Science Fair project and presenting in front of judges, some of which I have heard work at IBM. His teacher helped him set up an experiment where he rolled a car down an inclined plane 5 times adding more weight each time. Then he is to see if the car goes faster with more weight added. And then explain that due to research. I have read your other answers but I am so confused. Is gravity the only force working on the car to get down the incline? and if so does that mean that technically the car should always travel the same speed? Any help on where I can learn about this, or any information you may be able to help me with would be so greatly appreciated.

ANSWER:
The best place on my site to read is this earlier answer. There you will see that gravity is certainly not the only force acting on the car. Friction is very important. The normal kinds of friction, in axles, wheels, etc. should, according to the simplest approximation, not affect the speed of the car—all masses should take the same time. That, as I explain in that post, is because all these forces are approximately proportional to the weight of the car and, of course, gravity is also proportional to the weight. The other kind of friction is air drag. The air drag is proportional to the speed of the car squared, so the faster the car goes the more air drag it has. You know that this is true because you have a very different experience if you stick your hand outside the window of a car going 10 mph and 80 mph. If air drag is important, the heavier car wins. I have said before that this is not really a very good experiment for a science fair because there are too many variables and approximations. Friction can be a very sticky thing in physics (pun intended!).

If he did not find the heaviest car the fastest, I do not know what to tell you except it is difficult to understand. If he did find the heaviest the fastest, here is how he could justify it: Two cars, each going the same speed down the ramp, have masses 1 kg and 2 kg. Since they are identical in all respects except mass, each experiences the same air drag frictional force, let’s call that Fdrag. Newton’s second law says that the acceleration adrag is equal to the force Fdrag divided by the mass m and so adrag=Fdrag/1 for the 1 kg mass and adrag=Fdrag/2 for the 2 kg mass. Here the acceleration is the rate of slowing down due to air drag (you might call that deceleration) and you see that the heavier car slows down less than the lighter car. All the other forces result in the same acceleration for each car since those forces are all proportional to the mass; maybe I should elaborate on that a little: Forces due to friction and gravity can be written Fother=Cm where C is some constant. And so aother=Fother/m=C which means that all accelerations should be the same, so only air drag gives advantage to the heavier car.

Finally one more example to illustrate: Drop two balls exactly the same size, one is a nerf ball (very light) and one is a lead ball (very heavy). If there were no air, both would fall to the ground in the same time. But you can easily show that the heavier one wins if there is air.

FOLLOWUP QUESTION:
I'm still a little confused because his teacher is telling me it is not gravity that is pulling the truck down because it is a ramp and not falling and she says it is force and momentum moving the truck down the ramp. From what I have researched it looks to me like momentum is not really a force and only forces actually move objects? Also I think I understand that gravity is not the only force acting on the toy truck but it seems to be the only one pulling or pushing it downward and the frictional forces push upwards, opposite gravity?

ANSWER:
Your son's teacher has this seriously wrong if she really said that it is not gravity pulling it down the ramp. It most certainly is falling, it is just not falling straight down. And you are right, momentum has nothing to do with its going down the ramp; it has momentum but that is not what is causing it to go down the ramp. So, I guess I will have to give you a complete primer on the motion of an object on a ramp. I will draw some figures and use some equations but I will try to give you as many words as I can to qualitatively explain; then you can help your son create an explanation on his level. The basics are that an object which has no net force on it moves with a constant velocity (Newton's first law). To change the velocity (accelerate or decelerate) you must push or pull on it (exert a force) (Newton's second law). In equation form, Newton's second law may be stated F=ma where this says force=mass times acceleration.

  1. Let us first talk about the ideal case where there is no friction at all. The first picture shows the weight (gravity) as a red arrow. But you see, the weight actually does two things as indicated by the blue arrows—accelerate the car down the ramp (smaller blue arrow) and push the car through the ramp (larger blue arrow). The car does accelerate down the track but it does not go through the ramp because the ramp pushes on the car with a force opposite the blue arrow, shown as a black arrow. So we can agree that the reason the car accelerates down the track is gravity. Now how does the acceleration depend on the weight of the car? If you have two cars and one has twice the weight of the other, the force (smaller blue arrow) down the ramp will be twice as big for the heavier one. On the other hand, Newton's second law says that the acceleration a of some weight (mass m) due to a given force F depends inversely on the weight a=F/m. So even if F is twice as big, because m is also twice as big, they both have the same acceleration! This means that if there is no friction of any kind, the weight of the car should make no difference.

  2. Next let us add friction, the kind which results from one thing rubbing on another like wheels on axles, etc. The friction (shown as a purple arrow in the second figure) points up the ramp and so it tends to slow the car down. The important thing to know about these kinds of frictional forces is that they also depend on the weight of the car. If the car is twice as heavy, friction is twice as big. But if the friction depends on the weight, the acceleration will be the same just as it was for the weight force. Again, all cars should get to the bottom in just the same time again, it will just be a shorter time because the total force down the incline is smaller.

  3. Finally, let us add in an air drag force (the green arrow in the third figure). Air drag is, as I explained in the original answer, a force which depends on two things—the shape/size of the car and the speed of the car—but not on the weight of the car. So air drag gives the heavier car the advantage because there is less slowing down due to the drag force.


QUESTION:
This question should be fairly quick. Are there any know situations were momentum isn't conserved? I would say no, as momentum is always conserved if you make your system big enough. The only time momentum appears to be not conserved is when you put restrictions on the size of your system, and don't account for the momentum transferred outside your system. When you include the system " outside" your system, momentum is in fact conserved.

ANSWER:
First, what you learn in first-year physics is that linear momentum is conserved for an isolated system; an isolated system is one for which there are no net external forces acting. This is actually a result of Newton's third law which essentially states that the sum of all internal forces in a system must equal zero. This works really well until you get to electricity and magnetism where it is easy to find examples of moving charges exerting electric and magnetic forces on each other which are not equal and opposite (you can see an example in an earlier answer). In that case you would say that such an isolated system obeys neither Newton's third law nor momentum conservation. However, looking deeper, we find that an electromagnetic field has energy, linear momentum, and angular momentum content and, in the end, momentum conservation, because of the momentum contained in the field, is still conserved for an isolated system. Thus, Newton's third law is saved, but not always in the simplistic "equal and opposite forces" language. Finally, if linear momentum is p=mv, linear momentum is not conserved in special relativity. But, physicists so revere momentum conservation that in special relativity momentum is redefined so that it will be conserved but still reduce to p=mv, for low speeds: p≡γmv=mv/√[1-(v/c)2]; this then unifies energy E and linear momentum p into a single entity which is conserved, the energy momentum 4-vector: E2-p2c2=m2c4. Not as quick as you expected!


QUESTION:
What is the force that causes you to fall over when a moving bus comes to an immediate stop? I'm having an argument with my teacher over what the answer is, it would be great if you could explain!

ANSWER:
When the bus is stopping, it is accelerating and so it is a noninertial frame. That means that Newton's laws are not valid if you are riding inside the bus. But, if we watch you from the bus stop, Newton's laws do apply and we conclude that if you move with the bus, there must be a force which is causing you to accelerate also. Friction provides a force which, except under extreme circumstances, accelerates your feet along with the bus; but, unless you are holding on to something, there is nothing to provide a force on your upper body which therefore tends to keep going forward without accelerating. All this says that the reason you fall forward is not due to any force, rather it is due to lack of a force. There is, though, another way to look at this problem. If you are in an accelerating frame, like the bus, you can force Newton's laws to be true by adding fictitious forces. The best known example of a fictitious force is the centrifugal force in a rotating (and therefore accelerating) frame. In the bus which has an acceleration a you can invent a fictitious force Ffictitious on any mass m in the bus, Ffictitious=-ma; if you do that, Newton's laws become true inside the bus and the force Ffictitious may be thought of as being the force which provides your acceleration. Note that the acceleration is opposite the direction of the bus when it is stopping, and so the fictitious force is forward as you know if you have fallen over in a stopping bus. When the bus is speeding up you tend to fall backwards. Since there are two answers here, depending on how you choose to view the problem, so maybe you and your teacher are both right!


QUESTION:
if the earth had an orbit of 100,000 miles above the surface of the sun what would a person with a weight of 100 pounds weigh during the day and during the night and would there be any difference because of the gravitational pull from the sun.

ANSWER:
The radius of the sun is about 432,000 miles, so the radius R of the orbit would be R=532,000 mi=8.6x108 m. Since the diameter of the earth, about 12.8x106 m, is small compared to the radius of the orbit, there is only about a 0.1% change in the gravitational attraction to the sun if you change the distance by one earth diameter. The mass of the sun is M=2x1030 kg. The force of attraction of the sun on the m=100 lb=45 kg person would be given by F=GMm/R2=6.67x10-11x2x1030x45/(8.6x108)2=8117 N=1825 lb, much bigger than the 100 lb force which the earth exerts on the person. Now, to answer your question you need to define weight. I will assume that we mean the net sum of all forces on the person so that Wday=1725 lb upwards and Wnight=1925 lb downward. Or, maybe you mean what a scale on the floor would read (not actually what weight means) in which case Wnight=1925 lb and Wday=0 lb. These are really only noon and midnight weights since the forces due to the earth and sun would not be parallel at other times.


QUESTION:
If a bullet was shot through a window of a moving train and was to come out on the other side of the train through a window. Would it come out through window 2 on the same the exact opposite side as window 1 or would it look as if the bullets direction was bended?

ANSWER:
If the bullet is shot straight at the train with some speed V and the train is moving with some speed v, an observer on the train sees the bullet moving with a speed V toward the opposite side and, at the same time with a speed v toward the back of the train so that the bullet would be traveling, as measured on the train, in a straight line across and rearward with speed √(V2+v2)


QUESTION:
I've found the formula for gravitational attraction between two objects, but I can't quite "do the math" mainly because of the metric/english conversions...I want an answer that I can relate to in pounds or ounces, not dynes, ergs, or grams (I am aware of the distinction between mass and force) Here's my question: I am driving my 18-wheeler truck which weighs 80,000 lb. fully loaded. I am driving due west at sunset, heading straight toward the sun which is about 93 million miles away. What is the 'tidal force' of attraction between my truck and the sun?

ANSWER:
I have the feeling that you want gravitational force, not tidal force. Tidal force is the tendency for the truck to be stretched because the gravitational force on the front of the truck is a tiny bit larger than on the rear. The gravitational force is computed by F=GMm/R2 where G=6.67x10-11 N∙m2/kg2, M=2x1030 kg, m=8x104 lb=3.6x104 kg, and R=93x106 mi=1.5x1011 m. So I find F=213 N=48 lb. There is a very handy little free program called Convert which you can use to convert just about any units you might want to work in.


QUESTION:
Let's assume I have a magnet that can lift 100 tons. And I attach the magnet in a chain and attach the chain into roof for a system to magnetically lift items and then drop them to other places. Would the chain have to be able to take the 100 ton load or would the magnet take the 100 ton load because after all it is the one keeping the lifted item up?

ANSWER:
The magnet itself holds up the 100 ton weight. The chain holds up the 100 ton weight plus the weight of the magnet itself. The roof holds up the 100 ton weight plus the weight of the magnet plus the weight of the chain.


QUESTION:
In 1973 a Physics instructor explained via math that the sidewalls of a regulation tire need not be present if the velocity of the vehicle was above a speed of 65+ mph. I tried to explain this to family members at Christmas and was scoffed at and then ridiculed. The Physic instructor had been let go from the GMC/Chevrolet plant several years before and he took the educational retraining route. His job was to change out instruments on GM cars running around a track and in excess of 100+ mph and his driver advised him that they had had a blow out and he needed to get out from under the dash and safety belted in at which time the slowed below the critical speed with the result that they did not crash but came close to it. Can you provide a link or the math to show that the speed is somewhere about 70+ miles and then the centripetal force will hold up the outer part of the tire. He did the math as part of educating us on acceleration, speed and force as it involved that part of the class curriculum. The instructor was a good instructor in that he made the physics relevant to the real world if there is such a thing today and even at that time.  Also this is why tires need sidewalls as they won't hold up in gravity and below a specific velocity.

ANSWER:
This is nonsense. If there is no air pressure to connect the tire to the axle, which would be the case if there were no sidewalls, what is going to hold up the weight of the car?

FOLLOWUP QUESTION:
No not really if you get the tire up to speed as well as providing forward momentum the circumference and the center point about which the tire is rotating will hold the tire up even if there is a blow out as the forward speed or acceleration is sufficient to hold it up will prevent deflation aka collapse of the tire above a speed. Once the speed or acceleration drops below a key critical the tire will start to collapse and according to the GMC aka Physics instructor all hell broke lose on the track and only the drivers expertise kept the ensuing deceleration from causing him and driver problems. Once you reach velocity the outer rim of the circle/sphere need not have anything to hold it up if the instructors explanation and the math were correct. Key elements:

  1. Tires inflated to recommended PSI

  2. Vehicle an experimental test GMC product running in excess of 100+ miles an hour.

  3. According to the LAHC Physics instructor a knowledgeable and well trained driver at the wheel who on sensing the blow out got him out from under the dash and into the multi-point seat belt. Instructors job was instrument technician for the test bed aka the vehicle a GMC automobile with changeable instruments.

  4. He was alive to prove it to the class with not sure what Physic concept/principal that escapes me.

ANSWER:
Well, maybe I misunderstand something here, but let's boil this problem down to the simplest equivalent I can think of: imagine a tire with sidewalls and just an axle which is supported by the sidewall, shown on the left in my figure. Now, we would agree, I believe, that if the sidewall suddenly disappeared, the axle would fall because there would be nothing holding up that weight. How is that situation any different if the car is moving? So, let's agree that "
the sidewalls…need not be present" is wrong because there has to be some physical contact of the outer surface of the tire and the axle. So, my first answer was a knee-jerk response to the notion that the sidewalls were not needed.

THE ANSWER YOU WILL LIKE:
However, there is still a way that you might have a point. When the blowout occurs, the pressure inside the tire is lost; this pressure is typically 30 PSI=21,000 N/m2 above atmospheric pressure (which is about 100,000 N/m2). If the car is sitting still, this loss of pressure results in the wheel collapsing because the sidewalls alone are insufficient to hold up the weight of the car unless the force due to the pressure pushing on the outer part of the tire holds the sidewalls taut. Now, imagine that you are driving with some speed V and viewing a spinning tire from its axis, you see every point on the outer surface of the wheel accelerating with an acceleration V2/R where R is the radius of the tire. Therefore, every little piece of the tire with mass m experiences a (fictitious) force (called the centrifugal force) of mV2/R. That would be equivalent to there being a pressure P exerted on that little piece of tire of P=mV2/(aR) where a is the area of that little piece. But, every little piece behaves like this, so it is equivalent to a pressure of P=MV2/(AR) acting on the outer surface where M is the mass of the tire (assuming the sidewalls are a small fraction) and A=2πRW is the area of the outer surface and W is the tread width. So, if that pressure is equal to 21,000 N/m2, it will be like the blowout never happened! I took R≈16 in≈0.4 m, W≈12 in≈0.3 m, and M≈20 lb≈9 kg and solved 21,000=MV2/(AR)=MV2/(2πR2W) and found V=27 m/s=60 mph. (Incidentally, the "forward momentum" has nothing to do with it.)


QUESTION:
which team wins in a tug of war: the team that pulls harder on the rope or the team that pushes harder against the ground.Can you explain please?

ANSWER:
Focus your attention on the guy in the red shirt. There are two horizontal forces on him, the rope pulling to the left and the ground pushing to the right. If he is not moving, these two forces must be of equal magnitude. To win, he must accelerate to the right and so the ground must exert a bigger force on him than the rope exerts on him. To complete the answer, use Newton's third law which says that the force the rope (ground) exerts on the man must be the equal and opposite to the force that the man exerts on the rope (ground). So the winner must push harder on the ground.


QUESTION:
If a high jumper clears the bar, is it possible that the centre of mass of the body of the jumper passes below the bar? If so can you make me visualize the scenario by a video or image illustration or a vivid description? I think that the centre of mass can be below the bar during the jump, but it has come there after travelling above the bar.

ANSWER:
You can find dozens of pictures and videos on the internet. A nice one is shown to the right here. The path under the bar of the center of gravity of the jumper is shown. When the body is bent the center of gravity is outside the body. Going over with the back down is called the Fosbury flop after Dick Fosbury, the American high jumper who won the gold medal at the 1968 Olympics.


QUESTION:
I want to know why does torque is able to do work. I Mean Torque vector always acts perpendicular to the surface therfore meaningthat angle between torque vector and angular displacement vector is always 90 degree. Whereby meaning work done by torque is always zero but it is obviously not so. I know how to derive that work expression but still I am wondering why is it so.

ANSWER:
You are mistaken, angle and torque vectors are not always perpendicular. Shown in the figure to the right are the vector directions relevant to your question. The angle through which this cylinder rotates, θ, increases as the cylinder rotates counterclockwise as seen from above; the vector direction is seen in the black vector θ shown in the figure. If there is a force F acting at a distance r from the axis, the torque vector is given by the red vector τ shown in the figure, τ=Fxr. So, W=τθ.


QUESTION:
In a system where I have a 1600mm beam with I=1,300,000kg-mm^2 rotating about an axis (shaft) at the far left end of the beam where it's initial position is resting at 0 degrees horizontal then rotating 180 degrees counterclockwise (from 3 o'clock to 9 o'clock) about the axis (shaft) at the end of the beam described above by way of a cable wrapped around a 12" diameter disc (disc attached to beam & shaft) where the cable runs out to the right tangentially at 6 o'clock horizontally over a pully wheel then down to a weight, say 300lbs, that will fall to apply enough torque to effect the rotation of the beam and accelerate it. How do I account for role gravity plays in the falling weight keeping up with the rotation of the beam where the beam first has to overcome some amount of gravity to get from 3 o'clock to 12 o'clock but then gravity works in it's favor from 12 to 9. My goal is to get a point at the end of the beam to hit a plate at about 100mph at the end of the rotation (9 o'clock).

ANSWER:
I am not certain that I understand exactly what the configuration is and some of the numbers are missing (like mass of the disc) or not fully specified (like the axis for the moment of inertia of the beam). I will work it out in general and you can apply it to your situation. The situation as I understand it, before and after, is shown to the right. I will call the length of the beam L, its mass M1, the radius of the disc R, its mass M2, the hanging mass M3. I will assume the beam and disc are uniform so that their moments of inertia are M1L2/3 and M2R2/2 respectively. I assume that the mass of the pulley is negligible and that all friction can be ignored. In the final situation the beam and disc rotate with angular velocity ω and the hanging mass has a speed v=Rω. Note that the hanging mass has fallen a distance half the circumference of the disc, πR. If you are interested only in how fast everything is going at the end, you do not need to worry at all about what is going on at any other time because this is a classic energy conservation problem and the final energy is equal to the initial energy. I am not going to give you all the details, I will just give the initial equation and its solution for the end.

  • Energy conservation: 0=½[(M1L2/3)+(M2R2/2)+M3R2]ω2-M3gπR

  • Solution for ω: ω=√{2M3gπR/[(M1L2/3)+(M2R2/2)+M3R2]}

Once you get ω, you can get the speed V of the end of the beam by calculating V=ωL.

If I take your numbers, M3=300 lb=136 kg, L=1.6 m, M1=1.52 kg (assuming the I you have given me is about an axis through the end of the beam), M2=0 (assuming it is small since you did not specify it), R=6"=0.15 m, and g=9.8 m/s2 (acceleration due to gravity), I find ω=17 s-1 so V=17x1.6=27 m/s=60 mph.

ADDITIONAL THOUGHT:
You might think that you can increase the speed of the end of the beam by increasing M3. However, there is an upper limit. In the limit as M3—›∞, ω—›√{2/R}=20.3 s-1 giving V=72 mph. Assuming that the length of the beam is fixed, you would have to decrease the radius of the disc to reach 100 mph.


QUESTION:
Orbiting astronauts are weightless because they are essentially in free-fall and don't "feel" Earth's gravitational force. Earth is orbiting the Sun and similarly Earth should not "feel" the Sun's gravity. However, the Sun has an effect on the ocean tides therefore it follows that the ocean water "feels" the Sun's gravity. Please explain.

ANSWER:
An astronaut is not weightless but she is in free fall and therefore feels like she is. However, she is not aware of what are called tidal forces because they are too small because she is relatively small; I will try to clarify this. Let's take the case of the earth-moon interaction. Because the moon's gravitational force falls off like 1/r2, the forces felt on the earth are different at the side nearest to the moon, farthest from the moon, and at the earth's center; this is shown in the upper of the two figures to the left. The net effect, obtained by subtracting the central gravitational force, is as shown in the lower figure; that is why the tidal force raises the ocean level on both near and far sides of the earth. The astronaut will also experience a tidal force but since her size is so small compared to the size of the earth, she will not feel the force try to stretch her; if you think about it, you will see that the tidal force on an astronaut is greatest when she is standing on the ground. When an object falls into a black hole, tidal forces become huge. The above discussion contains nothing regarding how the moon and earth are moving relative to each other so all the same arguments apply to tidal forces the earth feels in the presence of the sun; these are, though, much smaller since we are much farther from the sun.


QUESTION:
Conventional automobile steering produces a weight transfer from the inside wheels to those on the outside of the turn. If, instead of deflecting the front wheels, one steers by turning the outside wheels faster than the inside wheels, will that also cause a weight transfer?

ANSWER:
There is an earlier answer about a bicycle making a turn. It would be helpful for you to read that first. I assume you want to understand why the weight is more supported by the outside wheels, so I will try to show that. The easiest way to do the problem of the car turning a curve is to introduce a fictitious centrifugal force which I will call C, pointed away from the center of the circle; the magnitude of this force will be mv2/R where m is the mass of the car, v is its speed, and R is the radius of the curve, although we do not really need to know that to answer your question. The picture to the right shows all the forces on the car: W is the weight and the green x is the center of gravity; f1 and f2 are the frictional forces exerted by the road on the inside and outside wheels respectively; N1 and N2 are the normal forces exerted by the road on the inside and outside wheels respectively; the center of gravity is a distance H above the road and the wheel base is 2L (with the center of gravity halfway between the wheels). Newton's equations yield:

  • f1+f1=C for equilibrium of horizontal forces;

  • N1+N2=W for equilibrium of vertical forces;

  • CH+L(N1-N2)=0 for equilibrium of torques about the red x.

If you work this out, you find the normal forces which are indicative of the weight the wheels support: N1=½(W-C(H/L)) and N2=½(W+C(H/L)). A few things to note are:

  • the outer wheels support more weight,

  • if C=0 (you are not turning), the inner and outer wheels each support half the weight,

  • at a high enough speed C will become so large that N1=0 and if you go any faster you will tip over, and

  • if the road cannot provide enough friction you will skid before you will tip over.

Now we come to your question. Regardless of how you cause the car to turn, the analysis done above will always be the same; the centrifugal force is trying to tip the car over and that is why the weight distribution shifts.


QUESTION:
If a gun was fired at the escape velocity of the Earth in a direction that it would not hit anything but not straightly up but instead more like in an angle of 45 degrees would the round still go to space? And if not what would the velocity needed be for a round that is fired at 45 degrees in to the sky be in order for the round to escape Earth?

ANSWER:
First, let's be clear that we are talking about an ideal situation where we neglect the effects of air drag (which will slow the projectile) or earth's rotation (which can add or subtract from the velocity you give the projectile). The easiest way to get the value of the escape velocity (ve) is to use energy conservation. Escape velocity is that velocity which results in the projectile being at rest at r=∞. Taking the potential energy U(r)=-GMm/r+C to be zero at r=∞ (i.e., choose C=0), energy conservation gives ½mve2-GMm/R=0 where M and R are the mass and radius of the earth, respectively; so, ve=√(2MG/R). You will notice that this derivation has been done without any reference to the direction of the velocity ve, so the direction makes no difference.


QUESTION:
This is one of the most baffling physics questions so far. Suppose a cyclist is pedalling briskly, thus accelerating forward. He exerts force on the tyres, the tyres exert force on the ground, and the ground exerts a reaction force which accelerates the cyclist. Right? But now the cyclist applies brakes. Now he starts to retards. But since the tyres are still moving in the same direction, hence friction must also be acting in the same direction. Then which force is responsible for the retardation? I suppose the brakes cannot retard the 'cyclist and the cycle' system, since it is an internal force.

ANSWER:
There is nothing "baffling" here. You are right, all internal forces should be ignored if you want to understand the motion of the bicycle. The only external forces on the cycle and its rider are its own weight, the normal force up from the road on the wheels, and the frictional force between the road and the wheels. (I am neglecting internal friction and air drag.) The first of these two add to zero (assuming a horizontal road) and can be ignored for our purpose. If the acceleration is forward, the friction if forward. If the acceleration is backward, the friction if backward. The direction of the friction is not determined by the motion of the wheels because the wheels are not slipping on the road (you hope) so static friction, not kinetic friction, is responsible for the acceleration. It is friction which speeds you up or slows you down.


QUESTION:
Consider a tug of war game. is the net work the ground on the two teams is negative, positive or zero?

ANSWER:
Just after one team has won, all players are moving. They have a kinetic energy. Therefore positive work was done on the whole system by friction with the ground. Once everyone stops moving, the net work done on the system would have been zero.


QUESTION:
Trying to understand physics concepts is an interest of mine. I was hoping you could shed some light to help my understanding? The equation for Kinetic Energy: KE = 1/2 m v2. I understand the value of velocity squared in that; the kinetic energy of an object is proportional to its speed squared:double the speed, quadruple the KE. But I am having trouble with why and what the 1/2 value is telling me about what is going on in the equation as to the motion of mass and speed. It is pretty elementary stuff but if I want to know about Energy, Work, Power and Heat I need to get the basics right.

ANSWER:
To do this on the simplest level, you need to know kinematics for uniform acceleration [x=x0+v0t+½at2 and v=v0+at where the 0 subscripts denote the position and velocity when t=0] and Newton's second law [F=ma]. Suppose that you have a constant force F that you exert on a mass m over a distance s; the work done is W=Fs and work changes the energy; that is, we say that the work done equals the energy given to m. I will assume that m begins at rest and at the origin, so x=s=½at2 and a=v/t; then F=mv/t and s=½(v/t)t2vt. Finally, W=Fs=(mv/t)(½vt)=½mv2. So, in the case of a constant force, the factor of ½ comes from the ½at2 part of the kinematic equation for position.

EXTRA ANSWER:
If you know calculus, this is much simpler. Write dW=Fdx and F=m(dv/dt) so W=0sFdx=0sm(dv/dt)dx=0vm(dx/dt)dv=0vmvdv=½mv2; and this does not require F to be constant.


QUESTION:
why do we need to have a new expression for calculating the kinetic energy of a body that is in rotational as well as translational motion?

ANSWER:
Whether you do or not depends partly on how the object is moving. For example, if the object is rolling on a surface without slipping, it is instantaneously rotating about the point (or line) of contact and you can write K2 where I is the moment of inertia about that axis; however, you could also show (using the parallel axis theorem) that KMvcm2Icmωcm2 where vcm is the speed of the center of mass,  ωcm is the angular velocity about the center of mass (which is the same as ω) and Icm is the moment of inertia through the center of mass. For an object not rolling, KMvcm2Icmωcm2 is still the correct expression for kinetic energy at any instant but you can no longer write the kinetic energy as pure rotation as you could for the rolling object because there is no relationship between vcm and ωcm.


QUESTION:
is work energy theorem valid in non inertial frames?

ANSWER:
The work-energy theorem says that the change in kinetic energy of an object is equal to the work all forces do on it. Imagine that you are in an accelerating rocket ship in empty space, a noninertial frame. You have a ball in your hand and you let go of it. You observe this ball to accelerate opposite the direction in which the ship is accelerating and therefore see its kinetic energy change. But, there are no forces acting on it so no work is done. Another way you could come to this conclusion is that the work-energy theorem is a result of Newton's laws and Newton's laws are not valid in noninertial frames. You can, though, force the work-energy theorem to be valid if you introduce fictitious forces, a way to force Newton's laws to work in noninertial frames. (Centrifugal force is an example of a fictitious force.) If you invent a force on the objects of mass m in the accelerating (a) rocket ship above of Ffictitious=-ma, this force will appear to do the work equal to the change in kinetic energy.


QUESTION:
First, how much energy (in the most basic sense) does a car expend driving at a moderate speed for one mile. Next, how big would a spring have to be (roughly) to store the energy equivalent of what that car expended during that mile.

ANSWER:
I can only do a rough estimate. Suppose your car has an internal combustion engine and gets 40 mpg. The energy content of 1 gallon of gasoline is about 120 MJ, so you would use about 120/40=3 MJ of energy per mile. But, a typical engine has only about 20% efficiency, so the energy supplied to the car is only about 0.2x3=0.6 MJ=600,000 J. Now, you want to store that much energy in a spring. For a spring, the energy stored is ½kx2 where x is the amount by which it is compressed (or stretched) and k is the spring constant which is determined by how stiff the spring is. (The constant k can be measured by k=F/x where F is the force you have to exert to stretch or compress it by x.) Clearly, the bigger k is the less the spring will have to be compressed to store a given amount of energy. The coil spring of a car, a pretty stiff spring, has a typical spring constant of k≈25,000 N/m, so to store 600,000 J of energy, 600,000≈½∙25,000x2 or x≈7 m! I do not see much possibility of having a spring powered car. You probably had in mind a spiral spring like a watch spring, but to store that much energy in any kind of spring is going to be impractical.

FOLLOWUP QUESTIONS:
Thanks for helping me with that. I will admit that I'm still a little unclear about the answer though. You are correct that I am wondering about a spring powered car but I definitely would not use a spiral spring. A compression spring would probably work best. I don't understand your final answer. I am getting lost in the spring conversion formula. What does the final answer mean in layman's terms please?

ANSWER:
The spring would have to be compressed by 7 meters. That would mean that it would need to be at least twice that long, about 40 feet! You could use a stiffer spring (see below) to get a smaller required compression, but imagine the force you would have to exert to compress it. Keep in mind that you would have to supply the energy to the spring in the first place, a lot of energy. I know this spring is not going to work, but how much force would it take to hold it compressed by 7 m? F=kx=25,000x7=175,000 N≈40,000 lb. Another issue is that as the spring unstretches, the force it exerts gets smaller, so you would need to have some kind of governor mechanism to deliver the energy smoothly.

CONTINUED
Also, there are two things that are working in favor of the feasibility of the spring car:

I know from researching that there are many other factors that affect the gas- to engine- to -transmission to -wheels to- road, etc cycle so that the total efficiency of the entire loop would be far less than 20%.

ANSWER:
This is taken care of by the fact that I have specified the miles per gallon for the hypothetical car. The miles per gallon you get is determined by how well you have minimized the effects other than engine efficiency, mainly frictional effects. You asked for an estimate of the energy which the car would use to keep going a constant speed, and this is the best way I can think of doing that. Here is another way to estimate the energy consumed: suppose that I push on the car to keep it going a constant speed. I could probably do that with a force of about 100 lb≈445 N; one mile is about 1600 m, so the work I do is the product of the force times the distance and the work I do is the energy I use: W=445x1600≈712,000 J. (Actually, I am pretty impressed by how close this is to my other estimate of 600,000 J! It's just an accident that they are so close, but good that they are of the same order of magnitude—it increases my confidence that the energy consumed is pretty well approximated.)

CONTINUED
Despite my very basic and incomplete understanding of the formulas that give exact measurements for potential energy in springs and the potential energy available in compressed air it is obvious to me that if a car can run 300 miles off of a bottle of compressed air and a compressed air driven motor (these cars already exist and are in production in places like holland, india, etc) then a large spring perhaps three feet long and two feet around with a wire diameter of over half an inch could at least do the same. Getting the potential energy out of the spring in a useful way to power the car is another ball of wax altogether though.

ANSWER:
If you want a 3 ft long spring, about 1 m, the most you could probably compress it is about a half meter. To store 600,000 J of energy, the equation would be 600,000=½kx2=½k(0.5)2=k/8, and so the spring constant would be k=4,800,000 N/m. Then the force to hold it at 0.5 m would be F=kx=2,400,000 N
≈540,000 lb. If you were not faced with such enormous forces, like if you just wanted to have a toy car go a few meters, you could certainly get the energy out of the spring with a cleverly designed gear box; but the structural problems you would face with forces of hundreds of thousands of pounds would be insurmountable, I believe.


QUESTION:
In the case of a block that's dropped some vertical distance onto a spring,  it's reasonably easy to compute this value & to find the KE of the mass just before it hits the spring, hence the velocity. What doesn't ever seem to be explained is what happens to the block's acceleration ("g" when it hits the spring). The motion of the block continues downward, but now the net force is Fnet = kx - Weight = ma, not Fnet = mg. This means that the acceleration continues to be positive as the block compresses the spring, but @ a slower rate. This also means that the velocity continues to increase until all the block's PE is converted to spring elastic energy (Eelastic). At maximum compression the block stops & motion ceases. Since the ideal spring is massless there are no losses due to friction, heat, sound, et. al. Where does the Vmax occur? It's hard imagine for Vmax to happen @ the instant it stops? Is this process for the spring's resisting force linear?

ANSWER:
OK, suppose the speed when it hits the spring is v0. Choosing +y to be up and y=0 at the end of the uncompressed spring, may=-mg+k(-y). (Note that when the spring is compressed its force is upward and y<0 so the force is in the +y direction; this is why I write the force of the spring as k(-y). Now, until y=-mg/k, the acceleration is negative (points down) and the mass is still speeding up; below that point, the net force is positive (points up) so the mass slows down. Therefore the greatest velocity will be at y=-mg/k. Note that this position would be the equilibrium position if you gently placed m on the spring. You could also get this answer with energy conservation; using the same coordinate system, E=½mv02mv2ky2+mgy. This can be rearranged to give v2=-(k/m)y2-2gy+v02; if you differentiate this with respect to y and set equal to zero, you find y=-mg/k for the maximum velocity position. At that position you can solve for the speed, v=√[(mg2/k)+v02].

The key here is that you need to be very careful with the coordinate system and the signs of forces and potential energies. If you do not choose +y up, the potential energy will not be mgy (it would be -mgy). If you do not choose the unstretched spring as zero potential energy for the spring, ½ky2 will not be the potential energy.


QUESTION:
How long would it take for a 50lb solid steel sphere to sink a quarter mile through ordinary water?

ANSWER:
I will assume that the sphere quickly reaches terminal velocity so that I can assume that it goes the whole half mile with that constant speed. This should be an excellent approximation. The terminal velocity may be written as vt=√[2mg/(ρACd)] where m is the mass, g =9.8 m/s2 is the acceleration due to gravity, ρ is the density of water (1000 kg/m3), A is the cross sectional area of the sphere, and Cd is the drag coefficient which is 0.47 for a sphere. There is one catch here, that this is without buoyant force and the buoyant force in water is equal to the weight of the displaced water which is not negligible here; when it comes time to put in the mass I will put in an effective mass of the mass of the steel sphere minus the mass of an equal volume of water. I will work in SI units, so 50 lb=22.7 kg and 0.5 mi=805 m. The density of steel is 7850 kg/m3 so the volume occupied by 22.7 kg is V=22.7/7850=0.0289 m3; the radius of a sphere with this volume (using V=4πR3/3) is R=0.0884 m and so the area (using A=πR2) is A=0.0245 m2. Finally, the effective mass would be m=22.7-1000x0.0289=19.8 kg. Putting all these into the equation for vt, I find vt=5.81 m/s. The time an object going this speed takes to travel 805 m is 805/5.81=139 s, 2 minutes and 19 seconds.


QUESTION:
A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration 'a' along a straight horizontal track, the string supporting the bob makes an angle with the normal to the ceiling whose tangent is 'a/g'. Suppose the train moves on an inclined straight track with uniform velocity. If the tangent of angle of the incline is 'a/g', the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerated on a horizontal straight track or it is going on an incline with uniform velocity? If yes, how? If no, is there a method to do so?

ANSWER:
First you should carefully read an earlier answer on the accelerated pendulum. (Actually, I see that you are the person who asked that question!) To answer your question, you can't tell by "looking" but you certainly can tell by measurements. For example, in the accelerated problem the tension in the string is m√(g2+a2); for the inclined track the tension is mg. Or, in the accelerated problem you feel youself being pushed back with a force (fictitious) Ma, but on the inclined track you feel yourself being pushed back with a force (real) Mgsinθ=Mg[a/√(g2+a2)].


QUESTION:
what would happen if someone were to HOLD the gun and shoot it. The person would not be tethered to anything and would be floating freely in space. If that person is around 160Ibs and shot a High power rifle how fast would it project the wielder in the opposite direction? I know it wouldnt be as fast as the bullet because of the mass of the person but I'm curious to what extent they would be projected into space... if thats would even happen at all.

ANSWER:
First, read an earlier answer to a question similar to yours. The concept you want to use here is momentum conservation. Momentum is the product of mass times velocity and the total momentum of a system must be the same before and after the gun has fired. The mass of a 160 lb man plus his gun is about 75 kg, the mass of a typical bullet is about 0.015 kg, and the muzzle velocity of a high-power rifle is about 250 m/s. So, since the momentum is zero before the rifle is fired, 0=75v-250x0.015=75v-3.75 or v=0.05 m/s which is about 10 ft/min.


QUESTION:
A rigid container filled with air is placed in vacuum. If a small hole is created on one side of the container, air leaks out and the container moves in the opposite direction. How would the container move if the situation were reversed, i.e. a rigid container of vacuum placed in air with a hole on one side of the container? I can't seem to apply Newton's 3rd law and momentum conservation to solve this convincingly.

ANSWER:
Let's just think of a cubical box with a hole in one face. Any molecule which finds the hole will enter the cube, go to the opposite face, collide with it, and rebound, thereby transferring some momentum to the box. The box is now moving in the direction in which the molecule was originally moving (I will call that the forward direction). The rebounding molecule will either go back out the hole or hit inside the cube and bounce back again. As long as the molecule stays inside the box, the net effect will be zero but eventually it will find the hole and so the net effect will be one collision moving the box in the forward direction. Now think of a huge number of molecules entering the hole. At the beginning, more will be coming in than going out so the net force on the box will be forward. Eventually, there will be the same density of molecules inside and outside the cube so the net force will become zero. Bottom line—the cube moves in the same direction as if air were being released from it. An intuitive way to see this is to note that each entering molecule carries a momentum in the forward direction, so that is the momentum available to be transferred to the cube.


QUESTION:
I have a question about force on a lever. I am building a set of oars for a whitewater raft. The industry is full of opinions but very few physicists. We all agree that the force is greatest at the oar lock (fulcrum) but nobody has any idea how much less force is present at the neck (the narrow part just above the blade). If you will do this one...here are the values. Using a 10' oar the fulcrum is at the 32-36" mark and the neck is at the 90" mark. The blade occupies the remaining 30". We taper the oars knowing they don't need to be as strong away from the fulcrum but nobody knows how much we can taper because we don't know how much less force they need to withstand. The taper results in the neck having anywhere from 65% to 80% of the wood on the shaft. I suspect 80% is overkill. Any thoughts.

ANSWER:
A disclaimer: I am not an engineer and an engineer would probably be a better person to ask this question. When you are rowing, the water exerts a force, call it F, on the blade of the oar. I am thinking that the thing we should be thinking about is the torque which this force exerts about the lock compared to the torque it exerts on the neck. This force may be thought of as being 15" from the neck, approximately in the middle of the blade; so the torque from the water force at the neck would be 15F. The torque at the lock would be, assuming the lock is 35 inches from the handle, about 55"; so the torque from the water force at the lock would be 55F. So the torque on the lock would about 55/15≈3.7 times bigger than the on the neck. Now we get into some pretty complicated materials engineering, see this link. It turns out that the stress σ is inversely proportional to a quantity Z called the section modulus. For a cylindrical shape of radius R, the section modulus is Z=0.78R3. So, I surmise that an estimate of how much smaller the neck would be than the lock would be Rneck3/Rlock3≈1/3.7 or Rneck≈0.65Rlock, about 2/3 the thickness.


QUESTION:
If I have a ramp that is 28 feet long, fixed at the upper end (shore) and weighs 400 lbs that has a 6 foot rise, how do I calculate the weight at the lower end (dock)? I am trying to determine how much floatation I need under the water end to support the weight of the ramp at that end. That rise varies during the course of the year from zero feet to a maximum of 7 feet.

ANSWER:
I am assuming that the ramp is a uniform plank, that is, that its center of gravity is at its geometrical center (14'). Refer to the picture on the left. Two equations must be satisfied for equilibrium, the sum of all forces must equal zero and the sum of all torques about any axis must be zero. The first condition gives us that F1+F2-400=0 and the second condition (summing torques about the center of the ramp) gives us that F1-F2=0. Solving these two equations, F1=F2=200 lb. Note that the answer, 200 lb, is independent of the rise.


QUESTION:
Two dice are suspended in outer space with no visible forces acting on them. Their center of masses are 10 cm apart, and they each have an identical mass of .0033 kg. How long would it take for the force of gravity between them to cause them to touch? (We will assume they are volumeless for more ease in calculation).

ANSWER:
This seems a very difficult problem because the gravitational force between them changes as they get closer and so it is not a case of uniform acceleration. However, this is really just a special case of the Kepler problem (the paths of particles experiencing 1/r2 forces) which I have done in detail before. You can go over that in detail. For your case, K=Gm1m2=6.67x10-11x(3.3x10-3)2=7.26x10-16 N∙m2/kg2, the reduced mass is
 μ=m1m2/(m1+m2)=0.0033/2=1.65x10-3 kg, and the semimajor axis a=2.5 cm=2.5x10-2 m. Now, from the earlier answer, T=√(4πμa3/K)=5.98x104 s. The time you want is T/2=2.99x104 s. This is only 8.3 hours and seemed too short to me. To check if the time is reasonable, I calculated the starting acceleration and assumed that the acceleration was constant and each die had to go 5 cm; this time should be longer than the correct time because the acceleration increases as the masses get closer. The force on each die at the beginning is K/r2=7.26x10-16/0.052=3.04x10-13 N; so, the resulting initial acceleration is F/m=3.04x10-13/3.3x10-3=9.21x10-11 m/s2. So, assuming uniform acceleration, 0.05=½at2=4.61x10-11t2. Solving, t=3.3x104 s. So, the answer above is, indeed, reasonable.


QUESTION:
I have a question concerning a magnet suspended inside a copper tube. Does the copper tube accumulate the mass of the magnet? In other words, does the copper tube now weigh more with the magnet suspended in the middle? or is it partial weight because the magnet does fall inside, albeit slowly.

ANSWER:
First, some terminology. Weight is the force which the earth exerts on something, so the weight of the copper tube is always the weight of the copper tube. If a magnet falls through a copper tube, it induces currents in the copper and these currents exert a force on the magnet which tends to slow it down. In fact, the force becomes strong enough that the magnet quickly reaches a terminal velocity—it falls with a constant speed. That means that the tube is exerting an upward force on the magnet equal to the weight of the magnet. But, Newton's third law says that if the tube exerts a force on the magnet, the magnet exerts an equal and opposite force on the tube. Therefore, if the tube is standing on a scale, the scale will read the weight of the tube plus the weight of the magnet, but that does not mean that the tube got heavier. It is just the same as if you put the tube on the scale and pushed down on it with a force equal the weight of the magnet; you would not say that the tube got heavier because you pushed on it. A good demonstration of this can be seen at this link.


QUESTION:
Suppose a constant force is acting on an particle, due to which the particle is accelerated. The the velocity of the particle is increasing at a constant rate. Now if I use the relation P=F.v, i get the power delivered to the particle different at different instants since the velocity is different at different instants. But it is difficult for me to digest that though the force applied is constant, the power goes on increasing. Am I thinking right?

ANSWER:
What is power? It is the rate of change of energy. In the example you give, a constant force in one dimension, the energy at any instant is ½mv2. Just to illustrate, let's let m=2 kg, a=1 m/s2, and the mass begins at rest at t=0. After the first second, E1=½x2x12=1 J; after the second second, E2=½x2x22=4 J; after the third second, E2=½x2x32=9 J; after the fourth second, E2=½x2x42=16 J; etc. So, the average power delivered over the first second is ΔEt=(1-0)/1=1 W; the average power delivered over the second second is ΔEt=(4-1)/1=3 W; the average power delivered over the third second is ΔEt=(9-4)/1=5 W; the average power delivered over the fourth second is ΔEt=(16-9)/1=7 W; etc. The reason you are not "thinking right" is that you deliver more energy to a faster-moving object with a given force over a given time because energy is proportional to the square of the speed. Another way to look at it is that average energy delivered by a constant force F acting over a distance Δx is ΔE=FΔx but, in any given time, the force acts over an ever-increasing distance adding an ever-increasing amount of energy; of course, that is where your power equation comes from, PEt=FΔxt=Fv.


QUESTION:
I've read about space habitat concepts for a while and I've ran into an interesting concept. The concept I've ran into is the McKendree Cylinder which is basically an O'Neill Cylinder made of carbon nanotubes. The O'Neill cylinder made of steel would be 32km long and 6km in diamter. The McKendree Cylinder would be 4600 km long and 460km in diameter. And the maximum length for MvKendree Cylinder is 10000km and diameter of 1000km. So McKendree one could be built a lot bigger than O'Neill one because the carbon nanotubes have greater endurance. But a habitat of thousands of km's seems to be really big when compared to what we can build from other materials. And as I recall we don't have any ways to produce Carbon Nanotubes in large quantities. Is it theoretically possible to build a habitat 10000km long and 1000 wide put of carbon nanotubes. And is the McKendree cylinder more of a theoretical design than a practical design that actually could be built?

ANSWER:
I presume that the issue is more a strength issue than anything else. To illustrate how the strength of the material and its mass determine the size the habitat can be, consider a rotating string of beads, each of mass m. The rotation rate must be such that a=v2/R=g  where v is the tangential speed of each bead. Therefore each bead must experience a force F=mg. This force can only come from the two strings attaching each bead to its nearest neighbors and, from my drawing to the left, F=mg=2Tsinθ. But, we will imagine many, many beads on this string and we will call the distance between them d; so we can make the small angle approximation that sinθ≈θ
=d/R. Solving for T, T=mgR/(2d). Now imagine that the beads are atoms; d will be about the same for steel or carbon, g is just a constant, msteel5mcarbon, and the Young's modulus of carbon nanotubes is about 5 times bigger than steel, Tsteel≈Tcarbon/5. So, Rcarbon/Rsteel≈(Tcarbon/Tsteel)/(mcarbon/msteel)≈25. Your numbers are Rcarbon/Rsteel≈460/6=77; I would have to say that my calculation is pretty good given that I have made very rough estimates and I am not an engineer! I do not know what considerations would limit the length of the habitat. (Of course, neither of these models is presently practical to actually build, so call them theoretical if you like. However, there is certainly no problem building them if resources and manufacturing capabilities were available.)


QUESTION:
If I mounted a gun against a force gauge that measured in pounds and fired it in a vacuum and then fired it in my back yard would it read any different? How much of the recoil is actually produced by the gas pushing against the air in front of the muzzle or is it all newton thrust? There are people who claim that brakes on guns actually pull the gun forward counteracting newton force from acceleration which makes about as much sense to me as moving a boat by standing on the deck and pushing on the mast. Others who claim that part of recoil is from newton force and part is from the gas pushing against the air in front of the muzzle.

ANSWER:
This is a little tricky as are most problems which involve air drag. First of all, look over an earlier answer about the recoil of an M4 carbine. To do my rough calculations, I will assume all the data refer to a gun fired in vacuum; muzzle velocity of 940 m/s, bullet mass m=0.004 kg, gun mass 2.77 kg, bullet diameter 45 mm. I will assume that the acceleration along the length of the barrel (0.37 m) is uniform. (Be sure that you realize that the muzzle velocity is the speed with respect to the gun, not the ground.) So, the bullet starts at rest and the equations which give its velocity and position at the end of the barrel are 940=at and 0.37=½at2 where a is the acceleration and t is the time to reach the end of the barrel. Solving these two equations, I find that t=7.87x10-4 s and a=1.2x106 m/s2. Using Newton's second law, F=ma, we can now estimate the force experienced by the bullet during its flight down the barrel as 0.004x1.2x106=4800 N=1079 lb. Now, suppose that there is air in the barrel. As the bullet flies down the barrel it will experience air drag which will be a force which will work against the force propelling the bullet and therefore the muzzle velocity will be smaller. As you will see from earlier answers, a fairly good approximation for air drag force on an object with cross sectional area A going with speed v is Fdrag≈¼Av2. For the 45 mm bullet, A≈1.59x10-3 m2, so taking the average speed to be 940/2=470 m/s, I find Fdrag≈88 N=19.8 lb. This will be, I believe, an underestimate because the bullet is not just plowing through the air as it would outside the rifle but compressing the air in front of it. Now the net force on the bullet is about 4800-88=4712 N. Now, the average acceleration will be about 4712/0.004=1.178x106 m/s2, slightly smaller; the corresponding time and muzzle velocity are t'=7.93x10-4 s and 934 m/s. In the earlier example I found the bullet speed v and recoil velocity V to be v=938.6 m/s and V=1.4 m/s; my (very rough) estimate for including the effects of air are v=932.7 m/s and V=1.3 m/s. There is slightly less recoil in air. The average force on the gun during the firing time is the same as the force on the bullet (Newton's third law) ≈4800 N=1080 lb for vacuum, ≈4712 N=1059 lb including air. That sounds like a lot, but keep in mind that it only lasts about 0.8 milliseconds. I have included a lot of details here, but you can understand it qualitatively: because the bullet has to accelerate through the air which causes drag, the muzzle velocity will be smaller so the recoil velocity will be smaller as well.


QUESTION:
In Halo video game series there are Magnetic Accelerator Cannons on orbit around planets that can launch a 3000 ton magnetic projectile to 4% lightspeed. These cannons use the principle of the coil gun. These projectiles have a kinetic energy of 216000000000000000000 joules which translates to around 51.6 gigatons of tnt. So these cannons seem to have really unrealistic velocities for these projectiles. What would be the problems in developing these cannons to defend the human species from possible alien invaders? I know energy is one but I've heard that that if you were to accelerate a projectile to these kinds of speeds they would turn into plasma from the sheer amount of energy being transferred into them.

ANSWER:
I think you will get the picture of why this is a preposterously impossible weapon if you read an earlier answer. There the speed was much higher but the mass much smaller. Here are the practical problems in a nutshell:

  • To accelerate it to this speed in a reasonable distance the force required would be so large as to totally disintegrate the projectile and the cannon for that matter.

  • Think about the recoil of the cannon. Unless its mass was much bigger than 3000 tons, much of the energy expended would be wasted, not to mention the disruption of the orbit. This would be a good reason to have it mounted on the ground rather than orbit.

  • Where are you going to get the necessary energy? I agree with your number for the kinetic energy of the projectile (½mv2 works fine for this relatively low speed and a ton here is a metric ton), ≈2.16x1020 J. Suppose it took one minute to get this much energy; then the power required would be 2.16x1020/60≈3.6x1018 W=3.6x109 GW. This is about 1,440,000 times greater than the current total power generated on earth of about 2500 GW. (Of course, that does not take into account the recoil energy of the cannon itself.)

  • Oh yeah, I almost forgot. There is no evidence whatever for alien bad guys.


QUESTION:
If I were to drop an empty wine bottle out of an airplane flying at say 35,000 feet above the ocean at 300 mph, would the bottle hit the surface of the water hard enough to break the bottle? I read somewhere something about terminal velocity being 120 mph, so would the resistance of the atmosphere slow the wine bottle to 120 mph by the time it made impact with the ocean? And would 120 mph be enough to shatter the wine bottle, or would it depend on how choppy the seas were versus a flat water surface?

ANSWER:
When I answer questions involving air drag and terminal velocity, I usually use the approximation that (in SI units) the force F of air drag is F≈¼Av2 where A is the area presented to the wind and v is the speed. So, as something falls, the faster it goes the greater the drag force on it so that, eventually, when the drag equals the weight, the object will be in equilibrium and fall with constant speed. Since the weight W is mg where m is the mass and g=9.8 m/s2, the terminal velocity can be calculated: ¼Avterminal2≈mg or vterminal2√(mg/A). So the terminal velocity depends on the mass and size of the falling object and your 120 mph is most likely not correct. Also, how it falls determines the terminal velocity since it has a much bigger area falling broadside than with the top or bottom pointing down. I figure that if it falls broadside there will be a bigger pressure on the fat side than the neck which will cause a net torque which will make it want to turn with its neck pointing down; so I will assume that is how it falls . I happened to have an empty wine bottle in my recycle bin which has a mass of about 0.5 kg and a diameter of about 8 cm. When I calculate the terminal velocity I get vterminal63 m/s=140 mph. The 120 mph number you heard was probably a typical terminal velocity of a human, and it is just coincidence that the wine bottle has a terminal velocity close to that.

It is hard to say whether it would break or not. I think probably not. Suppose that it took 1 s to stop. Then the average force on the bottle would be F=ma=(0.5 kg x 63 m/s)/(1 s)=31.5 N7 lb which the bottle should be able to withstand. I know that they say that at high speeds hitting the water is like hitting a brick wall, but if the stopping time were 0.1 s the force would still only be about 70 lb.

(Who would have thought that I would find a picture of a falling wine bottle? You can find anything on the web!)


QUESTION:
This question is regarding Newton's cradles. Let's say a particular one had 5 balls. We pull back 3 balls and let go. Now the other side must rebound with 3 balls as well. This means that the middle ball must carry on swinging. Does it ever theoretically stop for the briefest of moments? What about in real life with compression and other factors?

ANSWER:
I presume you have read the recent answer below. Usually, in collision problems, we ask what is going on before the collision and after the collision, but not during the collision. The details during the collision time depend on the details of the interactions among the balls which we generally do not know. So in cases like Newton's cradle, we approximate the collision to occur in zero time and therefore approximate all accelerations to be instantaneous; this is clearly not the case since it would require an infinite force to stop or start a ball instantaneously. So, to be more realistic, we must, as you have done by mentioning "compression", devise some model for the collisions. With steel balls we could assume that each was a very stiff but perfectly elastic spring which compressed during the collision. Then, assuming the collision were perfectly elastic (again an approximation), the collisions would happen in a very short but nonzero time. The two incoming balls would stop in a short time because the middle ball would exert a large backward force on them and the two outgoing balls initially at rest would get up to speed in a very short time because the middle ball would exert a forward force on them. Therefore, with this simple model, assuming all the balls are identical, the middle ball would experience a zero net force during the collision time and so it would proceed forward never changing its speed. I could imagine that the middle ball might slow down slightly and then speed back up to its original speed, but not stop. Even if the collisions were perfectly inelastic, the middle ball (and all others) would move with speed 3/5 the speed it came in with afterwards. Whatever happens, the middle ball would certainly never be at rest.


QUESTION:
This one has bothered me since seeing it on TV the other evening. The device called Newton's Cradle, that executive metal ball swing toy found on many desks - I understand the basic principles of the conservation of motion, energy, etc. My mind can also wrap itself around the fact that when you raise and drop one ball, one ball on the opposite side responds by moving. What I don't understand though is why when you raise and drop two balls, TWO balls on the other side respond by moving. It would seem that to conserve energy/motion, the kinetic energy of the two balls that was raised would be transferred into the last ball and it would swing out twice as much as the two balls that started the motion. How do the two balls on the end "know" that it was two balls that started it? Can you please explain this?

ANSWER:
Be clear that the approximation to use is that the collisions are elastic, both energy and momentum must be conserved; for steel balls, this is a pretty good approximation. OK, let's assume that two balls come in with speed v1 and one ball goes out with speed v2. Conserving momentum, 2mv1=mv2 or v2=2v1. Now, look at energy: E1=½(2mv12)=mv12 and E2mv22m(2v1)2=2mv12. Energy is not conserved, mv12≠2mv12. You can actually prove if that a mass M comes in, only if the mass going out is M will both energy and momentum be conserved. Here is the proof:

  • momentum: MV=mv => v=(MV/m)

  • energy: ½MV2mv2m(MV/m)2 => M=M2/m => m=M

  • since m=M, v=V from momentum.

Nature "knows"! You can't fool Mother Nature.


QUESTION::
I have searched for an answer to this question in vain. A dropped large ball, with a small ball on top, will bounce and transfer the upwards momentum into the small ball, launching the small ball very high into the air. (As far as I understand this, I saw a demo). Ignoring air resistance for now, how heavy would the large ball have to be, to launch a one kilogram weight into orbit, and what would the height of the drop have to be? As an example, a 100 tonne ball dropped from 10 metres, with a 1kg ball on top, does a bounce, what is the velocity result for that? How does the elasticity of the ball affect it? (eg, a large metal ball or weight might just impact the ground).

ANSWER:
This problem is fully explained and worked out at this link. The final result, if the mass of the big ball is much bigger than the mass of the little ball, is that the speed the little ball rebounds is given by v≈3√(2gh) where h is the height from which it was dropped. This means that the masses really do not matter. The height to which the smaller ball bounces is 9 times the height from which it was dropped. It is assumed that all collisions are perfectly elastic. The speed required for a near earth orbit is about 8x103 m/s, so h=v2/(18g)=3.6x106 m! To put this in perspective, the radius of the earth is about 6.4x106 m, so h is about half this. This is a really rough calculation because g will be considerably smaller (by a factor of about 0.4) at this altitude. Even if you did make this work, the ball goes straight up and you need it to go horizontally to go into orbit. I think it is not a very practical idea!


QUESTION:
What role does the force of friction play in the movement of a cycle....?? I have learnt that the friction acts in the forward direction on the rear wheel and backward direction on the front wheel.How can it be...?? I think I am missing a broader link to why a wheel actually rotates.

ANSWER:
In this discussion I will ignore all friction except that which occurs due to the contact of the wheels with the road. The two kinds of friction we normally learn about in elementary physics courses are static and kinetic friction. Kinetic friction is the frictional force between two surfaces which are sliding on each other; it is kinetic friction which stops a box sliding across the floor. Static friction is the force between two surfaces which are not sliding on each other; it is static friction which keeps your bike from skidding when turning a corner. A third contact friction force is called rolling friction; this is not terribly important for a bike but is the force which will eventually stop you if you coast on level ground and you are most aware of it if your tires are under-inflated—harder to pedal. Rolling friction on both wheels will always point backwards. First think about a bike which is not skidding. If you go in a straight path on level ground without pedaling, only rolling friction stops you and neither kinetic nor static friction are in play. Now suppose you start pedaling to accelerate forward. Think about what your back wheel "wants to do"; if the road were icy, the wheel would spin and the force which keeps it from spinning on a dry road is static friction. The road will exert a forward force on the wheel to keep it from spinning. A force is required to accelerate anything and this static friction force is what accelerates your bike forward. The friction on the front wheel is just the rolling friction backwards. Finally, suppose you brake: if both brakes are applied gently enough that you do not skid, static friction on both front and rear wheels will point backward; if both wheels skid during braking, kinetic friction on both front and rear wheels will point backward. You probably know that if you apply the brakes so that they are not quite skidding, you will stop in a shorter distance than if you skid. When you round a curve, both wheels have a static friction force which points perpendicular to your direction and toward the curve's center because you are accelerating when you move in a circle even if you are going with constant speed. More detail on rolling friction and on bicycle turning can be found in earlier answers.


QUESTION:
I had asked you a question and you had answered it. To illustrate your point, you had quoted an example that a pendulum hanging from the ceiling of a car moving with a uniform acceleration makes an angle with the vertical. Now we have been taught in our course as well. We have also learnt that tangent of the angle is equal to a/g. My question is that will that pendulum exhibit uniform oscillations if it is displaced from its normal position?

ANSWER:
I presume that by "its normal position" you mean its new equilibrium position off the vertical. The answer to your question is that it will exhibit oscillations about this position; the period, though, will not be given by T≈2π√(L/g) but rather by T'≈2π√[L/(g2+a2)]. This is most easily understood by introducing a fictitious force ma opposite the direction of the acceleration. Now you can see that the force exerting a torque on this pendulum has a magnitude of m√(g2+a2) rather than mg as in the unaccelerated pendulum. So it is just like the usual pendulum analysis with a somewhat larger acceleration due to gravity.


QUESTION:
Are the equations of motion v = u + at, S = ut + 0.5at^2 , and v^2 = u^2 + 2aS applicable if a particle travels at a speed close to that of light? I think that they are applicable, and only F =ma is not, but my friend says that those 3 equations are also not. Who is right?

ANSWER:
Your friend is right. These are the classical equations for uniform acceleration, and uniform acceleration is not possible in the theory of special relativity for a constant force. Constant force, where F=dp/dt, (which, incidentally, is the way Newton originally wrote his second law) can be constant but does not result in constant acceleration. I have treated this problem in detail in an earlier answer; there I find that
v=(Ft/(m))/√[1+(Ft/(mc))2] and S=(mc2/F)(√[1+(Ft/(mc))2]-1) (for your u=0, that is, the particle at rest at t=0) for constant force.


QUESTION:
This is a question about tractive effort of locomotives. I am a model railroader and we have been debating this situation for a while with no clear answer. If there is a locomotive on the track that has 4 driving wheels touching the rails, and the only change you make is to the number of wheels , now 6 driving wheels on the track, will tractive effort go up, stay the same or drop. The argument for staying the same is that each wheel now supports less weight, so tractive effort of each wheel would drop. The argument for tractive effort going up would be that there are more contact points on the rail, so it would pull more.

ANSWER:
What drives your train is the static friction between the wheels and the track. The nature of static friction is that a maximum amount of force may be achieved before the surfaces slip on each other. This force Fmax is determined, to an excellent approximation in many cases, by the nature of the surfaces (steel on steel for your case, I presume) and how hard they are pressed together (called the normal force N); Fmax=μN where μ is the coefficient of static friction, a number determined by the materials. You will note that this force does not depend on the area of contact. Essentially, adding wheels adds surface area but, as you note, the force pressing the surfaces together (namely the normal force which is determined by the weight) is simply distributed over a larger area. So, the simplest first-order physics answer is that the same maximum force from the wheels without slipping will be achieved regardless of the number of wheels. Friction, though, can be a tricky business and first-order physics does not always work for all situations. For example, there are instances where a car tire can have greater "road hugging" (read increased friction) if the road-tire contact area is increased. I believe in your case, though, since the wheels and track are so little deformed by contact, that increasing the number of wheels will not increase friction provided that the total mass of the locomotive remains the same. Of course, if you are adding wheels to an existing locomotive, you are adding mass so you will be increasing Fmax. There is a very easy way to test this. Put the locomotive (wheels not free to rotate) on a piece track which is attached to a horizontal board. Gradually increase the angle of the board until the train slides down. The tangent of the angle at which it starts to slide will be equal to the coefficient of static friction. If increasing the number of wheels does not increase the slip angle, there is no advantage.


QUESTION:
If a plane, while mid-flight, had an explosion from the wing or rear area, would the debris maintain the exact same momentum as the plane and go forward alongside it, would the debris go in front of the plane, or would the debris go behind the plane when the accident first happens? Me and two friends got on the topic after watching a movie dealing with a plane crash. One friend thinks debris would go in front, the other thinks the debris would maintain precise momentum with the plane, and I'm thinking it would go behind the plane. We'd really love for a physicist to clear up this question for us, thank you so much!

ANSWER:
What anything does is determined by the forces on it. A plane flying in a straight horizontal line with constant speed has four main forces acting on it: gravity (its own weight); lift which is the force which counteracts the gravity to keep it flying level, drag caused by the air, and the forward force exerted by the engines which counteracts the drag to keep it from slowing down. If a piece of the plane suddenly separates from the plane, it no longer has any foward force and it no longer has any significant lift; so, it will start dropping vertically and slowing down horizontally, falling down from and behind the plane. The falling is generally more prounounced than the slowing down (gravity usually a greater force than drag) so there would be a tendency for the piece to appear to just drop straight down as seen from the plane. (Be sure to realize that the forward velocity of the piece is approximately maintained so that someone on the ground sees it moving forward with about the same speed as the plane.) This is often seen in bombs dropped from a plane. On the left photograph the drag on the bombs is small so they keep pace with the planes as they drop. On the right, the bombs have little parachutes to increase drag and so they fall behind the plane.


QUESTION:
If some object, at some distance far from the surface of the Earth, but much closer to the Earth than anything else (meaning everything else in the universe is negligible) is at rest initially but then begins accelerating towards the Earth due to gravity, how long will it take to get to the Earth? I get that the acceleration is a=GM/r^2. And then you can write a as the second derivative of r which would make it a non-linear differential equation but I have no idea how to solve those really. I have taken a differential equations course but it was 3 years ago so I don't remember if we even solved non-linear ODE's ever.

ANSWER:
I have solved variations of this problem twice before, one very recently regarding Coulomb's law (simply another 1/r2 force) and the other where the object was falling into the sun instead of earth. You should read through these first since I will skip a lot of the detail here. In the spirit of those two answers, I will take you at your word that the object is "far from the surface of the earth", that is, r>>Rearth. Hence, it is just the Kepler problem with an orbit of eccentricity 1 and semimajor axis a=r/2 and we need to find half the period T. Kepler's third law states that T2=4π2a3/(GMearth). Therefore, T/2=π√[r3/(8GMearth)]. For example, if r=100Rearth (which would mean at an altitude of 99Rearth), T/2=9x105 s=250 hr.

You can always estimate how much error is made in this approximation (assuming the earth and objects are point objects going all the way to zero separation) by calculating the speed v the object arrives with at earth's surface. Energy conservation gives -GMearthm/rmv2-GMearthm/Rearth or v=√[2GM(1/Rearth-1/r)]. For the example I did, v=√[(198/100)GMearth/Rearth]=1.1x106 m/s. If the earth were a point mass, the object would continue speeding up. If it kept going into the earth at a constant speed, the time it would take to reach the center would be about 6 seconds, enormously shorter than the time to get to the surface, so the approximation is superb for r=100Rearth.


QUESTION:
i tried to find the time it would take for two charges to collide under electrostatic force,realizing simple kinematics won't cut it,tired to integrate but failed,how is the question done?and how does it differ from gravitational force??

QUERY:
You have to define what "collide" means. Since it is a 1/r2 force, if you use point charges the velocity will be infinite when they collide but they will do so in a finite time. Also, what are the initial conditions (velocities, positions), masses, charges.

REPLY:
My initial question was the time it would take for a 1/r^2,to collide,for example two bodies lets say 1g,and a charge of 1micro coluomb,initial at rest attract each other,and collide,I can't use simple kinematics to solve this question,what shall I do?

QUERY:
How far apart are they?

REPLY:
ok,for simplicity 1m apart,is there a general formula than can help?

ANSWER:
Whew! I finally have everything I need. This is the Kepler problem, the same, as you suggest, as the solar system with gravity. You may want to look at an earlier answer similar to yours. It is very lengthy to work out the whole problem in detail so I will refer you to a very good lecture-note document from MIT; I will just give you some of the necessary results to calculate what you want. First, a brief overview of two of Kepler's laws:

  • Kepler's laws refer to problems where the force is of the form F=K/r2 where K is a constant and the force is attractive. So it could refer to either two masses or two opposite charges.

  • The first law states that bound planets move in ellipses with the sun at one focus. This is really only true if the sun is infinitely massive but the generalization still leads to an elliptical orbit for each body, both of which move around the center of mass of the two. Still, the semimajor axis a of the ellipse (which we will later need) in the center of mass system can be found for any orbit from the simple equation a=-K/(2E) where E is the energy of the system.

  • For your case, the particles move in a straight line toward each other and then turn around and return to their original positions. This is just the most elongated possible ellipse with an eccentricity of 1. Of course this would never really be possible in the real world since the particles would be going an infinite speed when they "collide". That means we really should do the problem relativistically which would greatly complicate the problem. Keep in mind that you are asking an unphysical question requiring point charges and infinite forces and velocities. But the answer below should be a good approximation of the time if they have some finite size small compared to their initial separation.

  • The third law relates the period of the orbit T to a: T2=4πμa3/K where μ=m1m2/(m1+m2) is the reduced mass. In the gravitational problem, K=Gm1m2 and in the electrostatic problem, K=keq1q2 where ke=9x109 N·m2/C2.

For your case, the energy is given since the charges are initially at rest and separated by some distance S, so E=V(S)=keq1q2/S and so a=keq1q2/(2keq1q2/S)=S/2=0.5 m; the reduced mass in your case is μ=m1m2/(m1+m2)=10-3x10-3/(2x10-3)=0.5x10-3 kg; and K=kq1q2=9x109x10-6x10-6=9x10-3 N·m2. Finally, the time it takes for a complete "orbit" (which would correspond to the particles returning to their original positions) would be T=√[4πμa3/K]=√[4π(0.5x10-3)(0.5)3/9x10-3]=0.3 s. But, the time you want is just half a period, T/2=0.15 s.

To help you visualize the orbits, the figure below shows the orbits for the two charges when the eccentricity is just less than 1; imagine the orbits getting flatter yet, approaching two straight lines.

NOTE ADDED:
I got to wondering what the limits of doing this classically are, that is, how good an approximation my calculation above would be for some real system. This requires that I determine how close the two charges would approach each other before their speed v became comparable to the speed of light c. I will use the same notation as above and write things classically. If released a distance S apart, then when they reach a distance r apart energy conservation gives: keq1q2/S=keq1q2/r+½μv2 which results in v=√[(2ke|q1q2|/μ)(1/r-1/S)]. For the case in point, if I solve for r when v=c/10, a reasonable upper limit for a classical calculation, I get r=4x10-8 m, about 100 times bigger than an atom. Alternatively, we could ask what the velocity would be for a 1 mm separation, r=S/1000: v=√[(2ke|q1q2|/μ)(999/S)]=1.9x105 m/s=0.00063 c. In either case, I think we can conclude that the time remaining to complete the half orbit will be extraordinarily small compared to 0.15 s.


QUESTION:
does the terminal velocity in a parachutist's fall occurs two times? one before opening the parachute and one some time after opening the parachute Am i right?

ANSWER:
The terminal velocity is determined by (among other things) the geometry of the falling object. In the simplest approximation, what matters is the cross sectional area of the falling object. So, when the parachute is opened, the area gets much bigger and the terminal velocity gets much smaller. If the sky diver has achieved some larger terminal velocity, she will slow down to the new terminal velocity. However, this need not be the first time the terminal velocity changed. The sky diver can orient in a ball or like a down pointing arrow and have a relatively large terminal velocity or she can fall spread-eagle and slow down. So the terminal velocity probably changed several times before the parachute was opened.


QUESTION:
i am unable to understand a pulley problem

ANSWER:
This is a most peculiar problem! However, its solution ends up being very easy to do (if not to visualize). I will assume that all pulleys are massless and frictionless and that the string is massless and unstretchable. I always tell my students when attacking this kind of problem to "choose a body" upon which to focus and apply Newton's second law. There are two obvious choices here, the left-side m which results in -T+mg=maleft and the right-side m which results in -2T+mg=maright. So, we have two equations and three unknowns; the best you can do is to find the relation between the two acclerations, 2aleft+aright=3g. To generate a third equation, choose the left pulley which results in 2T-T=0=T. So, the only result which works is aleft=aright=g, both masses in free fall. I was having a lot of trouble visualizing how this happens, so you I made a little model and made a movie of it. To see it, go to the Ask The Physicist facebook page. (I know, it's not going to win an academy award, but it satisfied my curiosity!) The two masses, with equal accelerations, will move equal distances in equal times which then made it pretty easy for me to make a little model on the floor with strings and paint cans. Apparently the left pulley moves twice the distance the right pulley moves in any given time.


QUESTION:
I've been trying to write a action sci-fi screenplay, but there is one problem that I can't get my head around. During the climax, two characters fall from a 91 story building, with the first jumping off, and the second falling approximately 10 seconds later. The second character proceeds to catch up to the first, and they then brawl. My question is, how much time would pass before they would hit the ground?

ANSWER:
There is a good reason you can't get your head around it: the second character will not catch up to the first! I will give you a little basic physics tutorial on possible scenarios. For purposes of computation, I will assume that the height of your building is about 400 m (about 13 ft/story) and I will approximate the acceleration due to gravity to be g=10 m/s2.

  • First I will assume air drag is negligible, that they fall as if they are in a vacuum. Then the height y1 above the ground of character 1 is given by y1=400-5t2 where t is the time since he jumped. This tells us that when character 2 jumps, y1=400-500=-100 m; since y=0 is the ground, he has already hit the ground! The freefall time (the time when y1=0) from 400 m is about 8.9 seconds and the speed when he hits the ground will be about 89 m/s=200 mph.

  • So, maybe we just need to add some air drag. That certainly will be important for speeds on the order of 200 mph. When air drag is taken into account, you do not continue speeding up forever but eventually fall with some maximum constant speed called the terminal velocity; this happens when the air drag (up) is equal to your weight (down). If your mass is M, your weight is Mg or, with my approximation, 10M. Let's use M=100 kg (about 220 lb) so Mg=1000 N. A good approximation for air drag is FAv2 where A is the cross sectional area of the falling object and v is the speed. I will choose A=1 m2 so, if I set F=1000 and solve for v, I find a terminal velocity of vt=63 m/s=140 mph. The details of the math gets a little complicated here, so I will give you the results and spell out the details below for anybody interested. I find that the time that it takes to reach the ground now is about 10.8 seconds at which time the speed is about 59 m/s. So character 2 has less than a second to catch up to character 1, obviously impossible.

  • Finally, I should tell you that if the two characters are about identical, about the same size and weight, character 2 will never catch up with character 1 regardless of the height of the building. In the no-air-drag situation, both have the same acceleration and so their paths never cross. In the air-drag situation, both have the same terminal velocity so they will end up having that speed and separated by some constant distance. The only way to make it happen is to greatly change the cross sectional area of character 1, for example, give him a parachute!

I am guessing that, if you want to have any semblance of reality, you will want to rethink your climax!

ADDED DETAIL:
Vertical fall with quadratic air drag is a well-known problem worked out in any intermediate classical mechanics textbook. The two results which I used were v=vt√[1-exp(-2gh/vt2)] and t=(vt/g)tanh-1(v/vt) for an object dropped from height h.


QUESTION:
Was killing time on a slow day at work, and involved a coworker in the site. Some time later he ask a question, which might be up the physicist alley. Assuming we had a person seal in a vacuum with a piece of paper and a baseball, if he threw the baseball and then crumpled up the piece of paper and threw it. Would the paper fly as far as the baseball having no friction to affect it, or does gravity rule supreme and bring it to a halt just as well as it does outside a vacuum?

ANSWER:
Let's assume that you are throwing both as hard as you can which means you exert equal forces on them. Also assume that you exert that force over the same distance for each case which means you do the same amount of work (force times distance) on each. That means that they have equal kinetic energies when you release them. But, kinetic energy is ½MV2 where M is the mass and V is the speed. So, suppose the baseball has a mass 100 times the sheet of paper; then, if they have equal kinetic energies, the speed of the paper will be 10 times larger than the baseball and therefore go 10 farther. Your last sentence makes no sense because gravity does not "bring it to a halt", it is friction which does that.


QUESTION:
Suppose a bullet is fired parallel to the ground, due to perpendicular direction of work done with respect to gravity, no work is done against it. Then why does not the bullet fall immediately to the ground as it does when it is not in motion? I have speculated that motion in in dimension reduces the effects of forces of other dimensions. Is there really such a thing in physics? If not, then what is the correct explanation?

ANSWER:
First, do not worry about what was happening to the bullet while the gun was firing it; that is past history and has nothing to do with what happens to the bullet after it leaves the gun. Newton's first and second laws tell us that the only reason something will not move in a straight line with constant speed (or be at rest which is a constant speed of zero) is if there is a force acting on it. Further, the change in motion the object experiences is in the direction which the force points. Normally in elementary physics classes, we neglect the force of air drag (which is a pretty poor approximation for a bullet) such that the only force is that due to gravity which we call the weight of the object. So, all the bullet can do is change its motion in the vertical position because that is the direction that the weight points. So even though your bullet starts with no vertical motion, it will fall just as it would if you simply dropped it. In other words, it will take just the same time to drop to the ground whether you fire it horizontally or simply drop it. In an earlier answer, you can see a strobe photograph where a ball launched horizontally and one dropped fall vertically the same. In the real world, air drag is not negligible and is a force which always points in the direction opposite the velocity. This causes the bullet to slow down in the horizontal direction but to also speed up more slowly vertically. The figure shows an example where the conditions allow the whole range of scenerios to play out: because of the air drag, the horizontal motion is eventually stopped and, because of the interplay between weight and drag, the projectile ends up falling vertically with a constant speed called the terminal velocity.


QUESTION:
I am in my powerful rocket. If my orbital speed around the sun was zero and I was the same distance from the sun as the Earth, and my rocket was firing at the exact velocity to hold my position to keep the rocket from from falling towards the sun but no greater, what would my 200 lb Earth body weigh on the rocket ship scales? I am guessing a bit less than 800 lbs. What do you say?

QUESTION TO QUESTIONER:
What does that mean, "my orbital speed around the sun was zero"? If your speed is zero you are not orbiting. If you are at rest a distance of the earth's orbit from the sun, I would call that hovering. Then you have to calculate the sun's gravitational force on you to know your "weight".

REPLY:
You are right, I did wish to know my weight when hovering,

ANSWER:
The force W which the sun exerts on you (aka your "weight") is given by W=GMm/R2 where G=6.67x10-11 Nm2/kg2 is the universal constant of gravitation, M=2x1030 kg is the mass of the sun, R=1.5x1011 m is the distance to the sun, and m=91 kg is the mass of a 200 lb weight. If you do the arithmetic, the result is W=0.54 N=0.12 lb. Quite a bit less than 800 lb! You might be interested in a similar question where the questioner wanted to know the time it would take the earth to fall into the sun if it had no orbital speed.


QUESTION:
i have to know that why the mass of bob in a pendulum does not affect the time of oscillation.....i think it should as the more massive bob should have more potential energy when string is pulled for 10degrree but at the same time when its moving more air resistance would also act on it. My book of physics says that the number of oscilllations does not affect the final answer of time period per each oscillation. as we know that a pendulum should automatically stop moving after some time. This means that if 100 oscillations are taken then the time per one oscillation should be decreased. I know im wrong as my physics book of GCSE says so but i want to know why im wrong and what part of my staement does not make sense

P.S : please use simple language or formulas(if u wanna use) as i wont be able to understand complex ones.

ANSWER:
Let's get straight what we are talking about. You refer to air resistance and you note that the pendulum will eventually stop. Of course, the main reason it stops is air resistance. Any air resistance is assumed to be negligible in a treatment of the pendulum at your level (9th grade), so imagine that there is none. Second, your book should tell you that the pendulum only behaves in a simple way if the angle you start it at is small; but even 300-400 is sufficiently small for the standard treatment of the simple pendulum to be quite accurate. However, your expectation that the air resistance will affect your measurement of the period is wrong because, to an excellent approximation, the period does not depend on the amplitude and so one period when you start and the amplitude is, say, 200, will be almost exactly the same as one period when the amplitude has decayed down to 50. Given your level of school, I cannot give a very rigorous explanation for why the mass does not matter. But, if you can understand resolving a vector into components and know Newton's second law, that F=ma, then you should follow my explanation. In the picture to the right you see the forces, mg and T, which act on m. But we are interested in the motion of m along its arc and neither T nor the component of mg perpendicular to the arc can affect that motion, so all that matters is the component of mg along the arc which is mgsinθ and this is the force which causes the mass to accelerate along the arc. So, applying Newton's second law for motion along the arc, mgsinθ=ma; you will notice that m cancels out so that a=gsinθ. Clearly the acceleration will determine the motion along the arc and therefore the period will not depend on m. But, of course, a real scientest tests his hypothesis. Go to a park where there are swings and find a little boy and his father both swinging; you will see that they both have the same period even though they have very different masses. (Your argument about potential energy is not valid because, although the potential energy at the top is, indeed, proportional to m (U=mgh), the kinetic energy at the bottom is also proportional to m (Kmv2). So, again, mass cancels out.) Finally, I should tell you what the period does depend on: only on the length of the string and the acceleration due to gravity, T=√(L/g)]/(2π). A pendulum clock will not keep good time on the moon.


QUESTION:
Im trying to figure out what speed a sphere weighing 4,506 lbs will attain on a 3 degree grade from complete stop to 95 ft. [After communicating, it is a uniform sphere of radius 4', rolls, does not slip.]

ANSWER:
The easiest way to solve this is conservation of energy. Let's work in SI units, m=4506 lb=2044 kg, d=95'=29 m, and R=4'=1.22 m. The energy at the top of the ramp is E1=mgh where g=9.8m/s2, and h is the height above the bottom which is h=dsin30=0.052d. The energy at the bottom where the speed is v is E2mv22 where I=2mR2/5 is the moment of inertia and ω=v/R is the angular velocity. Putting this all together and solving for v, v=√(0.058gd)=4.07 m/s=13.4 ft/s=9.1 mi/hr. So, you see, the answer to your question does not depend at all on either the mass or the radius of the sphere, R and m cancel out as you can show for yourself.


QUESTION:
Figure skating - if you are doing a spin on ice and you leave your arms spread out it creates resistance, if you pull the in towards the body you create less resistance so you spin more easily. How do I express this with mathematical equation in physics? [The questioner is a 13 year-old high school student with no physics training.]

ANSWER:
While it is true that there is somewhat more air resistance if your arms are outstretched, you are barking up the wrong tree if you think this is the reason your spin speeds up when you pull your arms in. The principle involved here is conservation of angular momentum which is a topic usually covered late in an introductory physics course, so I have to give you a little background here. For translational physics (objects moving without spinning), Newton's second law is perhaps paramount: the rate at which linear momentum changes is equal to the force on the object. Linear momentum p is defined to be the mass m of the object times its velocity v, p=mv. If an object has no forces on it, the rate of change of linear momentum must be zero which means that it never changes; this is called conservation of linear momentum. For a single object, this means that it continues moving in a straight line with constant speed because the mass will not spontaneously change. In rotational physics, Newton's second law is analogous to translational physics: the rate at which angular momentum changes is equal to the torque on the object. Let's briefly discuss what the quantities are in translational and rotational physics:

  • Force is what causes an object to accelerate, change its velocity. Torque is what causes an object to spin faster (called angular acceleration). Note that it is possible to exert a force on something without causing a torque, for example, pushing on a door where it is hinged does not open the door.

  • Linear momentum is p=mv. Angular momentum is L=Iω, the moment of inertia (see below) times the angular velocity (for example, revolutions per minute).

  • Mass is the resistance a body has to being accelerated; a one pound force exerted one second will cause a much bigger acceleration for a baseball than for a locomotive. Moment of inertia I is the resistance of a body has to being angularly accelerated (increased spinning speed). But, monent of inertia does not just depend on the mass of the body, it also depends on how that mass is distributed. In the case in point, the figure skater has a larger moment of inertia with her arms outstretched than not; the same torque applied to each will result in a smaller spin rate for the outstretched arms.

  • If you exert no force on an object, its linear momentum must remain constant. If you exert no torque on the object which is spinning, its angular momentum must remain constant.

Are there any torques on the skater who is spinning? Yes, there is the friction of the skates on the ice and the air drag on her body, and if she just does nothing she will eventually stop spinning. However, those torques are small (it would take a pretty long time for her to stop), so we can say that angular momentum is approximately conserved. Now, suppose that her moment of inertia with arms outstretched is Iout and with her arms pulled in is Iin=Iout/3, and that she is initially spinning with ωout=2 revolutions per second. Then apply momentum conservation:

  • Iinωin=Ioutωout

  • 2Iin=Iinωout/3

  • ωout=6 revolutions per second.

That's the whole story. Because angular momentum is conserved, reducing the moment of inertia results in increasing the angular velocity.


QUESTION:
I teach a self defense course. I am trying to explain to my students that a 230 grain bullet moving at 850 ft per sec not stop a 200 lbs person running at them at full speed. If i am correct or not could ypu please put it into terms that i could use.

ANSWER:
The concept you want to use here is momentum conservation. Momentum is the product of mass times velocity and the total momentum of a colliding system must be the same before and after the collision. The one subtlety here is that if  the momentum of a 2 lb mass moving 300 ft/s north is 600 lb·ft/s, then a 2 lb mass moving 300 ft/s south  has a momentum of -600 lb·ft/s. I will use lb to measure mass and ft/s to measure velocity, so you have to be sure you get all the units the same to make any sense: 230 grain=0.033 lb. You say "full speed" so I will use 30 ft/s to be the person's velocity, which would correspond to a 10 second 100 yard dash. So, the momentum before the bullet strikes (and lodges in) the man is 200x30-0.033x850=5972 lb·ft/s. (Notice the - sign on the momentum of the bullet since it and the man are going in opposite directions.) After the bullet lodges in the man the momentum is 200.033xV where V is the velocity of the man (plus bullet) after he is hit. Setting the momenta before and after equal and solving for V, V=5972/200.033=29.9 ft/s. So, you see, the man loses almost no velocity at all due to the bullet. If you hit his head or his heart he might stop running but it is certainly not the impact with the bullet which stops him. A qualitative way to put this is that while the bullet has a speed about 30 times greater than the man, it has a mass almost 7000 times smaller. Related to that, when you see a movie in which a character is sent flying backward by the impact of a bullet, it is total nonsense and will not happen in real life.


QUESTION:
does the amount of force(i.e the amt. of RECOIL that a gun experience) require to hold a operating gun in position ,depends on the mass of it? consider the following example - there are two persons A and B.A is having a 10 kg gun and B is having a 5 kg gun. Both guns are shooting the same type and number of bullets per second. Now ,which of the two persons would require to exert a greater force to hold the gun in position? If the force require to hold both the guns in position is the same(which I think it is ),then is it not counterintuitive?? I mean ,imagine to exert the same force to the 50 kg gun and a 5 kg gun, to hold them in position??

ANSWER:
This problem is a little trickier than I expected. Here is the way I approached the problem: imagine that there is nothing holding a gun of mass M shooting a bullet of mass m. Then you can use momentum conservation to find the relative speeds of the bullet, v, and the gun, V, after being given an amount of energy (from the gunpowder in the bullet) E; V=(m/M)v. Usually, M>>m, so v>>V, so it might seem that the lighter gun would go faster and be harder to stop; but the catch is that it is easier to stop something with less mass, so it is not really clear which is easiest to stop until we can write what the speed V is. You might think I have already done that, but the speed of the bullet v depends on the mass of the gun. To get rid of v in my equation for V you need to introduce the constant energy which gets given to the gun and the bullet, Emv2MV2. If you now solve this equation for v, substitute that into the equation for V, you find V=√[(m/M)(2E/(m+M))]. Now, in order to stop this motion in a short time Δt you must exert a force F=MVt=√[(2mME/(m+M))]/Δt. If you examine this, you will see that if M>>m, as is almost always the case, F≈√[2mE]/Δt, independent of M. For smaller gun masses where the bullet and gun have comparable masses, I have plotted the force (divided by the constant √[2mE]) as a function of M/m. This is shown in the graph to the right and you can see that it is easier to stop a lighter gun.

For the usual case where the gun is much more massive than the bullet, it is easy to see intuitively that force is independent of mass. In that case, the velocity of the bullet will be essentially the same for each case so v is not (to an excellent approximation) a function of M. So, V is inversely proportional to M and, since F is proportional to MV, F must be independent of M.

NOTES ADDED:
It occurred to me that one could also define "recoil" as the force necessary to stop the gun in a given distance Δx (rather than in a given time Δt as above). Then the magnitude of the work done by that force must be equal to the kinetic energy the gun had, FΔx=½MV2M(m/M)(2E/(m+M))=E/(1+(M/m)). In this case, it is clearly easier to stop a heavier gun.

So, how recoil is defined is important in discussing this question. A little research reveals that, indeed, gun enthusiasts do not think of perception of recoil as what they mean by recoil. Rather, there are several types of recoil which are used to quantify the effect. Recoil velocity is what I have called V; recoil energy is ½MV2; recoil momentum is MV; recoil impulse is the time it takes the gun to accelerate from rest to V, essentially the time the bullet is in the barrel (not what I call Δt). This is eminently sensible since each of these is unambiguously defined. Perception is rightly considered to be qualitative, dependent on the person firing the gun.

I also should note that my analyses above assume that perception begins when V has been acquired, but this is not really what happens because you actually start feeling the recoil at the instant the gun is fired. So you can ask what force you need to exert during the time the bullet is accelerating to hold the gun at rest; so, when the bullet exits the barrel, you experience no more force. Clearly this force is independent of M since the gun never moved. Now all bullets to which a constant energy E has been added exit with a velocity v=√(2E/m) because by holding the gun still you effectively make its mass infinite. Approximating that the acceleration, and therefore the force, is constant while the bullet is in the barrel, it is pretty easy to show that F=E/L where L is the length of the barrel. Again, there is no dependence on M. Of the three kinds of "recoil perception" I have defined, I think this last one makes the most sense.


QUESTION::
Newtons Law states that to every action there is an opposite and equal reaction. If a cue ball is stuck hard enough and strikes another ball, the other ball moves off with the same speed at which it is struck, while the cue ball stops dead. This is providing the cue and cue ball and struck ball are in a dead straight line. Does this mean that the cue ball has transfered all of its energy to the ball that moves off therefore not being able to follow on? Have just found your site and find it very interesting and so give to you my thanks.

 

ANSWER:
This is an "action/reaction" example (Newton's third law) only in an indirect sense: because the balls, when colliding, exert equal and opposite forces on each other, their total momentum must not change (be conserved). Billiard balls are very elastic and therefore, to a good approximation, energy is conserved also. For the one-dimensional case where one ball is initially at rest which you cite ("dead straight line"), there are two equations to solve:

  1. momentum conservation, m1v1=m1u1+m2u2, and

  2. energy conservation, ½m1v12m1u12m2u22,

where m1 is the mass of the incoming ball, m2 is the mass of the other (at rest) ball, v1 is the incoming speed, and u1 and u2 are the outgoing speeds. If you solve these, you get the speeds after the collision:

  1. u1=v1(m1-m2)/(m1+m2)

  2. u2=2m1v1/(m1+m2).

So, if m1=m2 (as for billiard balls), u1=0, u2=v1, as you note. And, certainly, as you also note, the first ball has transferred all its energy (and momentum) to the second ball. Since I have gone to the trouble of solving the problem generally, we might as well look at a couple of other special cases. Suppose that m1<<m2; this would be like colliding your cue ball with a granite wall. In that case u1-v1 and u2≈0, where ≈ means approximately equal; this is just what you would expect, the ball bounces back (that is what the - sign means) with the speed it went in and the wall stands still. Similarly suppose that m2<<m1; this would be like colliding your cue ball with a BB. In that case u1 v1 and u2≈2v1; so the cue ball never knew anything happened and the BB ends up going twice as fast as the cue ball.

Incidentally, for completeness, there is a second solution when you solve the two conservation equations, sometimes referred to as the "trivial" solution. You should expect two solutions because one of the equations is quadratic. The second solution is

  1. u1=v1

  2. u2=0.

This is called trivial because it would correspond to missing totally and you would certainly expect the energy and momentum to be conserved if the collision never happened.


QUESTION:
would it be possible to have a particle accelerator as an engine for a spaceship? Where by rerouting particles that are moving near the speed of light and making them collide with the back of the spacecraft would give the spacecraft a push in the back?

ANSWER:
You are essentially asking "if I, standing in my spaceship, push on the front wall, will it go faster?" Whatever impulse you gain by pushing forward with your hands will be cancelled by the frictional force your feet exert backwards. Similarly, the impulse you get from the push from the particles forward would be negated by the force necessary to "reroute" them or to accelerate them in the first place. Anything you do inside a closed system cannot change the momentum of that system. If you were to fire the particles out the back like a rocket you could propel the spaceship; this is the idea of ion-drive thrusters.


QUESTION:
could centrifugal force force actually be used to simulate gravity like in so many sci-fi stories? one of my favorite sci-fi stories is the Ringworld by Larry Niven. The Ringworld of the title is a giant ring shaped structure the size of earth's orbit. It's centered on a star and has a habitable inside edge, gravity on this inside edge is simulated by the structure spinning fast enough to make objects feel as heavy as they would at 99.2% earth gravity. If I were stood on a real structure like the Ringworld, and I jumped up in the air, would I fall back down or fly off into space?

ANSWER:
Yes, as long as the radius of the ring is large compared to the size of the objects. In such a scenario, the centripetal acceleration should be set equal to g, so g=Rω2 where ω is the angular velocity in radians per second. So, the picture to the right shows the ring as viewed from outside. You now jump straight up with a speed v. However, note that you also have a tangential velocity of so your actual velocity is √(v2+R2ω2). So you see what will happen is that you will go in a straight line with constant speed (because there are no forces on you) along that velocity until you again strike the ring; it will seem that you jumped and came back down. You can calculate the time you were in the air and how high you went by doing some pretty straightforward geometry/trigonometry.

  • I find the two angles labeled θ above are the same so sinθ=v/√(v2+R2ω2) and cosθ=/√(v2+R2ω2).

  • From these you can find the length of the chord (the length of your flight) C=2Rsinθ=2Rv/√(v2+R2ω2),

  • the height you go above the surface h=R(1-cosθ)=R(1-/√(v2+R2ω2)), and

  • the time you are in the air T=C/√(v2+R2ω2)=2Rv/(v2+R2ω2).

Now, let's compare these with the time and height on earth. Remember that ω=√(g/R) and I will take R to be very large compared to compared to v2/g. So, now I find that

  • h=R(1-/√(v2+R2ω2))=R(1-1/√(1+v2/(R2ω2))=R[1-(1+v2/(Rg))]≈R[1-(1-½v2/(Rg)+…]≈v2/(2g) and

  • T=2Rv/(v2+R2ω2)=2R/(v(1+R2ω2/v2))=2R/(v(1+Rg/v2))≈2v/g.

These two approximate results are just the same as for a projectile launched straight up on earth. Keep in mind, though, that R must be large. Also, you do not want to jump with a velocity which has a component parallel to the axis of rotation. If you are too close to the edge, you will miss the ring when you "fall back".


QUESTION:
Balancing a glass on knives how does it work? The video.

ANSWER:
It seems to me that they show you how to do it in the video.

FOLLOWUP QUESTION:
i'M sorry I mean why does this happen? How come the three knives hold the glass?

                   

ANSWER:
Examine each knife. Each has a force A up on its handle from a glass; each has a force B up at the end of its blade from one of the other three knives; each has a force C down from the third knife; each knife has a force of its own weight W down at the center of mass of the knife (where you could balance it without any forces but the weight acting on it). On that knife, the sum of the forces is zero, A+B-C-W=0. On that knife, the sum of the torques about any axis is zero; for example, about the glass end the where A is applied, B(d1+d2)-Cd1-Wd3=0. These conditions are satisfied and the knives are all in equilibrium. There is another interesting thing you can note: the force C on this knife is the force one of its neighbors exerts down on it with the end of its blade; but, by Newton's third law, the forces forces these two knives exert on each other are equal and opposite. It follows, therefore, that C=B since the knives are symmetrically arranged and all three must have the same forces on them. You can also see that the three glasses must hold up the total weight of the three knives, so 3A=3W, A=W. If you solve for these forces you will find A=W  and C=B=W(d2/d3). Putting the beer glass on top simply increases the forces B, C, and A but the knives remain in equilibrium. Is this what you were looking for?


QUESTION:
"A tube in the shape of a rectangle with rounded corners is placed in a vertical plane. You introduce two ball bearings at the upper right-hand corner. One travels by path AB and the other travels by path CD. Which will arrive first at the lower left-corner?" My problem is with the velocity of the ball at the end of traveling down the first side, whether it be A or C. The problem states 'rounded corners'. Surely the ball does not come to a stop -i.e. zero velocity, as it turns the corner. So I would guess that there is a significant velocity after 'rounding' the corner and hence must be accounted for.

ANSWER:
This is a problem out of Halliday and Resnick, Chapter 3 on kinematics. I have been assured that it not homework, rather someone brushing up on introductory physics. I must admit that I am a little puzzled by it because, although I can solve the problem analytically as a function of the sides A and C and the angle θ which C makes with the vertical, none of these is given; this means to me that there must be some obvious qualitative reasoning for this picture which does not

require any details. In this picture, the ball going down C will have a much bigger acceleration than the ball going down A because it is steeper; since C is shorter, that ball will arrive at its turn earlier than the other ball. Also, because the first corner is lower along C than along A, the C ball will be going faster when it gets to its corner than the A ball will be. So, the C ball turns the corner earlier and with a higher speed. The C ball wins. That's the best I can do with qualitative reasoning. (I would be glad to have anybody write in with a better explanation.) However, my curiosity led me to solve the problem analytically so I could examine all the possibilities. I will not give the anaytical solutions for both of the two paths since they are a little messy, but they depend on A, C, and θ; click here to see these solutions. Plotted below are the times as a function of θ for 00-900. The case most closely resembling the given problem is where A=2C with θ≈300; The C ball is the clear winner as speculated above. One interesting thing to note is that when θ=450, the balls take the same time regardless of the lengths of the sides. This is easy to understand because both balls have identical accelerations of 0.707g the whole time. And, the ball starting on the steeper slope always wins the race, regardless of the lengths or the angle θ. Note also that as θ―>00 the time for A ―>∞, and as θ―>900 the time for C ―>∞. This is also easy to understand because a ball in a horizontal tube will not move at all.


QUESTION: 
how fast will i fall down on the moon if i way 1000 pounds?

ANSWER: 
Your weight is not relevant. The acceleration due to gravity at the surface of the moon is about 1.6 m/s/s, so if I drop you, you will have a speed of 1.6 m/s after one second, 3.2 m/s after 2 seconds, 4.8 m/s after 3 seconds, etc.


QUESTION: 
I have two friends that skydive. They got into an argument the other day about how they felt some time after they jumped, once they reached terminal velocity. One said he felt weightless, while the other said he didn't feel weightless. My question is this, what should they feel at terminal velocity? Should they feel 100% of their body weight? Does the force from air resistance become a pseudo normal force? And, just for fun, if they got crazy and jumped out of a plane with a large piece of ply wood so that they were laying on the piece of wood which was falling to the ground, would they feel their weight once they reached their new terminal velocity?

ANSWER: 
As my wife would say, there is no arguing with feelings. What it means to be "weightless" in the usual vernacular is that there are either no forces at all acting on you or else your weight is the only force on you but you have an acceleration equal to the acceleration of gravity. If there were no air at all your friends would be weightless as they fell; an astronaut in orbit has a centripetal acceleration pointing toward the center of the earth so, technically, he is free falling. It is important to bear in mind, however, that this use of weightless is, strictly speaking, incorrect because your weight is the force that the earth exerts on you and this does not go away if you happen to be falling. Now, I would say that neither of them should feel weightless by that definition because there is the weight force down and the air resistance force up which are equal in magnitude at terminal velocity so it is analogous to laying on the floor where the upward force equal to your weight is the contact force with the floor. Do you feel weightless when laying on the floor? I think you would not say so.


QUESTION: 
Does a scale with a box that contains a flying bird register the weight of both the box and the bird or just the box? Does relative size of box and bird make a difference? Does it make a difference if the box has perforations or no lid.(i.e., not a closed system). And is "how closed the system is" a critical component to this problem.

ANSWER: 
If the box is closed, the weight of the bird (assuming that it has no vertical acceleration) will register. The reason is that the air exerts an upward force on the bird equal to the bird's weight. So the bird, because of Newton's third law, exerts an equal (to the bird's weight) downward force on the air. So the forces on the air are its weight (likely to be negligible), an upward force, and a downward force equal in magnitude to the bird's weight; so, since the air is in equilibrium, the box exerts an upward force equal to the weight of the air plus the weight of the bird. Therefore, again because of Newton's third law, the air exerts a force down on the box which includes the bird's weight. To a first approximation, all of the above still applies even if the box is open, but things could get complicated in detail. You might be interested in a similar question I answered long ago about sand falling in an hourglass.


QUESTION: 
What happens if I jump vertically on a moving bus? Will I move back or will I stay in the same spot? And Why

ANSWER: 
Physics is an experimental science and this would be an extremely easy experiment for you to try. Even more impressive would be to jump straight up in an airliner which has a speed of about 600 mi/hr which is almost 900 ft/s. So, if the plane were to move forward and you not, if you were in the air for 0.2 s you would land about 180 feet behind where you jumped, obvious nonsense! The reason you come back to where you jumped from is that you and the plane both have the same forward velocity when you jump and it does not just disappear when you are not in contact with the plane, you keep moving forward with that same speed, just like the plane. An observer on the ground would see you as moving in a parabolic path as you went by. Another example is to look at WWII movies of bombers dropping bombs: the bombs all are strung out (approximately) directly below the airplane. All this assumes that the airplane (or bus) is traveling in a straight line with constant speed.


QUESTION: 
Does an object extremely far away from any other object has inertia in space?

ANSWER: 
Inertia is the property inherent of objects which cause them to resist accelerating when pushed on by a force. It has nothing to do with the environment. The object has no weight if in empty space.


QUESTION: 
How do I determine the ideal weight of an object with a given size to achieve maximum distance when put in motion? Example: I know a baseball made of styrofoam or lead would not go very far when thrown. An official baseball weight is 5 oz. If it were 4 oz or 6 oz would it go as far of a greater distance when hit or thrown as the 5 oz baseball?

ANSWER: 
Take first the baseball and the styrofoam ball. The reason the baseball goes farther is that air friction has a larger effect on the smaller mass. In particular, the terminal velocity, that speed which an object will end up having after a long time, is proportional to the square root of the mass. Hence the greater the mass, the greater the range will be. The reason you assume a lead ball will not go as far as the baseball is because you are thinking that you cannot throw it as far; that is certainly true, but the reason is that you do not have the strength to give it the same initial velocity as the baseball. For a meaningful experiment each ball should be projected with the same initial velocity; if that is done, the lead ball will go farther than the baseball which will go farther than the styrofoam ball.


QUESTION: 
Is a juggler, while juggling three weights or any number really, lighter at all times than she would be if she merely carried the weight about her person? If so then by how much, when and why? If not then what does happen to their weight while they juggle at the various times they are and are not in contact with the juggled objects?

ANSWER: 
First, I am a stickler for the use of the word weight. The juggler's weight is the force by which the earth attracts him and so it is always the same unless he overeats or goes on a diet. But, the apparent weight (what would be read by a scale he is standing on) depends what is going on with the balls. If all the balls are in the air at some time, his apparent weight will be his actual weight. If he is simply holding one ball, the scale will read his weight plus the ball's weight. If he is in the process of juggling one of the balls, he is exerting an upward force which will be larger than the weight of the ball (Newton's second law); but, because of Newton's third law, we can conclude that the ball exerts an equal and opposite force on him; and so the force read by the scale will be larger than the weight of the ball plus juggler. An interesting and related problem is that of what the apparent weight of an hourglass is, discussed in an earlier answer.


QUESTION: 
consider a space rocket in motion has ke=1/2 mv2. when the rocket accelerates to 2x its velocity, we would get 4x ke. would it really take 4x the amount of fuel to increase the velocity 2x? where is conservation in this example?

ANSWER: 
First, talking about fuel is a little tricky because fuel has mass and you have a more complicated rocket if its mass is changing. So let's talk about a rocket whose total mass is much greater than the mass of the fuel or else imagine an external agent pushing. Now, what is conserved in an isolated system? The two important things are total energy and total linear momentum. There is no such thing as conservation of velocity. To change linear momentum which is mv, the mass times the velocity, you must exert a force F over a time t and the change in momentum is Ft. Ft is called the impulse. If you double the impulse, you double the change in velocity if mass does not change. So push twice as hard or twice as long and change the velocity twice as much. Now, if you want to change the kinetic energy of the rocket, you must also push on it. But now the conservation principle says that the change in the kinetic energy is proportional to the work done one it. Work is Fx where F is the force which acts for some distance x. So, if you double the amount of work you double the change in kinetic energy, so you double the change in v2 if mass is unchanged. If you start from rest and do an amount of work W and end up with speed v, then if you do 2W of work you will end up with only (√2)v velocity. Burning twice the fuel will give twice the work (energy) so the kinetic energy increases by a factor of two and the velocity increases by a factor of √2.


QUESTION: 
Does adding mass to a toy car affect the force of gravity acting on the car, the acceleration of the car, or both as it rolls down a ramp? I want to do a lab in my middle school classroom where students investigate Newton's 2nd Law of motion by rolling a toy car down a ramp and measure the distance it rolls. They will use washers to increase the mass of the toy car. How would increasing the mass affect the distance the object rolls down the ramp? I found this lab on the Internet (without an answer key) and need to be sure that I'm right about what the outcome should be.

ANSWER: 
Let's first talk about what the results should be according to elementary physics. The force of gravity (called the weight) is proportional to the mass. If you increase the mass by a factor of 2, the force impelling it down the plane increases by a factor of 2. But, what does a force do? It causes an acceleration in the direction of the force. But the acceleration (from Newton's second law) is inversely proportional to the mass of an object; this is called inertia, mass is the property of an object which measures how resistant it is to an acceleration if you exert a force on it. So, if you increase the mass by a certain factor you increase the force by that factor and the net result is that the acceleration is unchanged! This should be the lesson learned by doing this experiment. By the way, this is the same idea as the old story of Galileo dropping a marble and a cannonball from the leaning tower and finding them to hit the ground at the same time. However, in the real world, there are other forces besides gravity, mainly frictional forces, acting on the car and these can confuse the issue when trying to do an experiment like you describe. As you will see when you look over earlier answers to just this question, sometimes the heavier car wins and sometimes the lighter one wins; many details can make this experiment a mess! If I were you, I would carefully do the experiment beforehand to make sure that it will teach what you want it to!


QUESTION: 
Is conservation of momentum proved practically if so what is the experiment?

ANSWER: 
Actually, momentum conservation is more a definition than something you verify. According to Newton's second law, the force equals the time rate of change of something called momentum. If the momentum of a system is not changing, it has, by definition, zero net force acting on it. In classical mechanics, the momentum turns out to be mass times velocity. In relativity, it is more complicated but still conserved for an isolated (no external forces) system.


QUESTION: 
Today on NPR's "Cartalk," someone called in a physics question. I would like to have a definite answer (very easy for you I'm sure). Here it is: A lady's car is stuck in the mud. She of course is alone with no phone and is a physicist. She ties a rope to her car bumper and a nearby tree. She then finds the mid-point of the rope and pushes with max effort which she estimates to be 300 Newtons. The car just begins to budge with the rope at about a 5 degree angle. With what force is the rope pulling on the car? Ray, co-host of "Cartalk," said to find the sine of 5 degrees and then multiply by 300. Then he changed it to cosine of 5 degrees and multiply by 300. If any of these is right, I don't understand why. I've done vector problems before but they were simple Pythagorean probs (like, two ropes pull on an object at 90 degrees to each other--find the the single vector--easy).

ANSWER: 
One of my favorite shows! Neither of the answers is right which is surprising since Tom and Ray are both are MIT grads. Here is how you do the problem: See the drawing (sorry for the quality) on the left. The point where she is pulling is in equilibrium, so the vector sum of the three shown vectors (her 300 lb pull and the tensions in the two halves of the rope) must equal zero. The components perpendicular to her pull must add to zero, so the tension (T) in each side of the rope is the same. This comes from T1 cos 50-T2 cos 50=0, so T1=T2=T. Similarly, the components parallel to her pull must sum to zero, so 300-T sin 50-T sin 50=0. So, T=300/2sin 50=1721 lb.


QUESTION: 
If I secure 2 hemispheres in a vacuum (ie in space), then I bring it back to Earth at around sea level, what would be the strength of the hemishperes against being torn apart?

ANSWER: 
I presume you mean how much force must be applied to pull them apart. The geometry of the sphere makes a quantitative answer to your question difficult (too mathematical) but I can give you an idea how big the force would be. I will assume we have two "hemicubes", two halves of a hollow cube which we put together. Atmospheric pressure is about 105 N/m2, about 2100 lb/ft2. The forces pushing on the cube (or sphere) depend on its size; imagine a 1 ft cube. The force holding it together, due to the pressure on the two ends opposite the seam, would be 4200 pounds.


QUESTION: 
picture a roof made made metal. stick a magnet to that roof. The magnet has weight naturally. so by keeping itself stuck to the roof it needs to oppose the force of gravity acting on it. It does so by exerting another force of attraction in the opposite direction, this force needs to be stronger than that of its own weight if it is to remain stuck to the roof. This is where my question comes in: that force is work is it not. It is actively working to oppose its own weight. But its a permanent magnet and the amount of energy used to magnatise it compared to the amount it expends holding itself aloft is disproportional. I read that you do the work for the magnet when you pick it up to sick it to the roof and you'll need to do the same amount of work to pry it off again so in essence you have done the work for it and you have simply transferred it to a different potential state. But im sorry it does not make sense. There is a constant force opposing its attraction force to the roof. the force of gravity. This is in violation of newton, to apply a constant force you need a constant supply of energy. Where does the magnet derrive that energy from?

ANSWER: 
You have this entirely wrong. First of all, the force necessary to hold the magnet to the roof must be exactly equal to the weight of the magnet. Second, the force holding it to the roof does no work because it does not act over a distance; if you used the magnet to move a nail across the table it would do work on the nail. Finally, exerting a force does not require energy, only if the force does work.

FOLLOWUP QUESTION: 
I am satisfied to a point with the answer but it still doesnt make sense If you look at it like this: Take an ordinary object and attach it to a pully system. pull on the chord until the object meets the roof. now keep it there. since the object is not exerting its own force keeping it there you have to do that for it. in doing so over a period of time you will begin to tire as you are burning calories, expending enegry to keep that onbject aloft. the magnet stuck to the metal roof is only kept there by the force it is applying to the roof. you said that force is not greater than its weight but just equal and opposite to gravity. but what about an industrial magnet. you can easily support its weight on your hand, but put the magnet on your hand when resting on a metal plate it will crush your hand. I want to know how a magnet can apply a constant force and not "tire".

ANSWER: 
Your question now verges on biology rather than physics. I do recording for the blind and recently read a discussion regarding just what you are asking, viz. how can you say I am not doing work when I hold a box when I know energy is required to do so? The gist of the answer is that muscles exert a force by individual fibers of the muscle continually slipping and then recontracting, so for this special case the individual componenets of the total force are all contiually pulling over a distance and hence doing work. This is not the case for a mechanical system like your magnet; or say you simply tied the rope in your pully to something (not a muscle)—the rope exerts the necessary force and does not use any energy to do so. I do not see the relevance of your "crush your hand" remarks. While your hand is being crushed the magnet is moving so work is being done; when it is all crushed, no more work is done.


QUESTION: 
Hi, I'm reading Hal Clement's Mission of Gravity set on an oblate Jovian sized planet. The planet spins at 8 minutes per rev to create such squashed sphere. The main special thing about the planet is that the gravity at the equator is about 2 g and the gravity at the poles is a massive few hundred g. Now, the question is, is Hal Clement's physics right? Would there really be a gravity disparity on a planet like this? And if so, what creates it?

ANSWER: 
The planet does not even have to be oblate for such an effect to take place. The reason is the rotation. If you stand on a pole, you are not rotating and so your apparent weight, the force the ground exerts up on you is equal to your true weight (the force the planet exerts on you). If, however, you are at the equator you are rotating around a circle of radius R with speed V and so you have an acceleration equal to V2/R pointed toward the center of the circle. The force causing this acceleration must be however much of your weight is needed. Thus, from Newton's second law, MV2/R=W-WA where W is your true weight and WA is your apparent weight. Hence, WA=W-MV2/R, much less than your true weight if there is a large rotational speed. Eventually, as you decreased the length of a "day", you would become "weightless". Be certain to note that this effect is one of dynamics, not of a gravity disparity. Incidentally, the true weight at the poles actually decreases as the planet becomes more oblate. This can be deduced from imagining a "maximally oblate" planet, a disk; because of symmetry, you would experience zero gravitational force at the center of a disk.


QUESTION: 
My friend and I are having a debate about a certain topic trying to come up with different theories of how the mechanics of this process works, yet with each theory we develop we get pulled deeper into the problem with more possibilities and factors affecting the issue. The question is, if an average car, with ideal conditions, (ideal tire pressure, gear ratios, weight, driving conditions, weather) is traveling at a rate of 60 mph, are the tires moving faster than than the car itself. I argued that the rpms of the tire are moving at a rate faster than the car, (obviously) but since the wheel is attached to the car and the car is moving at a rate of 60 mph, the wheel too (not counting rpms) is moving at a rate of 60 mph where rubber meets road regardless the size of the tire. I do not understand his theory, but he claims, that with rpm's aside the tire in and of itself is moving faster than the car. I told him if you took a snap shot of the car moving at 60 mph, everything about the car is moving at 60 mph. To be able to figure out the exact speed of the tire as a seperate entity, you would HAVE to factor in its rpm. Please help me win this battle!

ANSWER: 
Sorry, but you absolutely lose this battle! If the brakes are locked, every point on the wheel will be moving forward with a velocity equal to the velocity of the car. If the wheel is rolling, the point where "where rubber meets road" is momentarily at rest. The very top of the wheel is moving forward with a speed of twice the speed of the car. Points on the front half of the tire are moving both down and forward (with speeds and directions easily calculated) and those on the back half are moving both up and forward. You cannot talk about the speed the tires are moving since every point on the tire moves with a different speed. The best way to visualize it is that every point on the tire is rotating about the "where rubber meets road" point at any instant; the axel moves forward with speed equal to the speed of the car.


QUESTION: 
Can you provide a formula for the Earth's acceleration due to garvity and the speed of the Earth's rotation? Would like to know how g = 9.81 m./sec sq. would change if the Earth had 16-hour days instead of 24.

ANSWER: 
First, I must disabuse you of the notion that the acceleration due to gravity would change. What happens is that your apparent weight changes because you are accelerating in a circle. Your true weight is unchanged since weight is simply the force with which the earth's gravity attracts you. (The answer to your question depends on the latitude where you do the experiment; for simplicity, I will do it for the equator where the effect is largest.) From Newton's second law you may write N-mg=-m(2πR/T)2/R where N is apparent weight, mg is true weight, R is the radius of the earth, T is the period of rotation. Solving this, I find N=mg(1-(2/T2)) if T is measured in hours. Therefore the effect on apparent weight is about 200/T2 %. This is about a 0.35% effect for T=24 hours, a 0.78% effect for T=16 hours. If the length of a day were less than about 1.4 hours, you would leave the surface.


QUESTION: 
Would you be able to verify the reasoning in the four simple paragraphs presented below and find a fault in them?
1. It is a generally accepted fact that in a head-on collision between two balls of equal mass moving with the same but opposite speeds, the two balls will rebound with equal but slower speeds. (R. Feynman, Lectures on Physics, Vol. 1, p. 10.7.)
2. If the above is true, then the following also must be true: When one of the balls in the above example is at rest, this ball must rebound with a slower speed after a collision with the moving ball of equal mass, while the moving ball comes to a perfectly “dead” stop.
3. Because the ball initially moving must come to a stop after the collision, the slower speed of the ball initially at rest must be used in the calculation of the total final momentum, making it smaller than the total initial momentum. Hence, the total momentum cannot be conserved in this type of collision.
4. Proof: The above outcome is confirmed by the fact that the total final energy cannot be conserved in the above collision. However, the only way that energy will not be conserved is if the speed of the ball initially at rest is slower than the speed of the ball initially moving, as the ball initially moving comes to a perfectly “dead” stop after the collision. In other words, the same slower speed of the ball initially at rest must be used in the calculation of both the total final energy and the total final momentum. Because both energy and momentum are the function of the same factors (the same masses and the same speeds), neither energy nor momentum can be conserved in the motions of the two balls, contradicting the law of conservation of momentum. Thus, we have mathematical proof that momentum cannot be conserved in all collisions.

ANSWER: 
Statement #1 is far from true in general. This is true only for an inelastic collision, one in which energy is lost in the collision. Statement #2, the incoming ball being at rest after the collision, is true only for a perfectly elastic collision (in which case the ball originally at rest exits with the same speed as that of the incoming ball). Hence, since #2 is incorrect, your "proof" fails. Linear momentum is always conserved in an isolated system, i.e. a system which experiences no outside forces.

FOLLOWUP QUESTION: 
Thank you for responding to my query. However, I am not happy with your answer. I thought the conditions in my query were clear. However, on the second thought, it is better to stipulate them. Here they are: Conditions:
1. The two collisions described in paragraphs 1 and 2 are real collisions that could be performed in a laboratory. In these real-world collisions, energy is not conserved. Therefore, in order to avoid confusion, elastic collisions should not be considered or even mentioned in this case.
2. A made a survey among a dozen of reputable physicists, professors of physics and a well-known physics textbook whiter about the real collision in paragraph 2. The general consensus was that a glider initially moving on a frictionless air track in a laboratory will come to a perfect “dead” stop after colliding with a glider at rest of equal mass, as confirmed, according to them, by numerous already performed experiments. Therefore, the assumption in paragraph 2 is that the ball initially moving would come to the above-mentioned perfect ‘dead” stop. What would then happen to the ball initially at rest? You stated in your answer, that the ball initially moving would come to a stop only in a perfectly elastic collision, which do no exist in nature. This means that in a real collision, you assume that the ball initially moving will not come to a perfect "dead" stop. In what direction and at what speed will it move after the collision and what would happen to the ball initially at rest? Your assumption contradicts the result of my survey. Indeed, what exactly would happen in a real head-on collision where energy is not conserved. Suppose the two balls are 2 kg each, and one moves initially at 2 m/s, while the other ball is at rest. What will be the speeds of the two balls after the collision? Once again, let's forget elastic collisions.

ANSWER: 
Your conditions were crystal clear to me. All I said about condition #1 was that this is only true for an inelastic collision. I agree that real world, macroscopic collisions will always result in energy loss. Your dozen reputable physicists are either not competent or else they are telling you what they think you want to hear—the stock answer for elastic collisions. It is only for elastic collisions between identical particles, one initially at rest, that the other is at rest after the collision. So how could those physicists be wrong? Quite simply because the apparatus they describe is designed to have a very low (not zero, since we agree that is not possible) energy loss when the gliders collide and to have minimum friction (not frictionless as you state). Hence the collision between gliders is so close to elastic that the velocity of the incoming glider is so small that it is not noticable or is stopped by the (nonzero) friction. I will present two situations to you to try to convince you that condition #2 is flat-out incorrect if energy is lost.

  • Consider a perfectly inelastic collision, e.g. two balls of putty. In the head on collision, all kinetic energy is lost and the balls end up at rest, stuck together. If only one is moving before the collision, is it at rest after the collision? I think you must agree that it cannot be.
  • My second example is a little more complicated. I will take your choice of 2 kg balls moving at a speed of 2 m/s toward each other. Then the energy before the collision is 2x(½x2x22)=8 J. Suppose that ¾ of the energy is lost in the collision. Then after the collision the energy is 2 J. Therefore, 2x(½x2xv2)=2 J so v=1 m/s; each ball leaves the collision with a speed of 1 m/s. To find out how this same collision looks if one ball is at rest, I will not assume momentum conservation because you would object since you think you have disproved it. I will watch the original collision but while running alongside one of the balls, ok? Before the collision I see one ball at rest and the other approaching with speed 4 m/s. After the collision I see the struck ball moving (in the same direction as the other ball came in) with a speed of 3 m/s and the incoming ball moving (in the same direction) with a speed of 1 m/s. (Note that the same amount of energy, 6J, is lost in both scenarios.)
  • If you redo the second example for 10% energy loss (0.8 J) you will find the speeds after the collision are about 3.9 m/s (for the struck one) and 0.1 m/s for the incoming one. For a 1% loss (0.08 J) they are about 3.99 m/s and 0.01 m/s

QUESTION: 
I understand that "centrifugal" force is ficticious. Centripetal force is, however, a real force - but doesn't Newton's third law state that for every action there is an equal and opposite reaction? And wouldn't that reaction, in this case, be a "centrifugal" force? And, if the answer is that Newton's laws do not apply in a non inertial frame of reference, how are we permitted to use Newton's second law, F = ma (= v^2/r in this case), to quantify centripetal force?

ANSWER: 
Anybody being introduced to Newton's laws often gets things confused. Let us state Newton's third law carefully. If body A exerts a force on body B, then body B exerts and equal and opposite force on body A. Consider the earth going around the sun. The centripetal force is the force which the sun exerts on the earth. This is the only force on the earth. Where is the so called "reaction force"? Just read the law carefully and you will see that the earth exerts a force on the sun which is equal and opposite to the force the sun exerts on the earth. The "reaction force" is never on the same body as the "action force". We are permitted to use F=ma because what this equation means is: the total force on a body is equal to the mass of that body times the acceleration of that body. The force the sun exerts on the earth is the only force on the earth.

Let's take a look at what it means for there to be a "ficticious force". Suppose that you are in a car which is accelerating. What are all the forces on you? There is your weight, the force which the seat exerts up on you (equal and opposite your weight but having nothing to do with Newton's third law because they are both on you), and the force which the seat back exerts forward on you. Nothing mysterious, the seat back accelerates you forward. But suppose you want to do physics using the car as your reference frame. Then you are not accelerating (relative to the car) but the sum of the forces on you is not equal to zero; Newton's first law is not true in this reference frame. But, you insist on doing Newtonian mechanics in this frame. To do this you say, "hey, there is a force pushing me back in my seat". It feels that way but there is no such force. This added force is called a ficticious force added to make Newton's laws to be true in this accelerating system. If you are in a system which is spinning, imagine some carnival ride, you feel like you are being smashed back against the outside but what is really happening is that the outside is pushing in on you to provide your acceleration and you interpret that as your being pushed back. This is just like the car example in that there is no force pushing you outward, it just feels that way. And if you want to do physics using the spinning frame as your reference and using Newton's laws, you have to add a centrifugal force even though there is no such thing.


QUESTION: 
how much does a lacrosse ball (2 inch diameter) slow down (horizontal velocity only) if thrown at 80 mph from the instant it is released until it reaches a point 10 meters away. Taking into account air resistance.

ANSWER: 
I prefer to work in metric units so 80 mph is about v0=35 m/s and the diameter is about D=6 cm=0.06 m. I will also need the mass of a lacrosse ball which I looked up to be about m=0.15 kg. Now, for a ball of this size traveling through air with this velocity, the air resistance force is proportional to the square of the velocity. Therefore Newton's second law is of the form -Cv2=ma=m(dv/dt) where C is a constant which can be calculated approximately as C=0.22D2 for a sphere in air. Therefore we must solve the differential equation (dv/dt)+0.00079v2=0. (I completely ignore gravity because the ball starts with zero velocity in the vertical direction and flies for only a very short time.) If you know differential equations, then this is not particularly difficult to solve. I will do that later. For starters, however, it is instructive to make a reasonable approximation and see what we get. I am going to say that I expect, over so short a distance as 10 m and starting with such a large initial velocity, that the acceleration will not change much. So I will say that the acceleration at the beginning, a0=-0.00079x352=-0.97 m/s2, does not change much over the flight. So we have a uniform acceleration problem and we can say x=v0t+½a0t2=10 and solve for t; I find that t=0.29 s. Finally, we can get the estimated final velocity, v=v0+a0t=35-0.97x0.29=34.7 m/s. So the ball loses about 0.9% of its initial velocity.

For anyone interested in the exact solution of the differential equation, here it is. The solution to the equation is v=v0/(1+kt) where k=Cv0/m. And, x=(v0/k)ln(1+kt). Solving these I find that t=0.29 s and v=33.2 m/s. So, only about 5% of the velocity is lost.


QUESTION: 
Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an elastic collision with the floor and then rebounds to make an elastic collision with the still-descending small ball. Just before the collision between the two balls, the large ball is moving upward with velocity and the small ball has velocity . (Do you see why?) Assume the large ball has a much greater mass than the small ball.

ANSWER: 
I will not give all the details but, to understand this problem, you need to understand one-dimensional elastic collisions between two masses. There is a derivation in a Wikepedia article. If a mass with speed v collides elastically with a very much larger mass with speed V in the opposite direction as the smaller mass, the smaller mass rebounds with a speed approximately equal to v+2V. In the case you note, since the larger ball rebounded from the floor elastically, both balls have about the same speed when they collide, i.e. v=V, so the smaller ball rebounds with speed 3v.


QUESTION: 
Based on Physics, is a 90 MPH Fastball Slower or Faster than a 95 MPH. At work we are trying to determine if the 95 MPH fastball loses energy faster than a 90 MPH fastball. Your answer is greatly appreciated.

ANSWER: 
You are asking two questions; if a 95 mph ball loses energy faster than a 90 mph fastball (it does) you cannot conclude that it "is faster" (by which you mean, I presume, when it passes over the plate). For the details of the following, see the earlier lacrosse ball answer. Following the (exact) solution in that earlier answer, I find that the 95 mph ball reaches the plate in 0.47 s and arrives at the plate with a speed of about 80.8 mph. The 90 mph ball reaches the plate in 0.50 s and arrives at the plate with a speed of about 76.3 mph. So, each loses about 14 mph with the faster ball losing a bit more. This surprised me but I found another reference saying that something like 10 mph is what is lost, so my calculations are reasonable. So they do not lose energy significantly differently (the faster pitch lost more speed in a shorter time so its average rate of change of speed was indeed bigger). (I used 3 inches for the diameter, 0.145 kg for the mass, and 60'6" for the distance to the plate.) There is certainly no way that one could characterize a 95 mph fastball as slower than a 90 mph fastball.


QUESTION: 
Consider centripetal force of a tetherball. This would be provided by the tetherball's string. My physics teacher tells us that since kinetic friction is orthogonal to centripetal force, it may be ignored in centripetal force's calculation. However, it is obvious that centripetal force depends on the ball's velocity: F=m(v squared)/r, and that velocity depends upon kinetic friction. So if the tetherball were to rub against the ground, wouldn't it require are greater centripetal force?

ANSWER: 
The tetherball is a particularly tricky example because of the thickness of the pole. (First, ignore friction.) At any instant the ball is rotating about the point where the rope is tangent to the pole and the ball is moving perpendicular to the rope. But you would calculate centripetal acceleration relative to the center of the pole. Therefore, the tension in the rope is not the centripetal force, only its component Tr along the radius R is (see first picture). Similarly, the centripetal acceleration is not V2/R but Vt2/R where Vt is the tangential component of the velocity (see second picture). Finally, if there were friction present, it would contribute to the centripetal force because the the frictional force will be opposite the direction of the velocity and will therefore have a component fr which is along R (see third picture). So, your teacher is wrong because the velocity V is not perpendicular to the centripetal force Tr-fr (compare all three figures). On the other hand, you also are wrong because the reason is not because the friction is changing the velocity of the ball (which, of course, it is); note that all my arguments are made at an instant, not over a time when the ball will slow down because of friction.

A couple of other things about the tetherball (without friction): Angular momentum is not conserved because there is an external torque (-TtR). Energy is conserved because there is no work being done by the tension (T and v are perpendicular).


QUESTION: 
how much time will it take to travel down a 1000 ft , 30 degree ilcline with a 225 lb payload on a 200 lb soap box race car and if pay load was lighter would the time be shorter with lighter load?

ANSWER:
I have dealt with this kind of question before, but maybe it is time to revisit the whole thing with one answer. First, consider the ideal situation where there is no friction of any kind. Physicists do not like English units, so I am going to convert everything to SI units: 1000 ft=305 m, 425 lb=193 kg. I assume that you do not want all the details of my calculations, just the pertinent results. The time does not depend at all on what the mass is (if friction plays no role). There is an acceleration down the incline which is a=
½g. I find that the time to the bottom is about 11.2 s and the speed at the bottom is about 54.9 m/s=123 mph. I assume you are not crazy enough to be in a soap box car going that speed, so friction must play a role. There are two kinds of friction you have to consider:

  • Friction due to the moving parts, like bearings, like the wheels rolling on the ground, like wheels moving on axels, etc. Empirically, we find that this kind of friction increases proportionally with how hard the moving parts are pressed together and this, of course, is proportional to the total weight of car plus rider. But, the force impelling the car down the hill is also proportional to the total weight, so once again, the speed at the bottom is independent of the load. There is something called the coefficient of friction which tells you how much friction force there is for a given force pressing the surfaces together. For example, if this coefficient is 0.3 for a 100 lb box moving on a horizontal floor, you would have to push with a 30 lb force to keep it moving with a constant speed. If I take 0.3 to be your coefficient of friction, your speed at the bottom would be about 26.8 m/s=60 mph. Since the acceleration is about half what it was, the time is about twice as long, 22 s.

  • For objects which are going faster than a few mph, air drag becomes important. Air friction depends on two things, the shape of the object and the speed it is going. It does not depend on the mass of the object. Actually a pretty good approximation to the magnitude of the force of air friction is F≈¼Av2 where A is the area presented to the wind (this works only if F, v, and A are in SI units). So, this force, which points up the incline just like the moving parts friction, causes the car to slow down more. But, Newton's second law says that a=F/m so that if F does not depend on m, the acceleration (amount of slowing down due to this force) gets smaller as m gets bigger. It gets a little complicated to actually calculate the time and speed for this case, but the important part is that this is the only place I can find where the weight of the whole car makes a significant difference.

So, the bottom line is that if you go fast enough for air drag to be important (and I suspect you do), the heavier of two otherwise identical cars should win.


QUESTION:
Do you weigh more at the bottom or top of a large skyscraper? Does the mass over your head count for more than the distance? As an example, the tallest skyscraper currently is 828m tall and wieghs about 500,000 tons if it's empty, in Abu Dhabi. Assuming that you weigh more at the bottom, how dense or large would the mass above your head have to be to make you weigh less?

ANSWER:
There are two effects you need to think about:

  • Because you are a different distance from the center of the earth when you are at the top, your weight is less. This turns out to be a 0.03% difference.

  • The gravitational force due to the mass of the skyscraper is really small. I just took all the mass to be at the center to check this. Then, if you are at bottom, the gravitational force from the skyscraper mass would about 0.000003% smaller than your weight at the top which would increase your (0.03% smaller) weight at the top and decrease your weight by that amount at the bottom. So, the effects of the skyscraper are negligible compared to other effects.


QUESTION:
My husband when riding his vintage motorcycle had an accident when negotiating a bend (200m radius) in the wet. it was found that the road surface only had friction coefficient of 0.3. What formula and other known parameters (like weight of bike, lean angle etc) can I use to determine what speed he could have safely negotiated the bend? He was actually only doing about 40-43mph.

ANSWER:
I do not know where you got your coefficient of static friction μs=0.3, but it can be only a rough estimate at best. I find that rubber on wet concrete can have μs over the range 0.45-0.75 and on wet asphalt over the range of 0.25-0.75. But, often there is oil which was in the road which floats up when it rains and makes it slipperier yet. I am assuming, since you didn't mention it, that the road is not banked. In that case, the weight of the bike+rider and lean angle are irrelevant. It is easy to calculate the minimum possible speed from Newton's second law, μsmg=mv2/R, so v=√(μsgR). Putting in your numbers I find v=24.3 m/s=54 mph. Evidently the coefficient of friction was less than you assumed. "Discretion is the better part of valor" (Shakespeare, Henry The Fourth, Part 1 Act 5, scene 4); in other words, be very careful on wet roads!

FOLLOWUP QUESTION:
The friction coefficient of 0.3 (equivalent to wet snow on tarmac) was obtained by pendulum testing in the wet to reflect the conditions at the time of the accident. You are correct in your assumption that the road was not banked. In fact the road comprises a slight adverse camber. Given the low friction coefficient I am having great difficulty in understanding how the calculation can show that a motorcycle (at normal lean angle) can negotiate the bend at 54mph. I would not wish to attempt to negotiate that bend on the equivalent of wet snow at 54mph even in a car!

ANSWER:
This is a very standard calculation. I figure that at 54 mph it would take 13 s to traverse a 90 degree turn for this radius circle which does not seem unreasonable to be possible. Are you pretty sure of the 200 m number? The key to this may well be the banking of the curve; if the road is banked wrong by 100, I would calculate v=15.1 m/s=34 mph, quite a significant difference.


QUESTION:
I am falling from a plane at terminal velocity. I have a ball in my hand. I throw it towards the ground. What happens? Does the ball accelerate away from me, or does it simply fall beside me? What about if it was a very heavy ball such as a cannon ball compared say to a baseball or a plastic 'WalMart' pool ball?

ANSWER:
Terminal velocity is determined by the geometry of the object, the density of the fluid (air), and the mass of the object. The terminal velocity in air for an object of mass m and cross sectional area A can be roughly approximated by vt≈√(4mg/A) (only in SI units). You and the ball have some terminal velocity together. When you depart from the ball you both have different terminal velocities since the masses and geometries have changed. Suppose it was a cannon ball. Then your net mass gets smaller and your geometry does not change much, so you will have a lower terminal velocity and slow down; the ball has a considerably smaller cross sectional area than it did when it was "part of" you and therefore a larger terminal velocity, so it will accelerate down until it reaches its new terminal velocity. Suppose it was a styrofoam ball. Then neither your net mass nor your geometry change much, so you will have about the same terminal velocity and continue falling at about the same rate; the ball has a considerably smaller mass than it did and therefore a smaller terminal velocity, so it will experience an upward acceleration and slow down until it reaches its new terminal velocity, you will overtake it and it will appear to go up (but is actually just going down more slowly).


QUESTION:
My question has to do with traction and the movement of a wheel(a wheel alone). Traction is essential for its movement both linear and circular. But if we throw a wheel forward it rolls some meters and then it stops(and falls). Which force is responsible for the decrease in its velocity? Cause if traction is parallel to the ground facing backwards then linear movement 's negative accelleration is explained but not angular negative accelleration. If traction is parallel to the ground facing forward then angular negative accelleration is explained but not linear. If traction is zero then which force decreases both velocities linear and angular?

ANSWER:
One of the reasons I love doing Ask the Physicist is because I often learn things I did not know or had never thought about. You would think that a guy who has been teaching introductory physics courses for nearly 50 years would find this question simple. But, indeed I was puzzled by it because, as I have found by thinking about it and talking to some friends, I wasn't thinking beyond the friction force (which questioner calls traction) being simply the only force in the horizontal direction and obviously stopping the forward motion after some distance. I never addressed the angular acceleration of the wheel before. This answer will be long-winded because that is what I do when I have learned something which pleases me! What frictional forces are important to understand the rolling of a wheel? Most introductory physics classes talk only about the contact forces of static friction and kinetic friction. Kinetic friction is not applicable to this problem because the wheel is not slipping on the ground, and static friction might be important, but not necessarily. If we have a round wheel rolling on a flat horizontal surface (don't look at the figure yet!), there are three possible forces—the weight which must be vertical, pass through the center of mass, and (assuming it is a uniform wheel) pass through the point of contact; the friction, which must be parallel to the surface and pass through the point of contact; and the normal force which must be perpendicular to the surface and pass through the point of contact. If you now sum torques about the point of contact (as noted by the questioner), there are none! So, there can be no angular acceleration; if we have stipulated that the wheel does not slip, then there can be no linear acceleration either and the wheel will roll forever and no friction is required. But we all know better! A real wheel will eventually slow down. The key is that there is no such thing as a perfectly round wheel or a perfectly flat surface, one or both must be deformed. In that case, we have to think about a new kind of friction called rolling friction, the friction the wheel has because of the rolling. This is different from the static friction, and static friction may still be present still to keep the wheel from slipping. A perfectly round wheel cannot have rolling friction as I showed above, it must deform which means that there is no longer a "point (or line) of contact" but now an area of contact. Since the normal force is only constrained to act somewhere where the two are in contact, it is now possible (in fact inevitable) that this force will not act through the center of mass of the wheel. That is the whole key to answering this question. So, finally, the answer: refer to the figure where I have drawn the forces mg, N, and f. The weight is still constrained to be vertically down and pass through the center of mass (blue cross). The normal force is constrained to be vertical and act somewhere where the wheel and ground are in contact, drawn a distance d to the left. The frictional force (which now includes both static and rolling friction) is constrained to act at the surface and parallel to it. I choose a coordinate system with x to the left and y up; the axis (red cross) about which I will sum torques is at the ground directly under the center of mass and positive torque results in an angular acceleration which is positive when acceleration of the center of mass is positive (counterclockwise around the axis). All is now straightforward: ΣFx=-f=ma, ΣFy=N-mg=0, Στx=-Nd=Iα=Ia/L where I is the moment of inertia about x and L is the distance from x to x. Finally, N=mg, a=-f/m, and d=fI/(Lm2g).

Finally, a couple of real-world provisos. Of course, N is really distributed over the whole area, but the dynamics can be done by assuming it effectively acts all at one point just like we assume the weight acts all at the center of mass. And, the rolling friction might not really act at the surface of contact since it arises from the deformation of the wheel and it might not be purely horizontal since it is not directly a force due to the contact with the floor. So, there are still some idealizations in my analysis, but there are always idealizations when dealing with friction. And, the problem could have been equally well done assuming the ground, not the wheel was being deformed. One could also have done the analysis by summing torques about the center of mass and using the parallel axis theorem Icm=I-mL2.

I would like to acknowledge a very useful discussion over pizza with friends and colleagues Edwards, Love, Meltzer, and Anderson.


QUESTION:
If I could detonate a firework inside a contained vacuum void of any other mass objects (simulating the big bag) and watched (over time) would I see the particles eventually attract to each other? Would the results of this sort of experiment prove gravity exists in every particle ever created?

ANSWER:
No. The gravitational attraction is very, very small between the little pieces of your firecracker. So the speeds the pieces had when it blew up would have exceeded the escape velocities from their neighbors and they would continue moving apart forever. Escape velocity is the speed something must have to escape from the gravity of something else; e.g. the escape velocity from the surface of the earth is about 7 miles per second, but this is for a much stronger gravitational force. Just to give you another example where the mass involved is still much larger than your masses but much smaller than the mass of the earth, the escape velocity from the surface of a baseball is about 8 cm/hr.


QUESTION:
I am doing an experiment with dropping coffee filters. By changing the mass (adding more filters) and keeping the distance the same (4 meters) and recording the time it takes for the stack to hit the ground. By doing this I find the velocity. Wouldn't there be an asymptote at 9.8, for the velocity of the dropped object?

ANSWER:
Technically, you are not measuring the velocity, you are measuring the average velocity. But, since coffee filters have such small mass, you can probably assume that terminal velocity is reached almost immediately after dropping. What you are studying is the fall of objects for which air resistance is important. I do not know what you mean by the asymptote at 9.8, but it is certainly wrong because 9.8=g is the acceleration due to gravity, it is not even a velocity. What you want is to find out how velocity depends on mass. This will depend how the terminal velocity depends on mass. Now, the force of air drag will depend on two things, the geometrical size of the falling object (but all will be the same, so you cannot study that) and the speed the thing is moving. Usually this force can be parameterized by F=Cv2 where C is some constant determined by the fluid (air) and the geometry, but not the mass. So, the faster it falls the bigger this force is and when it gets as big as the weight of the object it stops accelerating and falls with a constant speed called the terminal velocity which you hope to be determining. So, drag=weight when mg=Cv2 where m is the mass and g=9.8 m/s2. Solving, v=√(mg/C); so, if you make a graph of v vs.m you should get a straight line if indeed the force is a quadratic function of the speed. And, from the slope, you could determine the constant C. If you do not get a straight line from your data, try using F=Cv; then if you plot v vs. m you will find a straight line.


QUESTION:
We're having this physics discussion at work (in the medical field). One of my co-workers insists that if you took a pound of anything and spread it out evenly on a scale, it would weigh less than if you piled it up in the middle of the scale. He said that the density of the mass of whatever you piled on the scale, would increase the gravitational pull moreso than if the object being weighed was spread out evenly, like say mashed potatoes, for example. What do you say?

ANSWER:
The stock answer to this question is that weight is the force that the earth exerts on a mass and that force is independent of how the mass is distributed. But, that is not quite right and your question is clearly a "hair-splitting" sort of question, so I will explain. The gravitational force between two objects depends on how far apart they are—the farther apart, the smaller the force. The weight of something is determined how far it "is" from the center of the earth. If you compare the mashed potatoes spread out with the mashed potatoes heaped up, those heaped up are, on average, farther away from the center of the earth by a few centimeters, and therefore weigh less, not more. But, don't get too excited about this—the difference would be about one millionth of 1%, less than you could ever hope to measure. You would get a much bigger difference if you weighed the same thing upstairs and downstairs in your home, again something you know from experience is, for all intents and purposes, the same. [Rereading the question, let me address the density argument of your colleague. It is true (again to a really small degree) that the weight of the potatoes on the top of the heap will press down on those on the bottom and make the potatoes slightly more dense. However, a scale does not measure density, it measures mass and therefore two objects with different densities but equal masses would weigh the same.]


QUESTION:
Why is it difficult to calculate the terminal velocity for a cat falling from a high roof top?

ANSWER:
I do not know what you mean "difficult to calculate". We can estimate it pretty easily, but certainly not do it precisely. First of all, any calculation having to do with air friction is going to have approximations and assumptions. For something like a cat, roughly 2 kg (4.4 lb), falling, it is a very good approximation to say that the drag force is proportional to the square of the velocity. It turns out that a fairly good approximation for the force is F=¼Av2 where A is the area the falling object presents to the onrushing wind and v is the velocity (this is only for SI units). Since it depends on A, it depends on how the cat orients itself: if in a ball he will fall much faster than if all spread out. Suppose we take the area of a falling cat to be about 20 cm x 40 cm=0.08 m2. Then the force will be about 0.02v2. Now, the cat's weight is about mg=2x9.8≈20 N. When the force of air friction is equal to the weight force down, the cat will fall with a constant velocity called the terminal velocity: 0.02vt2=20, so vt=√(20/0.02)≈30 m/s=67 mph. If you google "terminal velocity of a cat" you will find the number 60 mph, so my approximations were evidently reasonable. There, now, that wasn't so difficult, was it?

RELATED QUESTION:
I was asked what the terminal velocity of an unladen sparrow is. I read that the average weight of a field sparrow is .5 ounces. Approximately 5 inches in size with a wingspan of 7.9inches. Lets say falling from 50ft?

ANSWER:
Look at the answer above. If the sparrow falls straight down with wings stretched out, I would estimate his area to be ½x5x7.9=20 in2=0.013 m2. Following the same as for the cat, but using m=0.5 oz=0.014 kg, I find vt≈6.6 m/s=15 mph=22 ft/s.


QUESTION:
I want to ask about weight shift in vehicle. Weight shift is a very common terminology among car enthusiast (but most of them not physicist or engineer, some might not even study science well). Magazine always comment an accessory will help in weight shift of the car, etc. The common qualitative statements made by them in the use of weight shift are: (i) When a car turn, the weight shifts to outside wheels more; (ii) When a car accelerates, the weight shifts to the rear; (iii) when a car brakes, the weight shifts to the front. The above is just some statements to clarify what I meant by the term "weight shift".

Here is my question: Is "weight shift" a misnomer, or an ill concept of physics (vehicle dynamics to be specify)? The weight of an object is determined by the gravitational pull, regardless of what motion the object is doing, so how can the weight of different sections of the car change during maneuver? I think if one was to say force acting on different sections of the car during maneuver is different, then it is correct; but to say the weight shifted during maneuver, I can't get it.

ANSWER:
Technically, weight shift is a misnomer, but it is not too bad because it is a shift in apparent weight. Normally, we think of weight as acting at the center of gravity of something, but if that something is accelerating, it may appear to move elsewhere. I will try to explain. In all three of your examples, the qualitative observations are essentially the same. For the accelerating car, the "apparent weight" shifts toward the rear wheels, opposite the direction of the acceleration. For the braking car, it is just the same as the accelerating car except that the direction of the acceleration vector is toward the back, the "apparent weight" shifts toward the front wheels, again opposite the direction of the acceleration. For the car turning there is an acceleration also, this time toward the center of the circle (called the centripetal acceleration) and here the "apparent weight" shifts toward the outside wheels, again opposite the direction of the acceleration. So, if I understand one of these I will understand them all. In every case, the force which causes the acceleration is the static friction between the tires and the road. The figure to the right shows a generic case. The center of gravity is at the point marked x. The forces labeled are the weight W, the force of the road up on the left tire N1, the force of the road up on the right tire N2, and the static friction force of the road on the tire f. If the car is at rest or moving with constant speed, f=0, W=N1+N2, and N1=N2; this assumes the center of gravity is halfway between the wheels, but if the center of gravity were moved to the left, we would have found that N1>N2. The three scenarios you mention can all be represented by this figure. If the car is accelerating, it is moving to the right and speeding up and the left wheel is the rear wheel; here f=ma where m is the mass of the car and a is its acceleration. If the car is braking, it is moving to the left and slowing down and the left wheel is the front wheel; again, it is the friction which causes the car to accelerate (slow down). If the car is rounding a curve, the center of the curve is to the right, and the left wheel is the outside wheel. We still have W=N1+N2, but now the two forces N1 and N2 are not equal. Since the car is not rotating about its center of gravity, if we sum the torques about x it must sum to zero:(N1L/2)-(N2L/2)-fh=0, so N1=N2+2fh/L. So, N1>N2, it appears as if the weight (center of gravity) has shifted toward the left. So, weight shift is a pretty good name for it, I would say; apparent weight shift would have sounded a bit pedantic for car buffs, don't you think? (By the way, I drew the friction only on the left wheel but if there were friction on the right wheel instead or in addition, the conclusions would all be the same because the torques from the right wheel friction would have been the same sign as the left.)


QUESTION:
Theoretically, if you were traveling through space in an object e.g., a spaceship and going at an incredible speed in one direction what would happen to you, inside, if it made an immediate 90 degree change in direction? wouldyou smash against the inside of the craft?

ANSWER:
First of all, you cannot make "an immediate 90 degree change in direction". It would require an infinite force to stop your forward motion simultaneous with an infinite force to start your sideways motion. So, you have to say you make the turn through a curved path. As an example, I will choose "an incredible speed" to be 104 m/s≈22,000 mph, just a little faster than the shuttle goes. Now, at that speed let's make a 900 turn around a curve of radius 1 km=103 m. Then the spaceship and all its contents experience a centripetal acceleration of ac=v2/R=(104)2/103=105 m/s2. This is ten thousand times the acceleration due to gravity which means that it would take a force of 10,000 times your weight to move you in this circle. I would hate to see you after this maneuver, certainly you would not be recognizable.


QUESTION:
My musical instrument has been broken during an air travel, but its container, a hard case with thick foam padding inside and outside is absolutely intact! how is this possible ?

ANSWER:
Suppose that it is dropped from a height of 3 m, about 10 ft. It would hit the floor with a speed of about 8 m/s. Suppose the thickness of the foam is 5 cm, about 2 in. Then the instrument would stop in a time of about 0.013 s and would experience an acceleration of about 640 m/s2 which is about 64g where g is the acceleration due to gravity. What this means is that, to stop your sitar in this short distance would require an force of about 64 times bigger than its own weight. Even dropping from 1 m would result in an acceleration of about 20g. If the padding were well designed so that this force was distributed over the whole surface of the instrument, it might survive, but probably one part will take more than its share of the impact. So, the padding in your case should be thought of as protecting the instrument from normal bumping and jarring, not major drops.


QUESTION:
In strongman contests, the contestants pick up a telephone pole at one end, push it upright and flip it over. The other end of the telephone pole stays planted on the ground. Typically the telephone poles weigh 300 lbs and are 30 feet long.

  1. How would you calculate the actual weight of the pole being flipped at different points during the lift? For example, in the course of one flip we could say the pole travels in a 180 degree arc. The person is physically lifting for the first 90 degrees or less when gravity takes over. When the person picks it up off the ground it might be at 5 degrees, once it is over their head it could be at 30 degrees, once they take 2 steps forward it may be at 45 degrees, etc. - how do you calculate the actual weight of the pole along that arc at different points?

  2. As the person walks forward to flip the pole over their arms remain over their head (i.e. in this example once they're at or past the 45 degree mark in the arc), but it seems the weight would lessen along a curve as the pole gets higher and gravity begins to assist? That is - the person (fulcrum?) might be moving in - say - 3 ft. increments as they walk forward to gain leverage to flip the pole over. How is the weight being lifted adjusted to account for that variable?

ANSWER:
First, some terminiology: the weight of the pole is 300 lb, that is its actual weight. What you want, I believe, is the force you need to exert to hold it at any given angle. This is a problem involving torques—you say that, in order to hold it at some angle, the torque you exert is equal to the torque the weight exerts, a bit more to move it. (Torques are calculated here about the point where the pole touches the ground.) The torque due to the weight is easy, simply 300x15xcosθ=4500cosθ ft-lb where θ is the angle the pole makes with the ground. The torque you exert depends on where you push and the direction you push, an infinity of possibilities. I have calculated for the two extreme possibilities: you push vertically up or you push horizontally. For the vertical pushes I assume that your hands are 7' above the ground; for the horizontal pushes I assume that your hands are 5' above the ground. The graph above shows vertical and horizontal forces you must exert vs. angle. Clearly, you want to start your lift by exerting a vertical, not horizontal, force. When you first lift it, you need to exert an upward force of 150 lb and when you get to 13.50 your hands are at 7' and you can start moving forward. When you get to a little more than 500, it starts to become more advantageous to push horizontally, becoming much more advantageous as you near the end. If you go real slowly, you will have to exert a maximum force of about 500 lb, but when the force you need to exert is small at the beginning you should push harder, thereby giving it some added speed which will help it over the middle angles where the force is maybe bigger than you can exert. Also, you will, through practice, learn to shift smoothly from vertical to horizontal .


QUESTION:
A few days ago my friend & I were debating about how we could find the force with which a bullet of mass m and UNIFORM velocity v would hit a wall x m away. I said that the total energy would at the moment of impact would be equal to the kinetic energy of the bullet. So the total work done would be equal to the kinetic energy right at the moment of impact. So we could easily determine the kinetic energy by dividing the product of the mass and velocity by two. Then the result would be equal to the work done by the bullet. We could find the force by the formula W=fx. Please tell me if I am right.

ANSWER:
You are not right. You are right that the kinetic energy of the bullet is lost. However, x which you define as the distance from the wall to the gun, has nothing to do with it. Work is the force times the distance over which it acts and the force is not acting all the way across the room. What you might say is that if the bullet penetrates a distance x into the wall then the average force may be written as F=(½mv2)/x. But, that assumes that all the work done by the force results in stopping the bullet, but some, much maybe, of that kinetic energy is converted into heat and sound energy. (Incidentally, uniform velocity has nothing to do with it, all that matters is how fast it is going when it hits the wall.)


QUESTION:
I have struggled to get the simple explanation (non-mathematical form) for my troubling question from books. Still no luck. Whts is force ? (as a concept) What is the relationship between energy and force? I am not a physicist so could you please enlighten me and explain the concept in layman's term for me? Or with some analogy?

ANSWER:
A force is a push or a pull. It is one of those things you need to have an intuitive feel for to begin doing physics. It is that which causes an object to accelerate. In physics accelerate can mean two different things, either change speed (speed up or slow down) or change direction. Energy is a more subtle concept. To understand what energy is you need to understand the concept of work. In physics, work is done by a force only if the force is exerted over some distance. For example, if you push with a force of 1000 pounds against an unmoveable wall, you do no work, but if you push a box across the floor with a force of 10 pounds, you do work. When you do work on an object you increase its energy. So, when you push the box across the floor, it acquires what we call kinetic energy, energy by virtue of motion, because it started at rest and ended up moving. If you use a force to lift an object so that it starts out and ends up at rest, you have still done work but now the energy you have given it is called potential energy, energy by virtue of position. Different kinds of energy can transform to others. For example, if you drop something its potential energy is converted into kinetic energy as it falls.


QUESTION:
Is a golf ball going fastest after it leaves the club head striking it, or is there still some accelertion as it overcomes inertia?

ANSWER:
Nothing "overcomes inertia" on its own. Only a force causes an acceleration and acceleration is what is meant by overcoming inertia. The ball has its largest speed as it leaves the club because that force ends then. When the ball is in flight, only two forces act on it—air drag which always slows it down and gravity which slows it down on the way up and speeds it up on the way down. But gravity can never speed the ball up more on the way down than it slows it down on the way up.

FOLLOWUP QUESTION:
Sorry I didn't do a better job asking the question. My family and I are not looking to be told when peak acceleration happens, we want to prove it mathematically. We have put together a derivation to show that at its peak velocity a golf ball is still in contact with the club head. Starting out with the length of the drive we use the formula for projectile motion to find the peak velocity of the ball. We work backward from there to derive all of our unknowns but we run into trouble when we use the formula for uniform motion with constant acceleration to calculate how far the ball and the club travel during the time to accelerate. Because the distance the club travels is expressed as S = u * t and the distance the ball travels is express as s = u * t + a * t^2 / 2, there is always a few millimeters distance between the ball and club when the ball reaches peak velocity.

ANSWER:
There is a difference between peak velocity and peak acceleration. For example, a pendulum has its maximum velocity at the bottom when the acceleration is zero. What you are doing is wrong because the acceleration is not uniform as you assume. Let us, in accordance with my original answer, focus only on the time during which the club is in contact with the ball; the time after it leaves the club will always have a smaller velocity than when it left the club. The graph to the right shows what the force which the club exerts on the ball is likely to look like. The club first touches the ball at t1 and the ball leaves the club at t2. The force over the time of acceleration is not constant because the ball behaves like a spring. When the force is biggest, the acceleration is biggest because of Newton's second law, F=ma. However, the ball is speeding up over the whole time interval and will therefore be going fastest at t2. The area under the force curve is called the impulse and is equal to the change in momentum of the ball which, because the ball starts at rest, is mv where m is the mass and v is the speed at t2 of the ball. If you call Favg the average force on the ball, then mv=Favg(t2-t1). Putting in some numbers I found on the web, m=45 g, Favg=3000 lb, v=250 ft/s, I estimated that the time of contact is less than half a millisecond.


QUESTION:
Ok, so I got into a big discussion today with my physics Honors teacher and he didn't even say who was right or wrong. The problem was that there is what you could consider a clothesline with a cable (for purpose of the problem the weight of the cable does not matter) and a weight is suspended exactly in the middle of the cable. The weight of the object was 25 N. The angle at which the cable meets the object/block is 30 degrees. What is the tension in the cable? My theory behind this was that when you find the force tension it is for only half of the whole cable so you need to double the force tension and my teacher argued that you do not. So with this problem he said that the answer was 25 N of force tension and I said that it is 50 N of force tension. Who is right??

ANSWER:
I am sure that you will not be too surprised to find that your teacher is right. Maybe this would be easiest for you if you do not think of the tension in the cable but rather the tensions in the left and right parts of the cable. If the angles of the two parts of the cable were different, the tensions would be different. So, right away, the notion of doubling the tension to account for the two halves is seen to be faulty reasoning. If you call the two tensions TL and TR, you can see that the horizontal components of the two tensions must be the same: TLcos300-TRcos300=0, so TL=TR=T. Similarly, the vertical components of the two tensions must hold up the weight: TLsin300+TRsin300-25=0=T/2+T/2-25=T-25, so T=25 N. This means that the tension in each side of the cable is 25 N and you could say, in this case, the tension in the cable is 25 N everywhere.


QUESTION:
a body is moving with a velocity 'v' with respect to (w.r.t.) a frame of reference s1.It bumps into a very heavy object and comes to rest instantaneously,its kinetic energy(1/2*m*v^2) as seen from the frame s1 is completely converted to thermal energy.Now a man moving with a uniform velocity 'V' (in the direction of the body) w.r.t. s1 observes the body , he notes that its initial kinetic energy of the body is 1/2*m*(v+V)^2 and that after it rams into the heavy body as 1/2*m*V^2 and concludes that the thermal energy produced is m*v*V+1/2*m*V^2.Which of the two answers is correct?

ANSWER:
So, to answer your question we just need to find out how much the kinetic energy changed as seen by both observers, right? Where you have gone wrong is that you have not taken the energy of the wall into account and you assume the wall has infinite mass. Of course, no wall has infinite mass but if yours did, it would have, for the moving observer, infinite kinetic energy both before and after the collision but those two infinities would not be the same! What you need to do is assume the wall has a mass M and the body has a mass m. So, your first observer sees a speed after the collision (conserving momentum) of u=mv/(M+m) and a change in kinetic energy of ΔE=½(M+m)u2mv2mMv2/(M+m) which is the energy converted into thermal energy. In your second scenario, u=V+mv/(M+m) and ΔE=½(M+m)u2-(½mv2MV2)=½mMv2/(M+m), exactly the same. You can also note that as M-∞, ΔE-½mv2.


QUESTION:
gallile said that the force of gravity is the same on everything but the force of rubbing with the air molecule that change the speed of falling,so shouldnt 2 balls made from the same materiel but one is bigger than the other falling from the same heigth,the smaller touch the ground first cz her volum is smaller than the force of rubbing should be smaller than the big ball,but that is not the fact i dnt understand?

ANSWER:
First of all, Galileo never said the force of gravity is the same on everything because the notions of force and gravity were not even conceived before he was dead! Anyhow, that is not even true; the force of gravity is greater for greater masses. What is the same is the acceleration due to gravity, which Galileo supposedly demonstrated by showing that two objects of different masses hit the ground simultaneously. This is only approximately true because there is a force due to the air resistance as the balls fall. Galileo did not take air resistance into account and balls of different masses and sizes will not hit simultaneously. If you look at my FAQ page you will find lots of discussion of problems involving air drag and you will find that the force is proportional to velocity times the area presented to the oncoming wind (approximately). If the two balls are made of the same material, their masses are in the ratio (R1/R2)3 and their cross sectional areas are in the ratio (v1R1/v2R2)2. As each ball falls, it initially is accelerating down but this acceleration gets smaller as the speed increases because of the increasing air drag. Each ball eventually falls fast enough that the air drag is equal to the weight and falls with a constant speed called the terminal velocity. So, the weight (mg) and the drag (¼πR2v2) are equal. Taking the ratio of the equations for the two balls, (R1/R2)3=(v1R1/v2R2)2 so v1=v2(R1/R2). Suppose that #1 is the smaller ball and R2=4R1; then v1=½v2. Since the smaller ball has the smaller terminal velocity, the larger (more massive) ball will reach the ground first. So, in fact, your intuition was wrong.


QUESTION:
how long does it take a 1 kg mass with an area of 1 square meter to get to terminal velocity when dropped. actually any size will work but i need the mass and area.

ANSWER:
Technically, it never reaches the terminal velocity but approaches it. Also, you can only approximately do the calculation using an approximate expression for the air drag force which I often use, F≈¼Av2 where A is the area presented to the onrushing air and v is the speed (and this is only true in SI units). I cannot tell from your question whether you want a general solution for any mass (m) and area or just for this specific problem. So, when I solve the problem for the velocity as a function of time I get the solution v=√(g/c)tanh(t√(gc)) where g is the acceleration due to gravity (9.8 m/s2), cA/m, and tanh is the hyperbolic tangent function. For your specific problem, v=6.26tanh(1.57t), so the terminal velocity is 6.26 m/s because tanh(∞)=1. The graph of this is shown to the right; as you can see, the terminal velocity is reached, for all intents and purposes, by about 1.5-2 seconds. Sorry if this is more technical than you wanted; oh, and the tanh is defined as tanh(x)=[(ex-e-x)/(ex+e-x)].


QUESTION:
if a planet were suddenly stopped in its circular orbit ,show that it would fail into the sun in time (T into sq.rt. of 2 divided by 8) where T is time period of revolution.

ANSWER:
Kepler's third law states that the square of the period of an orbit T is proportional to the cube of the length of the semimajor axis A, T2/A3=constant. For a circle, the semimajor axis is the radius. For a straight line, the semimajor axis is half the length. So, T2/A3=T'2/(A/2)3 or T'=T/
√8.


QUESTION:
two massive particles m1 and m2 are released from a large distance.what are the speeds of particles when distance between them is 'r'?

ANSWER:
The total energy of the system is zero and the total linear momentum is zero. When they are a distance r apart, energy conservation and momentum conservation give the equations ½m1v12m2v22 -(Gm1m2/r)=0 and m1v1-m2v2=0. If you solve these you will find that v1=m2√[G/(r(m1+m2))] and v2=m1√[G/(r(m1+m2))]


QUESTION:
ok! Suppose we have to masses m attached to a line of length 2r rotating around their cm. Lets further suppose that there is a mechanism of some kind that when activated lets the line extend in length for example 2 times longer. The masses of the line and the mechanism are supposed to be very small compared to the masses m and the masses m are point masses so their individual rotations can be ignored. Initially the masses have speed v, so: total angular momemtum J = 2mvr total kinetic energy K = mv^2 after the line is allowed to become 2 times longer then, because of the conservation of angular angular momentum, the velocities of the rotating masses will be v/2 (J=2mvr= 2m*v/2*2r). But now the kinetic energy is 1/4*mv^2 So where did the energy go?

ANSWER:
Suppose we model your mechanism as a thread which connects the two masses which are beads which can slide frictionlessly on a long rotating rod. The two are each a distance r from the rotation axis, as you suggest, and have a speed v. Now, cut the thread. Each bead starts moving out the rod and both angular momentum and energy are conserved because there are no external torques or forces doing work. When each gets to a distance 2r from the rotation axis, it has a tangential component of its velocity, as you suggest, of vt=v/2 as required by angular momentum conservation. But, each also has a radial velocity; you can easily find the radial velocity vr from energy conservation: E1=mv2=E2=2x½m(vt2+vr2)=m(¼v2+vr2) so vr=(v/2)√3. Finally, to be sure we reproduce the problem you are interested in, we must stop those beads from sliding out, maybe just put a couple of little hooks to "catch" them when they get to 2r. The little hooks had to do work to stop the beads and that is where the energy went.


QUESTION:        
Why is it that if you blow a spider suspended by her web she floats out but then when this pendulum swings back it stops when the web is vertical and doesn't swing back and forth? Is it due to the air friction as it comes back to equilibrium or perhaps the dynamic structure of the web strand that absorbs energy that would have made the web swing back and forth?

ANSWER:
It is caused by air drag. This is called a damped oscillator. If there were no air, the spider would swing back and forth with constant amplitude, just like a clock pendulum (apart from the little friction from bending the thread she hangs from. A spider has so little mass that her terminal velocity is very small—drop her off the roof and she will not get hurt because she quickly comes to some constant velocity because the air drag, which can be approximated as being proportional to her speed, quickly becomes equal to her weight. If the air drag is not too big, the pendulum will swing back and forth with ever decreasing amplitude; this is called underdamped. For larger drag, as in the case of your spider, she never crosses over the equilibrium and just slowly approaches the bottom of her swing; this is called overdamping. There is a third possibility called critically damped, but it is qualitatively just like overdamping, so let's not go there. There is a neat web site (pun intended) where you can play with the damping for a mass on a spring, not a pendulum, but the idea is the same. (The constants there are k is a measure of the stiffness of the spring, m is the mass, and b is a measure of how big the drag is.) Two examples from that applet are shown below. The first might be a pea on a thread where it swings back and forth a few times before stopping. The second case is your spider whose swing just dies away.

                     


QUESTION:        
I am a lineman and a intriguing question has been brought to my attention and would love some help. How high would a wooden pole weighing approximately 1500 pounds need to be dropped in order to crush a steel toed boot with a crush rating of 2500 pounds?

ANSWER:
Usually I trash such questions (see my FAQ page). However in this case you tell me the force which the toe can withstand (which I will call  F=2500 lb) and the weight of the pole (which I will call W=1500 lb). We have to make one approximation, that the force is constant over the time it takes to crush it and you will also need to know how much the toe crushes (I will call that S). I will not bore you with the details, but I find that, if H is the height from which it is dropped, H/S=(F/W)-1. For your numbers, H/S=0.67; that is pretty small. For example, if the toe crushes by S=3", dropping the pole from H=2" would be enough to crush it. I would avoid dropping such a thing on my toes.


QUESTION:
Recently, I was involved in a 4-car pileup on the freeway. The roads were dry, and I was traveling at 40-45 mph in a 2003 Honda Accord (gross weight: 4300 lbs) when the car in front of me, at about 1 car length away, started braking. I braked and stopped 3-5 feet behind the car in front of me before the guy behind me rear-ended me. The CHP officer claims that I caused the accident by making an unsafe lane change. The guy behind me was driving a 2005 Toyota Sequoia (gross weight: 6500 lbs.) at 30-35 mph and he was roughly 1-1.5 car lengths behind me when I got in front of him. I argued that if I could come to a complete & safe stop going up to 10 mph faster than him, then there's no reason why he couldn't stop in time. The CHP officer says that since his car has more mass, he needed more room to stop. I think the difference in speed should account for the difference in mass. So, who's right?

ANSWER:
I cannot judge on this. If you made an unsafe lane change, I think you will not gain much by arguing. Certainly 1-1.5 car lengths is too small for the speeds you were going. To first approximation, though, if you are getting maximum braking (just barely not skidding) or if you are skidding, the distance to stop does not depend on mass, just the initial velocity (and reaction time).

FOLLOWUP QUESTION:
Would you mind explaining why distance to stop doesn't depend on mass, just velocity, if you are getting maximum braking?

ANSWER:
The force of friction if you are just about to skid on a level road is F=μsmg where m is mass, g is acceleration due to gravity (9.8 m/s2 or 32 ft/s2), and μs is a constant (called coefficient of static friction) determined by the surfaces (let's say rubber on dry asphalt). But, Newton's second law says F=ma where a is the acceleration. Therefore, ma=μsmg and so m cancels out and the acceleration (which determines how far you go) is independent of m: a=μsg. It is easy to show that the distance you go before stopping is sv02/(μsg) where v0 is the speed when you start braking. I find that, taking μs0.6, s≈110 ft for 45 mph and 70 ft for 35 mph. These are the absolute minimum (which you can approximate if you have antilock brakes) and assuming zero reaction time (which is impossible). Also, if the vehicle locks the wheels and skids, you go farther yet because the physics is all the same except the coefficient of friction is smaller if the surfaces slide.


QUESTION:        
ma+kx=0 →(A) is an equation which represents mass spring system. It is a linear equation of second order. If I want to make it non-linear I make it like ma+kx+k1(x^3)=0, i.e. I add another term k1(x^3). Why can’t I add k1(x^2) to (A) to make it non-linear?

ANSWER:
Because the force would be repulsive when x<0, that is, it would not be a restoring force. Your equation for the spring is essentially F=-kx. For F to be a restoring force, it must be an odd function of x.


QUESTION:
Suppose we have Bob the astronaut sitting without a space suit in a spaceship full of air on a mission to Mars. Bob is very fond of balloons and is holding on to a nice, big, red helium balloon via a piece of string. Bob is sitting facing the front of the spaceship. Mission Control decides to slowly accelerate the spaceship. In which direction will the balloon move relative to Bob? Why?

ANSWER:
I do not like this question because the reason a helium balloon floats is that the buoyant force, which floats it, arises because the pressure in the air is bigger underneath the balloon than above it; on a mission in empty space, the pressure everywhere in the cabin is the same and so the balloon would not go up! Let's just have Bob ride in an accelerating car right here on earth. There are three forces on the balloon, its weight W, the buoyant force B, and the tension in the string T. B and T are both vertical, and so, for the balloon to have an acceleration in the direction of the acceleration a of the car, T must have a horizontal component in the direction of a. All this is shown in the figure to the right. Therefore the balloon will move backward opposite the direction of the acceleration. In the spaceship where there would be no buoyant force and no weight, the balloon would appear to accelerate backwards until the string was straight and "horizontal".

"ENHANCED" ANSWER:
It has been pointed out to me (again, thanks to Michael Weissman at Ask the Van) that my answer would be correct only if there were no air in the spacecraft. What I failed to think about was that if there is air in the cabin, forward acceleration will cause the pressure at the rear of the cabin to be greater than at the front; therefore there would be a buoyant force on the balloon from back to front. So, if the mass of the balloon is less than the mass of an equal volume of air, the direction the balloon would move would be forward, not rearward. If you are familiar with the equivalence principle, it is even easier to understand. The equivalence principle states that there is no experiment you can perform which can distinguish between an accelerated frame and being in an unaccelerated frame in a uniform gravitational field with the same accelerateion (due to gravity). So, if the acceleration of the ship were g, the balloon would have to behave just the same as it would on earth except forward now would play the role of up on earth, so the balloon would go forward just like the balloon on earth would go up. (My example of the car on earth, above, would also have a "pseudo buoyant force" in the direction of a, so both this force and the horizontal component of the tension would be in the direction of a meaning that the balloon would always go the opposite direction of a here on earth.)


QUESTION:
I'm a big fan of SpaceX and space in general. So as you may know the Dragon Spacecraft is performing maneuvers to dock with the station. At this moment I'm watching the Dragon orbiting about 2 miles directly below the ISS. My question is: how can the spacecraft hold its position below the station when it should need to be going faster given that it's closer to the earth. Or is the difference so small that the effects of orbital decay are not even noticeable until many orbits later?

ANSWER:
Kepler's third law says that the ratio of the periods of two orbits is equal to the square root of the ratio of the cubes of their radii: T1/T2=(R1/R2)3/2. Suppose that R1=R2+δ, then T1/T2=(1+(δ/R2))3/2≈1+3δ/(2R2) where I have used the binomial expansion to approximate the root because certainly δ/R2 is very small. Now, put in some numbers: the radius of the orbit is about 4000 miles and δ is 2 miles, so T1/T2=1+6/8000=1.00075. This means that if T2 is 90 minutes, T1 is 90 minutes + 4 seconds. So, your last statement wraps it up pretty well.


QUESTION:
Would you give an example of how centripetal force (if I am correct in thinking and not confusing centrifugal force) helps keep the lacrosse ball from falling out of the net when cradling (semi circular motion) the ball. She is in 7th grade and used the example of the water bucket spinning Ina circle to explain to her friends why it works, but I would like a more thorough explanation.

ANSWER:
I will try to make this comprehensible to a bright 7th grader. Newton's second law says that what a force does is cause an object to accelerate. In everyday life, accelerate means speed up or slow down, but in physics it can also mean that the velocity changes direction without speeding up or slowing down. So, when something moves in a circular path, even with constant speed, there has to be some force present. Here is one example. Consider the International Space Station which moves in a circle about 200 miles above the earth with an approximately constant speed of about 18,000 mph. So it is accelerating because it is moving in a circle, always changing its direction. What is the force which causes that acceleration? It is simply the only force it experiences, its own weight which is a force which points toward the center of the earth. A force which causes a change in direction is called a centripetal force and must have a magnitude of mv2/r where m is the mass, v is the velocity, and r is the radius of the circle. Now, a more down-to-earth example. If you have a ball of mass m attached to a string of length r and you wish to swing it over your head, when it passes over the top of the circle it has two forces acting on it, its weight and the tension in the string, both downward forces. If it is going very fast, the tension in the string will be very big, maybe big enough that the string would break because the string was not strong enough. As it goes slower and slower, the required centripetal force gets smaller and smaller so the tension also must get smaller because the weight cannot change. Eventually you will be able to go over the top with the tension in the string zero, that is the string is not needed at that instant. But, that is the smallest speed you can go because if you try to go slower, the weight will be bigger that the centripetal force needed and so the string will go slack and the ball will fall out of its circular path. If you had used a stick instead of a string attached to the ball, you could go through as slowly as you wished because, unlike a string, a stick can push. There is no such thing as a centrifugal force, it is called a fictitious force and is added to make Newton's laws work in an accelerating reference frame. E.g., if you are sitting in a car speeding up, for example, you would swear that there is a force pushing you back but there isn't; you feel the back of the seat pushing you forward to accelerate you and your brain interprets this as being pushed back against the seat. See cartoon at the right.


QUESTION:
I am a professional juggler. One of the patterns I juggle has balls thrown to two different heights. The right hand throws across to the left and vice versa. Basically I know that the balls thrown from the right hand reach their peak in the same amount of time as the balls thrown from the left take to peak and fall down again (to be caught by the right at the same level as they were thrown). When I learned this juggling pattern I was still in school and I remember using Newton's laws to work out that the ball thrown from the right hand peaks at exactly 4 times the height of the ball thrown from the left. (If you google this, a ball dropped from 4m will fall to the ground in 9ms and a ball dropped from 1m will fall to the ground in 4.5ms) Please tell me how I worked out the relative heights using Newton's formulas and my knowledge of the relative time the balls take in the air! (If you need to see an example, you can google it, it's a standard juggling pattern known as "triple singles" because of the triple spins thrown if you juggle it with clubs.)

ANSWER:
This requires only that you know the equations of motion for uniform acceleration because the ball has a constant acceleration of g=-9.8 m/s2 when in flight (the minus sign meaning it is speeding up if falling, slowing down if rising. These equations are y=v0t-½gt2 and v=v0-gt where v is the velocity at time t and y is the height (at time t) above where it was thrown (at time t=0) with velocity v0. So, we have one pair of equations for the ball that goes higher (I will call it #1) and another pair for the other ball (#2):
   y1=v01t-½gt2 and v1=v01-gt and
   y2=v02t-½gt2 and v2=v02-gt.
Suppose #1 goes to a height h1 before coming back. The time t to reach h1 is found by knowing that v1=0 at that time, so t=v01/g. Put that time into the equation for y1 and find h1v012/g. Now, you tell me that you want #2 to make the full trip up and down in the same time; that means that it will make the trip up in half that time, tv01/g. If you now put this time into the equation for y2 (which we would call h2 at this time), you will find that h2=⅛v012/g=¼h1. Of practical interest to the juggler is that you must give #1 twice the initial velocity of #2 to achieve the desired heights.


QUESTION:
consider a water wheel with mass of 600kg, rotating @ 18 RPM, wheel diameter is 3 meters and the diameter of its axle is 0.1 meter. the wheel is fed with 10 liters of water per second from the top, causing it to rotate and eventually releasing the water at the bottom (3 meters lower). Neglecting efficiency losses due to friction, air drag etc..., the available energy from the water should be MGH: 10 x 9.8 x 3 = 294 joules. However, considering the wheel's diameter, its mass and its RPM, the wheel's kinetic energy (1/2 x moment of inertia x angular velocity squared) is 1200 joules. Where did the extra joules of energy come from? Secondly, the diameter of the axle is only 0.1 meter so this hypothetical wheel has a mechanical advantage of 30 (MA= radius wheel / radius axle). Does this mean the initial energy of water is magnified 30 times at the axle? If not so, then what has been magnified 30 times?

ANSWER:
Your analysis is taking the wrong approach on several counts. First, what you calculate as "available energy from the water" is available power from the water, J/s=watts. How much of that energy ends up being delivered to the wheel depends on the design of the wheel. To the right we see water entering the wheel at some speed; if it leaves the wheel at the same speed (which would happen if the circumference of the wheel were moving at the same speed as the incoming water for the wheel I use to illustrate), then your wheel is getting all that energy. If that were the case, then this wheel would be losing energy at the same rate to frictional forces (plus any work you were extracting from the wheel, maybe to grind grain or something). You do not say where you got the moment of inertia to calculate the kinetic energy of the wheel, but I figure that the biggest it could be, if all the mass were at the circumference (I=MR2), would be about 5400 kg m2 which, using 18 rpm=1.9 s-1, would give an energy of about 9700 J, quite a bit bigger than your 1200 J. If you were able to deliver 294 J/s to the wheel, it would take about 4 seconds to get to 1200 J if there were no friction. The bottom line is that a discrepancy between power being delivered to something and the energy it has is not a discrepancy at all since they are different things. Regarding your second question, it is force which is magnified 30 times; this is simply a lever.

FOLLOWUP QUESTION:
The formula I used for calculating the moment of inertia was 1/2 x mass x radius^2 = 1/2 x 600 x 1.5^2 = 675 kg-m^2. Is it possible to harness the available kinetic energy to do work? i.e can this energy be used to do more work (output) than what the original input by the water allowed (10 liters/second)? secondly, if force was magnified 30 times at the axle, is this magnification useful in terms of driving machinery such as water pumps, generators etc...? i.e can the original power of the water be used to drive a more powerful machine?

ANSWER:
First, note that I misread your original question and used 3 m as the radius, not the diameter. So, note that the water leaves the wheel halfway to the water below, so the available power is only 147 watts. The moment of inertia you used was for a uniform cylinder or disk and I would guess that the real moment of inertia is somewhere between our two answers (mine corrected to 1350 kg m2). You can never get more energy from a system than has been put into it. If you find a way to start using this kinetic energy, the wheel will slow down. You can think of the wheel as an energy storage device, having acquired that energy from the water. So, you could use that energy but it would quickly be used up. This is the principle of a flywheel on a mechanical device where you get it spinning when you have energy to store and then later use the energy.


QUESTION:
I've recently been having a heated debate over a scenario composed of very basic principles of physics, and I was wondering if you could help us figure this out. Both of us have presented different arguments, and neither of us can agree with the other. We have found very few relevant webpages, however, after scanning through some of your Q&As, we found one that I will mention. So here it is: If a man is on top of a moving object moving at a constant velocity x (in a vacuum, where air resistance is ignored) and he jumps straight up, will he land in the same position he jumped from, assuming that he doesn't flail his legs? I personally believe that he will, because first off, when released from the friction of the train, the man will still have the momentum of velocity x, and without air resistance will continue to go that velocity until gravity brings him back down. However, my brother believes that gravity will deter the motion of the man in some way that he will land off the mark. The other example that we found is the bullet on the train example. If a bullet is fired at a speed equivalent to that of the train's from the back of the train then relative to the ground , the bullet is at a standstill and will fall straight down. I think that this example supports my answer, but we are not sure. Could you please provide an answer?

ANSWER:
You have really not given me enough information. Is this happening on the earth? Or is it some idealized "flat" earth where the train is traveling in a straight line and gravity is always pointing in the same direction (perpendicular to the path of the train)? This (flat earth) is the scenario usually portrayed in textbooks because it is usually an excellent approximation.

  • In that idealized situation, you are right, you land where you went up; gravity acts only in the direction straight down and therefore there is no horizontal acceleration and you land where you jumped.

  • Now, suppose the earth is round but not rotating on its axis. When the man jumps, he has a velocity with two components, one away from the center of the earth and one equal to the train's velocity. The component parallel to the train will not change during the flight because there is no force in that direction. But, as he gets farther away, he covers less angular distance than the train so that when he finally gets back down the train has traveled farther and he lands behind it.

  • Finally, the real situation where the earth is rotating. Now where the man lands depends also on the direction in which the train is moving. If the train is in the northern hemisphere and is moving north, the man will not only land behind but also east of the train; in the southern hemisphere he will land west of the train. This is called the Coriolis force and results from the fact that the coordinate system is rotating.

Overall, you win the debate with your brother because, although only under ideal conditions does the man land where he launched, gravity has nothing to do with it. Your bullet on the train idea is an entirely different thing since the bullet which just drops was not fired vertically.


QUESTION:
I was watching the TV show, The Universe, the episode about alien planets. It talked about how planets and their stars revolve around their common center of mass and/or gravity, I forget which one. How does that work when you have more than one planet? Because, as we know, all of our planets do not revolved with the sun in a straight line. So where is the octuple fulcrum that creates the revolutions of our sun and the planets? And how did it occur? Please give me the most detailed answer you can, as well as fact checks with my question.

ANSWER:
For simplicity, suppose the solar system is isolated, the forces it experiences from the rest of the galaxy or the rest of the universe are negligible; this is an excellent approximation over times long compared to times associated with the motion of the planets. You can compute the center of mass of the whole solar system just about as easily as for two bodies. Since you say you want "the most detailed answer" I can give, I will do the details. For anybody reading the answer, you can just skip the math and read the narrative and understand things qualitatively. Suppose we choose some point in space as the origin of our coordinate system; each body in the solar system has a vector position Ri in this system and a mass Mi. Then the position of the center of mass is defined to be Rcom=ΣMiRi/ΣMi where the symbol Σ means to sum over all the objects. For example, the term ΣMi is the mass of the whole solar system which I will call M. Notice, for example, that if the mass of the sun is much greater than all the other masses, the position of the center of mass is approximately the position of the sun. Now I am going to find the velocity and acceleration of the center of mass by differentiating with respect to time once and twice respectively: Mvcom=M(dRcom/dt)=ΣMi(dRi/dt)=ΣMivi and Macom=M(dvcom/dt)=ΣMi(dvi/dt)=ΣMiai where vi is the velocity of the ith object and ai is its acceleration. Now, we can apply Newton's second law to the acceleration equation: Macom=ΣMiai=ΣFi where Fi is the net force felt by the ith object which is simply Σj≠iFij where Fij is the force on the ith particle due to the presence of the jth particle and this sum runs over j but not the term j=i because an object does not exert a force on itself. So, Macom=ΣΣj≠iFij. But now look at the double sum: for every term Fij there is also a term in the sum Fji and Fij=-Fji because of Newton's third law (the force of the ith on the jth is equal and opposite of the force of the jth on the ith). So, finally we have acom=0, the center of mass moves with constant velocity and if we happen to have chosen a coordinate system at rest with the center of mass, the center of mass never moves. In other words, the whole solar system, no matter how complicated its motion, orbits around the center of mass. If the total solar system experiences a force F from the rest of the universe, the only difference is that the center of mass now has an acceleration acom=F/M; but you can still say that the solar system orbits around its center of mass, it is just that the center of mass is now "orbiting" around some other point. Maybe that is more detail than you really wanted, but it is a standard derivation in any introductory physics course. The bottom line: any system of interacting objects orbits around its collective center of mass.


QUESTION:
My friend and I had a drunken argument. I would like independent council to weigh in (there's $300 on the line) I was given a unique bottle opener by a friend who is a brewer for craft brewery in the northeast. It is a flat piece of wood with a smooth screw embedded in one end ( think ______T_ ) Its measurements in 1/16ths of inches are as follows: 88 long. screw centered 11 from one end. side of screw to end is 9 short side. 75 long side screw lip is 4 from bar bottle cap is 4 high 17 across. The argument is as follows.
Person A: There is less force required to open the bottle pressing down with the cap positioned _________XXXT___ (between the screw and the user)
Person B: There is less force required to open the bottle pulling up with the cap postioned __________TXXX .
With my horrible description of the problem. Can you prove either arguement successfully?

ANSWER:
I have redrawn your attempt to picture this to make it clearer. The picture to the right shows the two situations with person A below and person B above. To answer the question I will compute the force which the nail exerts on the bottle top for equal forces by the user. Whichever of these is the biggest is the winner. Doing this is a simple first-semester physics statics problems, most easily done by summing the torques in each case about the point on the bottle cap just opposite the nail; that point is a distance R from the end where F is applied for person B and a distance R-2d for person A. I find that the nail exerts a force of A=F[(R/d)-2] for A and a force of B=F(R/d) for B; B is the winner of the bet. For your numbers, d=17/16" and R=(77+17)/16"=94/16", so R/d=5.53 and the ratio of the forces is B/A=5.53/3.53=1.57, making option B 57% bigger, quite definitive. (If a $300 bet is really on the line, don't forget to reward The Physicist!)


QUESTION:
If I am towing a vehicle from a standing start is there an equation for calculating the amount of force I would be using for example if I tow a vehicle that weighs twelve tonnes because it has wheels and is therefore not a "dead weight" how do I work out how much force I would be exerting on the tow rope/towing vehicle and also how would I factor in different gradients as it would obviously require greater force on an upslope. This came up in my workplace where our towrope are rated to three tonnes and I was trying to explain that it does not mean you could not tow a vehicle over that weight.

ANSWER:
This is a good question to illuminate elementary Newtonian physics. Your referring to "dead weight" really has no meaning in physics, but you apparently mean that the object can move with little friction. So, let's assume there is no friction; this is, of course, never true, but it puts an upper limit to anything I do. On level ground, any force will move the vehicle if there is no friction. What matters is how quickly you start it moving, in other words what the acceleration is. For example, suppose you have a 4 lb fish hanging on a 5 lb test line; if you pull it up slowly you will land it, but if you try to jerk it up really fast the line will break. The physical principle in play here is Newton's second law, F=ma where F is the force, m is the mass, and a is the acceleration of m due to F. So, in your case, m=12 t=12,000 kg; the maximum force you can apply is 3 t=29,420 N because the 3 t rating means that it can hold up a 3 t mass which has a weight of 3000x9.8 N. So the maximum acceleration is amax=29,420/12,000=2.45 m/s2=5.5 mph/s. This means that if you speed up to 5.5 mph in one second, the rope will almost break. Of course, there will be friction and so to be safe I would recommend a factor of roughly two, an acceleration of about 3 mph/s would probably be safe. Here is an equation you can use (which does not include any safety factor): amax=22(MT/MV) mph/s where MV is the mass of the vehicle, and MT is the mass rating of the towrope. If you are trying to tow up a hill which makes an angle θ with the horizontal, you need to apply a factor of sinθ to the equation above,  amax=22(MT/MV)sinθ, because some of the vehicle's weight is now directed down the hill instead of straight down. If the grade is 300, for example, sinθ=½.

SIMILAR QUESTION:
watching a tv commercial showing how mighty a pickup truck is-it's towing the space shuttle, which weighs (according to the announcer), 292,000 lb (146 tons). Now I know that it's not as if the pickup is lifting 146 tons-I figure the load on the little pin hooking the shuttle to the pickup will be (initially) 146 tons times the coefficient of friction for the tarmac upon which both vehicles are riding-am I right?

ANSWER:
First, read the question and answer preceding yours. Assuming there is negligible friction in the bearings of the carriage for the shuttle, it is not hard to get the shuttle moving with a small acceleration. In the other question, though, the numbers were much smaller than in your case where there is what appears to be a steel towing bar which would far exceed the strength of a tow rope to tow things with weights of several tons rather than several hundred. So, with such a strong "tow rope" you might think that you could have as big an acceleration as you like. For example, if the breaking strength of the pin (probably the weakest link) were 100 tons, my little formula above would say that you could have an acceleration up to about 22x(100/146)≈15 mph/s (0 to 60 in 4 seconds)! This will obviously not happen. There are two considerations you need to think about. First, the force which provides the acceleration is actually the static friction between the truck wheels and the road; the biggest this force can be is f=μW where μ is the coefficient of static friction and W is the weight of the truck. For rubber on dry concrete, μ≈0.7 and the weight of a Toyota Tundra pickup is about 3 tons, so f=3x0.7≈2 tons; so, the maximum acceleration is only about 0.3 mph/s. The second consideration is how rapidly the truck can deliver the energy needed to move the load, in other words its power rating of about 300 hp. I calculate that the maximum acceleration with a 146 ton load would be about 4 mph/s. So, it appears that the main limiting factor on the acceleration is the possibility of the tires spinning. Keep in mind that these are all just rough estimates, but they give the general picture.


QUESTION:
Why when a cyclist is turning round a bend, he tends to lean inwards with the bike?

ANSWER:
The picture at the right shows the forces (real and fictitious) on the cyclist. (I copied this from the Wikepedia article on bicycle dynamics.) The circle represents the center of mass of the system. Since he is moving in a circle of radius r and with speed v, he experiences a centripetal acceleration ac=v2/r to the left. The forces on him are his own weight mg, the normal force N up from the road, and the frictional force Ff which is the force providing the acceleration. If you want to apply Newton's second law in the frame of reference of the cyclist, which is not an inertial frame, you must add the fictitious centrifugal force -mac as shown in the figure. Note that if he were not leaning, there would be an unbalanced torque about the point where the tire touches the ground, τ=mLv2/r where L is the distance to the center of mass, which would cause him to rotate clockwise, that is to fall over. When he leans, though, the weight also exerts a torque, so the two torques can balance if the angle is just right: mgLsinθ=mLv2cosθ/r, or θ=tan-1(v2/rg).

FOLLOWUP QUESTION:
Can it be explained without invoking the fictitious force because it may be too technical? Can I say the bike has a tendency to move in a tangential direction to the bend, this causes the bike to topple outward, so leaning inward will prevent the bike from toppling?

ANSWER:
You can say "the bike has a tendency to move in a tangential direction to the bend, this causes the bike to topple outward" but that does not explain why it "topples outward", does it? You certainly can do this problem using Newtonian mechanics without introducing the centrifugal force, but you must calculate any torques about the center of mass, nowhere else will work. The new figure looks just the same but with the centrifugal force removed. Now, the sum of all the vertically directed forces must add to zero, -mg+N=0 which tells you that N=mg. And, the sum of all the horizontally directed forces must equal mass times acceleration, Ff=mv2/r. So, given m, v, and r, you now know all the forces. But you still need to know the angle of lean for the cyclist to not topple over. This is achieved by summing all torques (about the center of mass) and setting it equal to zero (so that it does not start to rotate in the plane of the page), Στ=0=NLsinθ-FfLcosθ=mgLsinθ-(mv2L/r)cosθ. And so, solving for θ, we get the same answer as above, θ=tan-1(v2/rg).


QUESTION:
Is it the orbital velocity of an object that enters the earth's atmosphere that causes that object to typically burn up and disentigrate? Could you, in theory, freefall from the International Space Station safely if you had zero orbital velocity?

ANSWER:
It is certainly high speed through the air which causes an object to get hot. The effect of high speeds on the body and the effectiveness of space suits in protecting it at high speeds will be tested in a jump from about 23 miles any day now. Felix Baumgartner is expected to exceed sound speed before his parachute opens. However, the altitude of the space station is about 250 miles and I suspect that falling from that height through almost no atmosphere would give the jumper too much speed when he encountered the atmosphere to be able to survive.

RELATED QUESTION:
I have a question about Felix Baumgartner and his skydive attempt. How is it possible for him to break the speed of sound in his jump? My thinking is that objects reach a terminal velocity - jumping from a higher distance from earth would therefore not make any more difference in the maximum velocity. Also, I heard that he won't be able to breathe for a certain period of time. Is that because of how fast he's going, or because of the atmospheric conditions at such a distance from earth?

ANSWER:
The reason there is a terminal velocity is that the air drag on a moving object increases approximately like the speed squared (if you go twice as fast, you experience four times the drag). But what the terminal speed is depends on how much air there is and at high altitudes, there is almost no air. There is some terminal speed, but it is much bigger than it is at lower altitudes. The terminal velocity is inversely proportional to the density, 1/ρ; the density of air at 23 miles is less than 1/100 that at sea level, so the terminal velocity is ten times bigger. The speed of sound, though, also changes with altitude; it is about 740 mph at sea level and about 670 mph at 20 miles up. I do not really know what your question about "won't be able to breathe" means. If he did not have his "space suit" he would certainly not be able to breathe until he got down to an altitude around 5 miles up (just not enough air).


QUESTION:
Consider two satellites in orbit around earth following each other in the very same orbital path, but say 2000 km apart. And consider that they be connected together by a very thin string. Here is the question: Can this string be gently pulled until it makes an absolutely straight line? Or will it follow the so-called curvature of space that is supposed to exist around earth, between the two satellites’ positions?

ANSWER:
This is a tricky question with lots of possible secenarios for what might happen for various conditions.

  • First, suppose the string is not a string but a rigid stick but of negligible mass which we miraculously make appear between the two satellites happily orbiting in a circular orbit, one following the other. Now, we no longer have two satellites but one instead. One of the laws of classical physics says that if you want to know how a rigid body behaves under the influence of external forces, find the net force on the body and the center of mass of the body will move as if it were a point mass experiencing that force. Now, the center of mass of the pair is below the orbit they are in and therefore this single satellite is moving too slowly to move in a circular orbit and the center of mass will change to moving in an elliptical orbit.

  • What if you put a string of negligible mass between them in a straight line? Since it has no mass, it experiences no force and everything will behave like it was not there.

  • Now, if you connect them with a string which has mass, it also experiences gravity and each piece of string is really wanting to be at a larger orbit so, I believe, the tendency will be for the string to curve to fit the circular orbit of the two heavier satellites. This is an approximation assuming the string has a much smaller mass than the satellites. In reality, the string will exert forces on the satellites and alter their orbits.

  • Under your "gently pulled" secenario, the string would pull gently also on you and the other satellite thereby altering both orbits.


QUESTION:
My question is about tension: what exactly is it? Now, I know how tension is defined, but my problem comes when trying to reason it out with force diagrams or drawings. Take this classic example: There is a tug-of-war match going on between 2 contestants. One contestant pulls to the right with a force of 300 N. The other pulls to the left with a force of 300 N. What is the tension in the rope? Now, I know the tension in the string is 300 N just by definition/recognition...but why? The example proposes a situation in which the net force is zero, so there is no acceleration of the rope. The "force of tension" resists the pull from the contestant on the right with 300 N, and the same goes for the contestant on the left. So, if we represent these forces of tension with vectors, they would both be pointing toward the "midpoint" of the rope. That being said, why don't these tension vectors add to make 600 N net force? Or conversely, why don't these vector forces subtract to make a 0 N net force. Also a side note: In a lot of classical mechanics problems, we always see problems involving "massless" ropes. What exactly does that mean and why do we need to define a rope as "massless" in order to carry out problems involving strings/ropes?

ANSWER:
The tension is nothing more than the force which a rope exerts on something it is tied to. Tension has some important properties. It is always tangent to the rope and a rope can never push, it only pulls. Of course, ropes are never massless in the real world, but we can often approximate them as if they were if they were if their mass is much less than everything else. I can see that your main problem is that you have not been properly instructed on how to attack these simple statics problems. The first thing you have to do is to choose a body on which to focus and look only at that; you are attempting to look at the rope and both men all at once.

  • Let me start with a simpler example, a 1 kg mass hanging from the ceiling on a massless rope. (I will use 10 m/s2 to approximate acceleration due to gravity). So, I will choose a body, the 1 kg mass. What are all the forces on it? Its own weight, 10 N straight down and the force which the rope exerts up. That force by the rope is the tension. Pretty simple, the tension at the bottom of the rope must be 10 N.

  • Now I will choose a different body, the 1 kg mass plus half the rope. What are all the forces on the body? The weight of the body, still 10 N, and the force which the upper half of the rope exerts on the body (at the point where the upper and lower halves of the rope touch). The unknown force again must be 10 N and so the tension in the middle of the rope is 10 N.

  • Now I will choose the 1 kg mass and the whole rope as the body. What are all the forces on it? The weight of the body, still 10 N, and the force which the ceiling exerts up on the top end of the rope. So now we find that the ceiling must exert an upward force of 10 N on the top of the rope. But, Newton's third law says that if the ceiling exerts a force on the rope, the rope exerts an equal and opposite force on the ceiling. So, once again we find the tension in the rope, now at the top, is 10 N. (This is very much like your tug-o-war problem if you think about it: the weight pulls down on the rope with a force of 10 N and the ceiling pulls up on the rope with a force of 10 N, but the tension in the rope, I have just shown, is 10 N everywhere, not 20 N.)

Next, do the same problem but now let the mass of the rope be 1 kg also, clearly not even approximately massless.

  • So, I will choose a body, the 1 kg mass. What are all the forces on it? Its own weight, 10 N straight down and the force which the rope exerts up. That force by the rope is the tension. Pretty simple, the tension at the bottom of the rope must be 10 N.

  • Now I will choose a different body, the 1 kg mass plus half the rope. What are all the forces on the body? The weight of the body, now 15 N, and the force which the upper half of the rope exerts on the body (at the point where the upper and lower halves of the rope touch). The unknown force must now be 15 N and so the tension in the middle of the rope is 15 N.

  • Now I will choose the 1 kg mass and the whole rope as the body. What are all the forces on it? The weight of the body, now 20 N, and the force which the ceiling exerts up on the top end of the rope. So now we find that the ceiling must exert an upward force of 20 N on the top of the rope. But, Newton's third law says that if the ceiling exerts a force on the rope, the rope exerts an equal and opposite force on the ceiling. So, now we find the tension in the rope at the top is 20 N.

Finally, let's do your tug-o-war problem. Part of your confusion is saying that each man exerts a force of 300 N. It is much better to simply say that it is a tie right now so the men must be exerting equal forces in magnitude.

  • I will choose man #1 as the body. What are all the (horizontal) forces on the man? The rope pulls him with an unknown force and some other force must pull him in the opposite direction so that he remains in equilibrium; that other force is simply the friction between his feet and the ground which we will take to be 300 N. Therefore the tension at end #1 will be 300 N. Of course, since the rope exerts a force of 300 N on the man, the man must be exerting a 300 N force on the rope as was originally stated in the problem, but that is not a force on the chosen body.

  • I will next choose man #2 as the body. What are all the (horizontal) forces on the man? The rope pulls him with an unknown force and some other force must pull him in the opposite direction so that he remains in equilibrium; that other force is simply the friction between his feet and the ground which we will take to be 300 N. Therefore the tension at end #2 will be 300 N.

  • I will choose man #1 plus half the rope as the body. What are all the (horizontal) forces on the body? The friction pulls one direction with 300 N and the #2 half or the rope pulls on the #1 half of the rope with the same force, so the tension in the middle of the rope is 300 N.

  • Suppose we choose both men plus the rope as the body. Then friction pulls to the left with 300 N and to the right with 300 N and the rope can be totally ignored because it does not exert any net force on the body being an internal force.

 FOLLOWUP QUESTION:
I see where I got confused. I didn't understand how to apply Newton's 3rd Law to the problem. So the force of the man on the rope (his pull on the rope) is an action-reaction pair with the force of the rope on the man (tension)? Also, do we always consider action-reaction pairs as forces that "cancel" each other out or sum to zero? Lastly, if the forces of friction between their feet and the ground are the ones that add up to a zero net force, what would happen to the tension in the case that one man was able to exert a greater force of friction in one direction? How could we calculate the tension in the rope then?

ANSWER:
Yes, Newton's third law states that if A exerts a force on B, B exerts an equal and opposite force on A. There is never an occasion to add the action and reaction forces because they are forces on different things. They do sum to zero, but that means nothing because one is on the man and one is on the rope, so why would you add them? If the men pull with different forces, the system is no longer in equilibrium and accelerates in the direction of the net force. But still, the tension in the rope is the same throughout its length. Here is how that goes, using Newton's second law (N2) (it is a bit more complicated). In the picture, the man on the left in the blue shirt has a frictional force of 400 N to the left and the man on the right in the red shirt has a frictional force 300 N to the right. Each man has a mass of 100 kg and the rope is massless.

  • First, choose both men plus the rope as the body. All the forces are 400 N to the left and 300 N to the right, so the net force is -400+300=-100 N to the left. N2 says this is the mass of the body (200 kg) times the acceleration a, so -100=200a and a=-0.5 m/s2. The acceleration of both men will be 0.5 m/s2 to the left.

  • Next, choose the left man as the body. All the forces are 400 to the left and the force the rope exerts on the blue shirt man, call it Tblue. Now N2 says that Tblue-400=100a=-50 so Tblue=350 N.

  • Next, choose the right man as the body. All the forces are 300 to the right and the force the rope exerts on the red shirt man, call it Tred. Now N2 says that -Tred+300=100a=-50 so Tred=350 N.

So, the tension in the rope is still the same everywhere (you could do the above exercise of choosing one man plus half the rope to show that it is also 350 N in the middle) and, for this problem, it turns out to be just the average of the pulls at each end.


QUESTION:
My co-workers and I have been in a very heated discussion. They tell me I am wrong and my theory is stupid. Will a bullet fired (perfectly horizontal) and a bullet dropped hit the ground at the same time? My answer is yes, because gravity is constant.

ANSWER:
One picture is worth a thousand words. The figure to the right is an actual strobed photograph of a ball (red) launched horizontally and another (yellow) dropped. If the red ball had been going faster, both would still have moved together in their vertical positions. The reason is that each starts with the same vertical velocity (zero) and experience the same acceleration in that direction (32 ft/s/s). Because a bullet has such a high velocity, air drag will have a significant effect on it but this will affect its motion in the horizontal direction, almost not at all in the vertical direction. This is not a "theory", it is an experimental fact as the photograph shows.


QUESTION:
I am a writer putting together a science fiction screenplay. Those who know me say I have an attention to detail--to a fault. There is one particular element I would like to be as accurate as possible. I'm hoping you might be able to help me. Here is the scenario: A spacecraft leaves earth on course to the moon. In order to create an Earth-like gravity inside the ship; the ship accelerates at a constant rate exerting a force on the occupants equal to one G. Half way through the trip the craft will flip, then decelerate for the remainder of the journey. This would give the same sensation of false gravity to the occupants of the craft. So here is the question: If this were possible; how long would it take to actually reach the moon?

ANSWER:
Since you are such a stickler for detail, I will give you detail, probably far more than you want! Your scheme of having an acceleration with "constant rate" would work in empty space but not between the earth and the moon because the force causing the acceleration is not the only force on you, the earth's and moon's gravity are also acting. As you go away from the earth the earths gravity gets smaller like 1/r2 where r is the distance from the earth's center, and the moon's gets bigger as you get closer. So, it becomes a complicated problem as to how much force must be applied to keep the acceleration just right for where you are. The picture to the right shows you in your rocket ship. Let's call your mass M. Then there are two forces on you, your weight W down and the force the scale you are standing on exerts on you, F. W gets smaller as you get farther and farther away and you want F to always be what your weight would be on the earth's surface, Mg. So, Newton's second law says that F-W=Ma=Mg-W where a is the acceleration you must have. Note that, for the time being, I am ignoring the moon; that would just complicate things and its force is much smaller than the earth's, at least for the first half of the trip. I want you to understand the complication caused by the fact that W changes as you go farther away. Now, how does W change? W=MMEG/r2=Mg(R/r)2 Where R is the radius of the earth and ME is the mass of the earth. We can now solve for the acceleration the spacecraft would have to have: a=g(1-(R/r)2). I have plotted this in red on the graph above. (The distance to the moon is about 60 earth radii.) Note that for most of the trip the acceleration is just about g. I also calculated the effect the moon would have, blue dashed line, and, except for the very end of the trip, it is pretty negligible. Now that we have taken care of the always-important details, we can try to answer your question. To calculate the time exactly would be very complicated, but, since the required acceleration is g for almost the whole trip, it looks like we can get a real good approximation by just assuming a=g for the first half and a=-g for the second half; your perceived weight (F) will just decrease from twice its usual value when you take off to about normal when you get to about 5 earth radii in altitude. The symmetry of the situation is such that I need only calculate the time for the first half of the trip and double it. The appropriate equation to use is r=r0+v0t+½at2 where r0=R is where you start and v0=0 is the speed you start with. Halfway to the moon is about r=30R=R+½at2 and so, putting in the numbers, I find t≈1.69 hours and so the time to the moon would be about 3.4 hours. You can also calculate the maximum speed you would have to be about 140,000 mph halfway.


QUESTION:
Lets say you have a tennis ball machine. The tennis balls must travel though a 5ft tube before leaving the machine. Lets say that the machine is spitting out 60 balls a second at 50 mph. The machine is on a swivel and a person changes the position of the machine by moving the 5ft tube left and right by pushing or pulling with his or her hand. If the speed of the balls were increased to 500 mph would it be harder for the person to move the tube? And if so why?

ANSWER:
This question is very similar to one I answered a long time ago. If you are interested in the mathematical details, you can link there. Here is the old question: Suppose that a mass m is tied to a string of length R0 and is going in a circle with angular velocity ω0. Now, the string breaks. A force which has no radial component (the direction of the force is always perpendicular to the line drawn from the original axis of rotation to the mass) is applied to keep the mass's angular velocity constant; what is the force? How is this like your question? As your ball travels down the tube, assuming the barrel has no friction, the tube exerts a force like the one described above if the tube rotates with constant angular speed. How is it different from your question? The original question has the ball at rest in the radial direction at the beginning but your ball starts out with some radial velocity (your 50 or 500 mph). The earlier solution has two constants a and b which will be different for your problem. Suppose we call the initial ball velocity v0; then where the earlier solution says "…where a and b are constants to be determined…" we will get different values because dr/dt=v0 instead of zero at t=0. If you really care and are good at math, you can find the new values of the constants yourself, but I suspect you just want an answer. The answer for the force the tube exerts on the ball (which is the force you must exert on the tube) as a function of time turns out to be F(t)=202[R0sinh(ω0t)+ω0v0cosh(ω0t)]. (Don't lose heart, this will get much less technical soon; it is just that I want to include details for my many readers who like the math!) To finally answer your question, I need to put in some numbers. Since physicists do not like English units, I will use comparable metric system units:

  • I will choose ω0=1 s-1. This corresponds to about 500 sweep of your gun in one second.

  • The two velocities I will use are v0=20 m/s (about 45 mph) and v0=200 m/s (about 450 mph).

  • The tube length will be 2 m (about 6 ft).

  • I will use R0=0, that is, the bullet is launched from the point where the gun is pivoted; hence F(t)=203v0cosh(ω0t).

  • I will choose m=0.5 kg, a little heavy for a tennis ball, but it really doesn't matter because I am just interested in the relative forces for the two v0 values.

  • As you stipulated, I will choose 60 s-1 as the rate that the balls are launched, but first of all I will just look at what happens for a single ball.

  • Note that the slower ball (20 m/s) will take 0.1 s to reach the end of the tube and the faster ball 0.01 s

So, putting this in, we get F(t)=v0cosh(t). Now, the hyperbolic cosine function (cosh) at t=0 is 1, at t=0.1 is 1.005, and at t=0.01 is 1.0001. This is what I meant when I said things would get simpler. For all intents and purposes the force is a constant because the cosh does not vary significantly for the time the ball is in the tube. So, finally, F≈v0. The force you have to exert while a single fast ball is in the tube is 200 N; the force you have to exert while a single slow ball is in the tube is 20 N. But, that is not the whole story because there may be more than one ball in the tube. The distance between fast balls is (200 m/s)/(60 s-1)=3.3 m, so there is only one ball at a time in the tube. But, the distance between slow balls is (20 m/s)/(60 s-1)=1/3 m so there are 6 balls in the tube at a time. The final answer: you need a force of 200 N for the fast ball gun, 120 N for the slow ball gun, the fast ball gun is harder to rotate.

NOTE ADDED:
While the fast ball is in the tube, the required force is 200 N, but there is only a ball in the tube (4/3)/2=2/3 of the time; so the average force over many balls is more like 133 N, still bigger than for the small balls but not so much bigger. It also occurs to me that for the real-world application we should be thinking of the torque we have to apply which changes as a ball moves down the tube; oh well, I'm getting tired of this problem!


QUESTION:
Assume there is a buoy in the ocean, anchored to the ocean floor. If the tide is low, there is little to no pressure on the chain holding the buoy to the floor. When the tide is high (the depth of the water = the length of the chain) the chain is completely taut, so let's say the pressure on the chain is X. But what if there is a storm and the water level rises over the buoy. Is the pressure on the chain still X or is it greater than X? Does the pressure continue to increase as the water level rises or does it stop at X? In other words: does buoyancy increase as you dive deeper or does it have a set value?

ANSWER:
(We do not talk about "pressure" in the chain, rather the force by the chain on what it is attached to, usually called tension; pressure is force per unit area.) The buoyant force is equal to the weight of the displaced fluid. To make things clear, let's choose a specific example. Suppose the buoy has a density which is 1/10 the density of water. Then, if untethered, it will float with 1/10 its volume submerged so the buoyant force just counteracts the weight. Now, at low tide, attach a chain to the ocean floor which is straight but slack. As the tide rises, more and more of the buoy is under the surface. For example, when half the volume of the buoy is submerged the buoyant force is 5 times the weight of the buoy. So all the forces on the buoy are the weight W down, the tension T in the chain down, and the buoyant force B=5W up: B-W-T=0=5W-W-T=4W-T, so T=4W. Suppose the water keeps rising until the buoy is totally submerged so the buoyant force is now 10W; so now T=9W. Now, if the water keeps rising, the same amount of the buoy is always submerged and so the tension stops increasing no matter how deep the water becomes. (Although we think of water as incompressible, at extreme depths the density does get a tiny bit bigger so the buoyant force at the bottom of the ocean will be a tiny bit bigger, but a trivially small difference.)


QUESTION:
Why do rifles not slightly fire high? The split second a bullet explodes the recoil force should start to spin the gun slightly so that the muzzle lifts up. This is because the recoil action of the gun is not directed at the center of gravity of the gun but above it. (I just checked my rifle and the center of gravity is almost a inch below the center of the barrel).

ANSWER:
You are certainly right, an unrestrained rifle would both recoil backwards and acquire an angular velocity about the center of mass. However, and I will not do any calculations here, because the rifle has such a large mass compared to the bullet, I am confident that the angle through which the rifle would rotate during the extremely short time the bullet is in the barrel would be trivially small. Even so, the sight mechanism is always adjusted for a certain distance and the effect you site would be included in the calibrations of the sight if it were not totally negligible.

NOTE ADDED:
The following occurs to me: the angular momentum of the bullet plus gun before firing is zero and must be zero afterwards. The bullet is confined to move along the barrel and the center of mass (of the bullet plus gun) does not move significantly during the time the bullet is in the barrel. Therefore, I do not expect the gun to acquire an angular velocity until the bullet leaves. Again, this is contingent on the bullet having much less mass than the gun.


QUESTION:
If you  have two objects with exactly the same volume and surface area, and if you apply an equal force on both of them, the one with less mass will have a  greater acceleration. This is because F=ma. But, if you do this experiment  in any fluid, which one will go farther? The surface areas are the same so  neither will have a greater form drag. But because acceleration due to  force, the object with less mass will slow down much more than the other It  is also known that the drag acting on an object is directly proportional to  the square of its speed. So, even if the object with less mass has a greater  acceleration in the beginning, it will slow down much more than the one with  greater mass.  So which one will travel farther?

NOTE ADDED:
After several communications, the question is this: If the same force F acts on both objects for the same time and then stops, which object goes farther? There is no gravity, so this is a one-dimensional problem. Both objects start at rest. This is one of those quite technical answers although, without understanding all the math, you can see the qualitative behavior. This is going to be a lengthy answer, so maybe you just might want to skip it!

ANSWER (CORRECTED, 11/11/12):
I will assume that the drag force is given by Fdrag=-cv2 where c is the same for both objects (as stipulated by the questioner).

  • For the first part of the problem, Newton's second law is mdv/dt=F-cv2. At large times the velocity will become essentially equal to a constant because F≈cv2; this speed is called the terminal velocity vt=√(F/c). Without going into any detail the solution is v=vt tanh[(cvtt/m)]. This is plotted on the first figure above; the larger the mass, the longer it takes to reach the terminal velocity so, as noted by the questioner, the lighter mass will go farther in a given time. To get the position as a function of time, we need to solve v=dx/dt=vt tanh[(cvtt/m)]; the solution is x=(m/c) ln[cosh(cvtt/m)]. This is plotted in the second figure above; the smaller mass goes farther but after a long enough time, the two keep a constant distance apart.

  • For the second part of the problem, Newton's second law is mdv/dt=-cv2. Now, the velocity will begin with a speed v0 which is the speed at the end of the time interval in the first part. Without going into any detail the solution is v=v0/[1+(cv0t/m)]. This is plotted on the third figure above; the smaller mass clearly loses velocity more quickly as expected. To get the position as a function of time, we need to solve v=dx/dt=v0/[1+(cv0t/m)]; the solution is x=(m/c) ln[1+(cv0t/m)]. This is plotted in the fourth figure above; if the two start with equal speeds at the same place (which they do not), the larger mass would always be ahead of the smaller mass. However, there is a curious result if the drag force is of the form -cv2—because the position is logarithmic, the objects never stop moving so the question "which one will travel farther?" has no answer, each will go infinitely far. The "fix" for this situation is that the nature of the drag force changes for very low velocities and is better described as Fdrag=-bv which results in a solution like x=D(1-e-bt/m) which does have the maximum range of D for large t. I am not going to make this a three-step problem because at large enough times the velocities become so small that we might as well consider the objects to have stopped, knowing that they will certainly stop due to the changed nature of the drag.

  • To solve this problem analytically from here is probably not the most illuminating thing to do. I am going to just calculate a few numerical examples where I will choose

    • F=1 N, c=1 N·s2/m2, vt=1 m/s.

    • For the first part of the problem I will choose two different times, t1=0.5 s and t1=10 s; the first will be before terminal velocity is reached, the second has terminal velocity nearly reached by all.

    • For the second part of the problem I will choose a time such that v=0.001 m/s; these times turn out to be approximately equal numerically to the 1000m for my choice of parameters, e.g. it takes t≈1000 s for a 1 kg mass to reach v=0.001 m/s regardless of initial velocity.

    • I will calculate for three masses, 1 kg, 2 kg, and 5 kg.

The results are shown in the following table:

m (kg) t1 (s) v (m/s) x1 (m) t2 (s) x2 (m) x1+x2 (m)
1 0.5 0.462 0.120 1000 4.796 4.916
1 10 1 9.307 1000 6.909 16.216
2 0.5 0.245 0.062 2000 11.011 11.073
2 10 1 8.614 2000 13.818 22.432
5 0.5 0.0997 0.025 5000 23.061 23.086
5 10 0.964 6.625 5000 34.361 40.986

 

My earlier answer had an error, namely the time it takes to reach v=1 mm/s is not  tm but rather t≈1000m. Now the answer to the question seems to clearly be that the more massive object goes farthest. Plotting the three cases for different masses below (choosing a t1 large enough that all masses have the terminal velocity), it is clear that at long enough times the largest mass will always be the farthest ahead at any given time. This conclusion is independent of the values of c or vt.

ALTERNATIVE SOLUTION:
An alternative solution, with F=-bv, is presented below. For a linear drag force it is shown that the total distance traveled is independent of mass.

 

ANSWER:
I will assume that the drag force is given by Fdrag=-bv where b is the same for both objects (as stipulated by the questioner).

  • For the first part of the problem, Newton's second law is mdv/dt=F-bv. At large times the velocity will become essentially equal to a constant because F≈bv; this speed is called the terminal velocity vt=(F/b). Without going into any detail the solution is v=vt (1-e-t/τ) where τ=m/b is the characteristic time. This is plotted on the first figure above; the larger the mass, the longer it takes to reach the terminal velocity so, as noted by the questioner, the lighter mass will go farther in a given time. To get the position as a function of time, we need to solve v=vt (1-e-t/τ); the solution is x=vt[t-τ(1-e-t/τ)]. This is plotted in the second figure above; the smaller mass goes farther but after a long enough time, the two keep a constant distance apart.

  • For the second part of the problem, Newton's second law is mdv/dt=-bv. Now, the velocity will begin with a speed v0 which is the speed at the end of the time interval in the first part. Without going into any detail the solution is v=v0e-t/τ. This is plotted on the third figure above; the smaller mass clearly loses velocity more quickly as expected. To get the position as a function of time, we need to solve v=dx/dt=v0e-t/τ; the solution is x=v0τ(1-e-t/τ). This is plotted in the fourth figure above; if the two start with equal speeds at the same place (which they do not), the larger mass would always be ahead of the smaller mass. Here, unlike the quadratic drag case, the objects reach a maximum distance and so we do not need to make approximations regarding how far each goes. It now turns out that the problem can be done analytically quite easily. The maximum distance for the second part of the problem is for t=∞, so x2=v0τ with v0=vt(1-e-t/τ) where t is the time in the first part until the force is stopped. Now, x1=vt[t-τ(1-e-t/τ)] and so x1+x2=vt[t-τ(1-e-t/τ)]+vtτ(1-e-t/τ)=vtt. The distance traveled is independent of the mass.


QUESTION:
I am teaching a high school physics class and we are using the range equation for projectile motion. Why isn't there two angles that would work each time for this equation? The author only has us solving for one angle but wouldn't there always be two angles that would work for the same range? 

ANSWER:
The range equation, R=2v02sinθcosθ/g  is not usually used to solve for the angles. However, given R and v0, you could solve for θ. A little manipulation leads to cos4θ-cos2θ+R2g2/(4v04)=0 which is a quadratic equation in cos2θ; there are therefore two solutions for cos2θ and two positive solutions for θ as you have correctly deduced. Note that if R2g2/v04>1 there are no real solutions.


QUESTION:
If you drop an egg and it crashes, it breaks. If the egg is dropped and there is a platform going down below the egg, at the same speed as the egg, very slowly decreasing speed, until they touch (while falling), will the egg break?

ANSWER:
By making the platform move just a little slower than the egg you are achieving the same thing as if you were to carefully put the egg on a stationary surface—causing the force between the two to be small. What causes an egg to break? When it collides with the floor, the floor exerts a force on it to cause it to stop. The more quickly the egg stops, the greater the force it must experience. Of course, if the force gets too big, that is if it has to stop too quickly, the shell will not be strong enough to withstand it. To keep the egg from breaking, you may cause the time of the collision to be longer. For example, drop it on a pillow and the egg will not break because it takes longer to stop and so it experiences a smaller force. But, there is another way to get a smaller force—if the egg is moving more slowly, it does not to have such a large velocity change to stop. This is what happens when you place the egg on a table and that is what happens if the table is moving down with just a little less speed than the egg.


QUESTION:
Please settle a debate in my family: would it be possible for a human being to execute a one-armed iron cross on the still rings (gymnastics)?

ANSWER:
Technically possible, physically probably impossible. The physics is that the torque exerted by your hand grasping the ring would have to be equal and opposite the torque exerted by your weight. Let's say the ring has a thickness of 5 cm (about 2"), the distance from your hand to your center of mass is 1 m, and your weight is 650 N ( about 150 lb). Then the torque due to your weight is (Tw=1 m)x(650 N)=650 N·m. Your hand has a moment arm of 2.5 cm=0.025 m, so the torque your hand exerts must be Th=650 N·m=Fhx0.025 where Fh is the force your hand exerts on the ring, so Fh=650/0.025=26,000 N=5845 lb. Your gripping hand would have to exert a force of nearly 6000 lb on the ring to keep you from rotating about your hand. I think that even if you were strong enough to grip it tightly enough for there to be a 6000 lb force, that force would rip ths skin off your hand or perhaps the squeeze would break bones in your hand.


QUESTION:
When you put a candle on a spinning turntable, why does the flame point inwards?

ANSWER:
You can find a good explanation at Big Site of Amazing Facts. The answer is very similar to an earlier answer of mine.


QUESTION:
I teach Jr. High science. I have a question that I need help with when explaining it to some of my SPED students. If you were playing football on the moon, would it be as hard to stop a 150 kg lineman as it would be on the Earth?

ANSWER:
I would say that it would be easier to lift him but just as hard to stop his forward motion. On the moon you weigh less; the 100 kg man would weigh about 330 lb, but because gravity is about 6 times weaker on the moon, he would only weigh about 55 lb. But to stop him, you are not stopping his weight, you are stopping his mass and his mass is still 150 kg. You need to impress on your students the difference between weight and mass. Weight is how hard gravity pulls on a mass but mass is a measure of inertia, how hard it is to stop something from moving (or to start it moving).

(The following is probably too abstract for your students, but I include it for completeness. The force which stops the advancing runner is actually the friction between your feet and the ground. For example, if you were standing on very slippery ice it would be much harder to stop him. The friction between you and the ground is proportional to your weight, so you have less friction on the moon resulting in less ability to stop the runner.)


QUESTION:
Hello, my teacher and I had a bit of a bicker. A man is driving a truck and is on the highway going 70 MPH. Another driver is going 70 miles per hour with the same truck. They hit each other and my teacher believes it will be as if the truck hit a stationary object at 140 MPH, and I said and argued that it would still be as if it hit a stationery object at 70 MPH. Who is right? I saw Mythbusters (The Discovery TV show) do test on it and that is how I got my answer.

ANSWER:
The problem is that the situation is not fully enough described—"…as if the truck hit…" is not defined, what does that mean? The important idea here is Newton's second law, F=mvt) where F is the average force experienced by mass m during the time Δt which the collision lasted and Δv is the change in velocity during that time. Also useful is conservation of momentum for a collision (which is actually just Newton's second law in disguise); momentum is the product of mass times velocity and the sum of all the momentum after the collision must equal the sum of all the momentum before the collision. I will give you two scenarios:

  • The collisions are perfectly inelastic (which means colliding objects stick together) and the "stationary object" is a wall of essentially infinite mass (like a really thick stone wall or a really big tree). This means that all objects come to rest after the collision. The force experienced by anything depends on how long the collision lasted, so we need to stipulate that the collisions last the same time. For this scenario, the force felt by the 140 mph truck would be twice as large as the force felt by the 70 mph truck. This is a situation where you are right.

  • The collisions are perfectly inelastic and the "stationary object" is the other truck. Again, assume that the collisions last the same time. In this situation, the trucks are not at rest after the collision if one truck is parked before the collision. You can (neglecting any friction) easily calculate their speeds after the collision using momentum conservation: 140M=2Mv, so v=70. But, note that the change in speed for either truck for either collision is 70 mph, so this is a situation where your teacher is right.

Remember, in science you can only argue intelligently if the subject of the argument is well-defined!


QUESTION:
The earth orbits around the sun because it has angular momentum. If we stopped the earth in orbit and then let it fall straight towards the sun, how long would it take to reach the sun in seconds?

ANSWER:
The questioner sent me a bunch of data about the masses of the sun and earth, the radius of the earth's orbit, and Newton's universal constant of gravitation. But, you do not need any of that stuff—all you need to know is Kepler's first and third laws and the fact that the period of earth's (approximately circular) orbit is one year.

  • Kepler's first law states that the orbit of a planet is an ellipse with a semimajor axis a and with the sun at one focus of the ellipse. A circular orbit has a semimajor axis equal to the radius of the circle, so a1=RE where RE is the radius of the earth's orbit; the eccentricity of a circle is 0. The other extreme is an ellipse with eccentricity 1 which is a straight line from the sun to the earth and so the semimajor axis for a "dropped earth" is a2=RE/2. Of course, this is not an orbit you will actually see in nature because there is no such thing as a point mass and the speed of the earth when the two point masses coincide would be infinite, but if we can cleverly deduce the period of this orbit, one-half that period will be the answer to your question.

  • Kepler's third law states that the square of the period T of an orbit is proportional to the cube of its semimajor axis, so T12/T22=a13/a23=RE3/(RE/2)3=8. So, T2=T1/√8=0.354 years. So, the time to go half a period is 0.177 years=64.6 days. I will let you put that into seconds.

I am not sure I would say that the earth orbits the sun is because it has angular momentum. True, it has angular momentum, but the reason is that the sun exerts a force on it. Technically the straight line path is an orbit, but it has no angular momentum.


QUESTION: 
I am curious about something that I just read in an earth science book about the shape of the Earth. Since the polar circumference is less than the equatorial circumference, the Earth is an oblate spheroid. The authors attribute this shape to centrifugal force caused by the Earth's acceleration. However, I thought that centrifugal force was not a real force, but was instead due to a mass's inertia resisting the acceleration. Am I misunderstanding the concept? Thanks for helping me resolve this conundrum.

ANSWER: 
(This is sort of a long-winded answer, but I want to have an explanation of centrifugal and fictitious forces I can later refer to.) Centrifugal force is what we call a fictitious force. In an accelerating system, Newton's laws are not true. For example, if you are inside a big rotating drum you feel a force pushing you into the wall, right? Well, actually you do not; what you feel is the wall pushing you toward the axis of the drum. Your brain is trained mostly in nonaccelerating systems (moving with constant velocity) and so it wants to use Newton's first law which says that if an object is at rest the forces on it must add to zero; to make Newton's first law true (you are at rest in your system which is accelerating), your brain invents a force equal and opposite the centripetal force. This is called the centrifugal force and it does not really exist. A fictitious force is one we invent to force Newton's laws to be true in an accelerating system. Just because it does not exist does not mean that we cannot do physics with Newton's laws using it. If the accelerating system is a centrifuge, it works, doesn't it? The heaviest stuff is "pushed out" by the centrifugal force is true even if centrifugal force is just something we made up to make calculations easier. The rotating fluid earth is a little complicated, but I think that it is pretty intuitive: a ball of putty (or pizza dough) made to rotate faster and faster will stretch out into a pancake and you can say it is the centrifugal force which is causing it. A simpler case of "centrifugal stretching" is the following. A spring is attached to a nail at one end and a mass at the other. If you make it rotate about the nail, the spring will stretch out just the right amount so that the force of the spring will provide the necessary centripetal force. But if you make it go faster, a larger force will be required and so the spring will have to stretch more to do that. That example is easy to see without centrifugal force, but if you view it from the perspective of the rotating mass, you will say the centrifugal force pushes you out.


QUESTION: 
In a newton's cradle(which has usually 8 bobs or balls)if i let 5 balls bang on the other 3 balls then after collision 5 balls would move on the other side.Why 5 balls are moving and not three balls?

ANSWER: 
Both energy (½Σmv2) and linear momentum (Σmv) must be conserved where Σm is the mass. Suppose that 5 come in and 3 go out. Then momentum conservation says 3m0u=5m0v (where v is the speed of the balls coming in, u is the speed of the 3 balls going out, and m0 is the mass of one ball). So, u=(5/3)v. So now, the energy of the incoming balls is ½(5m)v2=2.5v2 and the energy of the outgoing balls is ½(3m)u2=½(3m)((5/3)v)2=4.17v2; so energy would not be conserved. The only way to have both energy and momentum conserved is to have the same number of balls going out as went in.

QUESTION: 
Two balloons that have the same weight and volume are filled with equal amounts of helium. One is rigid and the other is free to expand as the pressure outside decreases. When released, which will rise higher? Why?

ANSWER: 
On each balloon there are two forces—its weight (which makes it want to fall) and the buoyant force on it (which makes it want to rise). The weights are the same and never change. The buoyant force is proportional to the volume, so whichever has the greater volume has the greater buoyant force. The expanding balloon wins.

QUESTION: 
In case one I accelerate an apple to one mph. It requires x amount of work to accelerate the apple by one mph. Next I accelerate the same apple by another one mph. This requires the same amount of energy as in the first example, ie the energy necessary to accelerate one kg by one mph. Thus I have expended twice as much energy to accelerate from one mph to two mph. But the apple now has 4 times as much energy. I would so much appreciate an explanation in laymen's terms. I must be really stupid. I just cannot see the logic.

ANSWER: 
No, you're not stupid; it is a little subtle. Energy is not what we call an invariant quantity, that is the kinetic energy something has depends on the frame from which you observe it. What your thought experiment does is shift frames for the second acceleration; you have put yourself in a frame where the object starts at rest. When you do this, you move into a frame where the apple is at rest, that is has zero energy. Therefore, your second experiment is nothing more than a repeat of your first experiment. If you calculate the work you have to do in the original rest frame to accelerate the object from speed v to speed 2v you will find W=½m(3v2) which then leads to the correct total energy and work of ½m(4v2).

FOLLOWUP QUESTION: 
This is what I have trouble understanding. Assuming no friction or wind force, etc. it requires x amount of fuel to accelerate a car by 1 mph, and x amount of fuel for each additional 1 mph. Thus change in velocity is directly proportional to energy input. But change in kinetic energy is proportional to the square of change in velocity. Energy input should equal the change in kinetic energy of the car, but apparently does not. I know there is a basic flaw in that reasoning, but where is it?

ANSWER: 
It does not take the same amount of fuel to accelerate the second 1 mph. It is a little tricky to talk about a car where it is actually the force of friction of the tires which push the car forward to accelerate it. That is were the work is done on the car, a forward force F which is constant if you are accelerating at a constant rate. Now, if the car starts at rest and increases its speed at a constant rate to speed v, how far, d, has it gone? Well, simple kinematics tell us that d=½(F/m)t2 and v=(F/m)t, so d=½(F/m)[mv/F]2=[½mv2]/F. Therefore, the work F does to bring it to v is W=Fdmv2. Now, let's calculate the work F must do to increase the speed from v to 2v. The new distance traveled, d', is d'=vt+½(F/m)t2 with 2v=v+(F/m)t => v=(F/m)t; note that the time the force acts is the same but the distance over which it pushes is larger, so more work is being done. If you do the algebra here (put t=mv/F into the d' equation), you will find d'=3d. It takes three times the energy to increase the speed by another 1 mph in your example. I suspect you are still bothered! And I think I can say why. If you burn fuel at a constant rate you will not accelerate at a constant rate; to accelerate at a constant rate you have to increase the rate at which you burn fuel as you go along. If you burn fuel at a constant rate, the time for the second 1 mph gain will be longer than for the first 1 mph; and guess how much—exactly 3 times longer! (I always go that extra mile for folks nice enough to make a donation!)

QUESTION: 
if a water-filled container's weight with an object in it is equal to the container plus the buoyant force on the object if it has not sunk, Why then does the buoyant force point opposite to the direction of the weight of the container? Would you then not subtract the buoyant force from the weight of the container to get the total weight?

ANSWER: 
Aha, you make the classic mistake made by thousands of physics students going before you—you are not focusing on one body at a time. Let me run through the various possible scenarios here and see if you understand.
  • Look at the object. What are the forces on it? Its weight w down and a force b which the water exerts on it, up. (We call b the buoyant force.) Newton's first law (N1) requires b-w=0, so b=w.
  • Look at the container and water. What are the forces on them? Their weight W down, a force N from the scale they are sitting on, up, and a force f which the object exerts on the water. Newton's third law tells us that f=b and points down. N1 now tells us N-W-f=0=N-W-w, and so N=W+w. There is the answer to your question—the scale reads the total weight. But let's look at another possibility.
  • Look at the container and water and object all together. The forces are the weight, W+w down, and the force of the scale N up. The buoyant force does not come into it at all because the forces the water and object exert on each other are internal forces and cancel out (or, b-f=0). So N1 tells you that N=W+w. Again, there is your answer—the scale reads the total weight.

QUESTION: 
Tie an object such as a pencil eraser to a string. Use the string to swing the object around your head at a constant speed. Can the object be swung so that the string is parallel to the horizontal, flat, and even floor on which you stand?

ANSWER: 
No. The reason is that the object must in equilibrium in the vertical direction since it has no vertical component of its velocity (or acceleration), so the forces in the vertical direction must add to zero. One force in the vertical direction is the weight, straight down. There must be another force which points up and is equal in magnitude to the weight; this can only be the vertical component of the tension in the string and a horizontal string has no vertical component of its tension.

QUESTION: 
a co-worker of mine was describing a time where he threw a box weighing about one hundred pounds out of a window on the eighth floor of a building. In an effort to dramatize the story a little, he wanted to express to us how forceful the impact was of the box on the ground below. he mentioned what I considered to be a rather dubious rule-of-thumb: an object's weight doubles for every foot it falls. At home in the evening I relayed the story to my roommates, and we discussed how unlikely it was that this was true--just for being such a simple, neat multiplier. Our subsequent research on the internet (involving many formulas, calculators and conversion charts) seemed to (roughly) confirm his statement. Being poor physicists, we are open to believing the calculations, but, perhaps as good scientists, we continue to doubt. Can you help? As a general rule-of-thumb, does his statement hold any water: does an object's "weight" double for every foot it falls?

ANSWER: 
I must have written this a hundred times—the weight of something is the force that the earth pulls on it. The weight of the box is the same no matter where it is dropped from (unless the height is not small compared to the size of the earth). So, what might he be trying to say? Suppose we ask how much force would be needed to stop the box in a distance of, say, one foot. The work to stop an object with an energy of 500 ft-lb is equal to 500 ft-lb; so, to stop a 1 lb box which has an energy of 500 ft-lb, you would have to exert a 500 lb force to stop it in a distance of 1 ft (work is force times distance). To acquire this energy, it would have to be dropped from 500 ft up; the energy something has at the ground is proportional to how high you drop it from, so it has twice as much energy than if you had dropped it from 250 ft. Clearly, the energy does not double with every foot, so I don't know what he means about doubling every foot of fall. With every foot of fall a 1 lb box increases its energy by 1 ft-lb.

QUESTION: 
I am in the 8th grade. I am doing a project for my science class on the physics of arm motion in swimming. Can you give me some information? I get extra credit if I get information from a professional.

ANSWER: 
I do not quite know what you want. Here is something: what is it that pushes you forward? Is it your arm? No, it is the water that pushes you forward. Here is how it works. When you move your arm backwards under the water toward your feet, you are pushing on the water, pushing it backwards. Newton's third law says that if one object (hand) exerts a force on a second object (water), the second object (water) exerts an equal and opposite force on the first object (hand). Therefore, the water pushes you forward.

QUESTION: 
This question may sound trivial, but please bear with me. Assume a person is standing on Earth and is going to be hit by a very large object. Let's say there was a large object (with a large mass) and it started accelerating towards you from a foot away at a rate of about 1 foot/second. The mass of the object is such that the gravitational force between the person and the huge object is negligible compared to the gravitational force of Earth. If the acceleration is extremely small, will the huge mass of the object coming in contact with the person be enough to send the person flying?

ANSWER: 
The acceleration is irrelevant, it is the speed at the instant it hits you that matters. "Send the person flying" is not a very quantitative statement. Let's just talk about your speed after the collision. Let your mass be 100 kg (about 220 lb) and you are hit by a freight train with a hugely bigger mass. Two extremes:
  1. Perfectly inelastic, you "stick" to the locomotive so your speed is equal to the speed of the train, v. The collision will happen in a relatively short time, call it t. Your average acceleration is then v/t so the force you experience is mv/t. If v=0.1 m/s and the time of collision is 0.1 s, you would experience a force of about 100 N, about 22 lb. No big deal. But, as v gets bigger or t gets smaller the force gets bigger.

  2. The other extreme is a perfectly elastic collision where you are moving on your own after the collision (and have done so conserving energy). In this case your final speed will be about twice the speed of the train, 2v. So the force over the collision time will be about twice as large.

The bottom line of my answer is that if you are hit by a big thing going very slowly you will probably not get hurt badly.


QUESTION: 
My question is about fluid dynamics. What is the relationship between the drag coeffient and the reynolds number? Based on my research I've been able to establish that for low reynolds numbers (which is what I'm interested in) the relationship is inversely proportional and defined as: Cd=24/Re

ANSWER: 
This took some research on my part! The Reynold's number is Re=ρvL/μ where ρ is fluid density, v is speed of an object relative to the fluid, L is a length characteristic of the situation, and μ is the viscosity of the fluid. Drag coefficient is CD= 2FD/(ρv2A) where A is the area presented normal to the flow and FD is the drag force. To relate these, we must choose a specific example, typically a sphere at relatively low velocity for which the drag force may be written as FD=6πμRv and the characteristic length is the diameter, L=2R. Putting in that A=πR2, simple algebra leads to ReCD=24.

QUESTION: 
Could you explain why the driver of a car must keep her foot on the accelerator to maintain a constant speed and therefore why energy is needed to maintain the cars speed??

ANSWER: 
Wouldn't it be great if we could have a car which had no energy loss? Unfortunately, the world has forces which we cannot avoid which take energy away from something moving along. These fall into the category of frictional forces: a spinning wheel has friction in its bearings which will eventually cause it to stop; an object moving thought the air has air resistance which will eventually stop it as it moves along; the tires are not perfectly elastic and as they roll they are being continually deformed and undeformed and energy is lost. Without all these forces, we could accelerate up to speed and disengage the engine from the wheels and turn it off and just cruise. However, one can work hard to minimize these forces in the design of cars; making the cars aerodyamic, reducing the weight, and other tricks can minimize the energy we lose.

QUESTION:
Do you know why do golf balls have dimples? (does it make them more aerodynamic and if so why?)

 ANSWER:
I am The Physicist, I know everything! Well…maybe not everything. I do know why golf balls are dimpled and the reason is also why tennis balls are fuzzy. Aerodynamics can be very complicated, but I can explain it qualitatively. Air drag on a projectile is caused mainly by the turbulent wake behind the object; the pressure is lower in this turbulent region so there is more force on the front pushing back than there is on the back pushing forward. As shown in the figure to the left, if the air moves smoothly over the ball, the turbulent wake is very broad. But, if you break up the boundary layer of air with some kind of roughness on the surface, you get a much smaller turbulent wake and therefore less drag, as shown in the figure to the right. It is counterintuitive that a rough ball encounters less air drag than a smooth one, but true.


QUESTION: 
I dont completely understand newton's third law of motion.. It says every action, there is an equal and opposite reaction, but when we apply force to a book, why doesnt the book apply the same force to us? And why are we able to push the book wherever we want, if, according to third law of motion, book should also have an equal reaction force?

ANSWER: 
The book does exert an equal and opposite force on you. So, why does the book move and you do not? There are several reasons. First, you have much more mass than the book does and so your force on the book has a much bigger effect than its force on you (Newton's second law, a=F/m). Second, there is friction between you and the floor which can be much bigger than the force the book exerts on you, so you do not move; but the friction between the book and the table is normally less than the force you apply to it, so it does move. Imagine that the book is nailed to the table and you are standing on a floor which is very slippery ice. Now, when you push on the book, it will be you who moves. Or, if you and the book were in empty space and you threw it (by exerting a force on it), you would recoil (with a much slower speed) because of the book's force on you.

QUESTION:File:Drum vibration mode01.gif  File:Drum vibration mode02.gif  File:Drum vibration mode21.gif
An old type car antenna, often called whip aerial , moves back and forth when the car is moving but when you speed up it seem to oscillate in figure 8 motion. Why is that?

ANSWER:
Vibrating systems have different modes of vibration. The prettiest example I could find is some of the modes of a drum, shown above. A real drum will have all its infinite modes going on simultaneously but some will play a more important role than others depending on how the drum is excited. The leftmost one above is called the fundamental and would be the primary one if the drum were struck right in the center; the center of the drum in this case is called an antinode, a place which has maximum amplitude. The right-most one above is another higher mode which, you will note, has the center of the drum not moving; this is called a node and would probably not contribute much if the drum were struck in the center. Your antenna, when just going back and forth, is mainly in its fundamental mode. But when you go a different speed you are doing the equivalent of striking the drumhead somewhere other than its center and exciting a higher mode.

ANSWER ADDED LATER:
I found modes of a beam which should be analogous to your specific problem. Is the second mode similar to your "figure eight"?

 


QUESTION: 
A friend of mine and I have an argument over what is a faster sport tennis or badminton. The criteria is how fast it would take to serve a tennis ball/shuttlecock from one side of an olympic sized tennis/badminton court to the player waiting on the other side assuming that both are standing on the out of bounds line. We are assuming ideal conditions and that the players in both cases are equally strong and fast.

ANSWER:
You may not realize it, but your question is mostly about air drag on projectiles. I seem to get more questions about air drag than just about anything else except maybe variations of the twin paradox. Maybe that is because it is perhaps the most important phenomenon mostly swept under the rug in most elementary physics courses. There are several instances of earlier questions involving baseballs and lacrosse balls which are very similar to this one. For high speed projectiles, air drag is very important; e.g., a 100 mph baseball loses about 10 mph by the time it crosses the plate. Approximations have to be made to quantify the situation you are interested in, but I feel the results I will present are pretty close to what happens on the court. (The following is probably way more detail than you want, but I put it here for anybody who might be interested.)
The approximations are:

  • I neglect gravity because the times involved are are sufficiently short that the ball/shuttlecock will not fall far or very much change its vertical speed.

  • I assume that the drag is proportional to the square of the speed—twice the speed, four times the force of drag. This is an excellent approximation for these speeds, these objects.

  • The form of the force I use is F≈¼Av2 where A is the cross sectional area presented to the wind. Here A=πR2 where R is the radius of the ball or the outer circle of the feathers. This probably slightly overestimates the force for the tennis ball (whose "hairs" have the function of decreasing the drag) and underestimates it for the shuttlecock (whose "feathers" are designed to increase drag).

  • Data for tennis:

    • v0=73 m/s=163 mph

    • R=0.032 m=1.26 in

    • m=0.057 kg=2 oz

    • back line to back line distance: 24 m

  • Data for badminton:

    • v0=92 m/s=206 mph

    • R=0.025 m=1 in

    • m=0.005 kg=0.18 oz

    • back line to back line distance: 13.4 m

I used the fastest recorded serves for the velocity off the racquets, v0. If you integrate F=ma, you get the following solutions: v=v0/(1+kt) and x=(v0/k)ln(1+kt) where k=¼Av0/m. Here are the results:

  • The tennis ball takes 0.39 s to travel the distance, arrives with a speed of 62 m/s (139 mph), a loss of 11 m/s (24 mph).

  • The badminton shuttlecock takes 0.30 s to travel the distance, arrives with a speed of 25 m/s (56 mph), a loss 67 m/s (150 mph).

I will leave it to you to argue about what these numbers tell you about which "is the fastest sport". The shuttlecock starts off the fastest because is has a smaller mass and can therefore have a larger acceleration from the force from the racquet. But it slows down very rapidly mainly because of its small mass. The graph shows the speeds as functions of time.


QUESTION: 
A rollerblader leans into a turn as he moves in a circle. The net torque about any point in the rollerblader must be zero if the rollerblader is not to fall over. If the torque is computed about his center of gravity, then the torque caused by the normal force of the ground up on his wheeels is balanced by the torque due to the centripetal force of static friction of the ground on his wheels. If these torques are computed about the contact point of the wheels on the ground, however, there seems to be a non-zero, net torque due to his weight since the normal and static friction forces act through the contact point and would not contribute to the torque about this point. The moment of inertia of the wheels negligable. How can there (seemingly) be a net torque about the contact point and none about the cg? P.S. I submitted this question to the "Mad Scientists Network" and the engineer who answered it said there was an outward, "righting" force that acts through the cg. He did not state the nature of this force but it sounded suspiciously like he was thinking of centrifugal force. This didn't seem right to me. I want to know how a physicist would explain this.

ANSWER:
Hey, guess what: the engineer got it right! He just didn't give you a complete enough explanation, and you deserve one! Ask the Physicist goes where other Q&A sites only dream to go! Enough silliness, here it is. Newton's laws are not valid in accelerating frames of reference and you are attempting to apply Newton's first law (N1) (the sum of all torques equal zero is the rotational equivalent of N1) to the skater who is accelerating because he is moving in a circle. If he were not accelerating, i.e. moving in a straight line and leaned over he would fall. The best way to deal with accelerating systems like this is to do a trick which inserts forces which do not exist to force Newton's laws to be true; these are called fictitious forces. In this case you must add a fictitious force which points away from the center of the circle, acts at the cg, and has a magnitude equal to the centripetal force; this is called the centrifugal (Latin root center fleeing, fugo is I flee) force. Then everything works out and he is in "equilibrium" in his accelerating frame.


QUESTION: 
I just recently had an argument with an engineering friend of mine and was hoping you could help settle it as we tend to end up at an impasse when it comes to disagreements like this. The argument involves human body falling at terminal velocity and landing back first on various materials. He is of the opinion that landing on cement will ultimately cause less damage to the body than landing on soft loosely packed soil. He bases this on the idea that the speed of sound through something as dense as cement is faster than the speed of sound moving through something as dense as flesh and that this somehow imparts a greater ability to absorb an impact and therefore will do less damage to the body than loosely packed soil. I am of the opinion that the soil is able to shift its position under pressure and able to be compacted into a denser form than its current state and that this allows it to better absorb the impact (than concrete would) which reduces the damage done to the object hitting it. He argues that this is only the case in low speed impacts and that in a high speed impact at terminal velocity the concrete would be the better material to land on. So, which material would you expect to do the least amount of damage to an object with roughly the same density as water that hits it at terminal velocity and why?

ANSWER:
That is just about the craziest thing I ever heard! What hurts you when you stop is the force you feel. If you feel a large force, you are hurt; if the force you feel is small, you shrug it off. So the trick is to minimize the force. One way to write Newton's second law is that mΔv=Ft where m is the mass, Δv is the change in speed, F is the average force, and t is the time during which F acts. (The physics way to say this is that the change in momentum is equal to the impulse, mv being momentum and Ft being impulse.) So, letting your mass be about 100 kg and your terminal velocity be about 50 m/s, F=5000/t. Clearly, the bigger you make the collision time t, the smaller F will be. If you hit concrete you stop very quickly, say a hundreth of a second, so the force you feel is 500,000 Newtons, about 112,000 pounds. If you could make the collision last 1 second, for example by landing in a swimming pool full of shredded foam rubber, the force would be 5000 Newtons, about 1100 pounds, which you would probably survive.


QUESTION: 
If your riding a motorcycle, going 55 miles per hour, and it is 75 degrees outside, what is the temperature of the air that you feel on your skin? I ride a motorcycle and I know that there is a formula but I just don't know what it is and I hope that you have the answer.

ANSWER:
You must be referring to the wind chill index which attempts to quantify how cold it feels at a certain temperature and wind speed. I find that it is not defined for the situation you want 750F just never feels cold, I guess. It is only defined for temperatures between -450 and 450. Read all about it at NWS.


QUESTION: 
if you shoot a gun towards the back of the train and the bullet goes 2000 miles and hour in a train thats going 2000 miles does the bullet stop? what would happen?

ANSWER:
For questions like this I always start with "…you must specify the velocity relative to what…" You would see the bullet travel toward the back of the train with a speed of 2000 mph but somebody standing by the side of the tracks would see the bullet standing still.


QUESTION: 
I am a 8th grade physical science teacher and I want to be sure about my understanding of action/reaction forces before explaining it to my students. My students recently completed a ‘balloon rocket activity’ where they calculated speed and identified and described the action/ reactions forces involved. Here are my two possible explanations. Which one is correct?
#1 action force-the air in the balloon exerts a force on the air in the room as it is released, reaction force-the air in the room pushes on the air coming out of the balloon causing motion
#2 action force-balloon pushes on the air in the balloon (forcing it out), reaction force-the air in the balloon pushes back on the balloon causing motion.
I do understand that action/reaction forces involve two objects and do not cancel. I tried to relate the balloon’s behavior to that of a rocket when launched to space.

ANSWER:
The balloon would work perfectly well in a vacuum. Hence, the room air has nothing to do with it. Your #2 is the correct view. The balloon pushes the air and the air pushes with an equal and opposite force on the balloon. When you say that they "do not cancel", that means that they do not cancel if you look at one body or another. But if you look at the two or more bodies as a whole, all the Newton's third law forces do cancel out. That is why when an artillery shell explodes the center of mass continues moving in a parabola even as the pieces fly apart. It is particularly important not to consider any pair of equal and opposite forces to be due to Newton's third law. For example, my weight is a force 200 lb down and the floor exerts a force up on me equal to 200 lb. These are due to Newton's first law, not the third; you seem to understand that when you say the forces "involve two objects". Any two forces on an object which are equal and opposite cannot be Newton's third law forces. For me on the floor, the floor exerts a force up on me and I exert an equal and opposite down on the floorthat is Newton's third law.


QUESTION: 
Alright - hopefully you can help me clear up my issue with this concept. I made up this question for some students and I understand how the math works, but why it does is beyond me - hope you can help! A man who jumps straight up with an initial velocity of 3.70 m/s (enough a person with a 2.36 m reach to jump and touch the rim on a 10 ft hoop: 3.05m high) will travel up the 3.05 m. Easy math, easy understanding. What is hard to understand is how if that man was to run horizontally with that same speed (3.70 m/s) how he would also be able to reach that same height? This is PE=KE stuff and also easy, just hard to believe. Do you have a way you could better explain this? It just seems even in a perfect world with no wasted energy that some of the energy would not be enough compared to just jumping straight up. Instead of all of the velocity being concentrated in one direction (up), its broken into 2 components going up and out. I know energy does not have vectors, but it still baffles me - thanks!

ANSWER:
The problem, I think, is that PE=KE simply does not get it except for the simplest problems. The way to view it is (PE+KE)before=(PE+KE)after. So, choosing the ground as PE=0, the first problem has PEbefore=0 and PEafter=mgh=9.8mh, KEbefore=½m(3.7)2=6.85m and KEafter=0, so 6.85m+0=0+9.8mh or h=0.70 m; add to that your 2.36 m reach and you get 3.06 m. In your second (more puzzling to you) problem, you still need to get to h=0.70 m, but you start with a horizontal component of your velocity vx=3.7 m/s. But to go up, you must give yourself a vertical component of your velocity of vy. When you get to the top you do not have zero kinetic energy, but the kinetic energy due to your horizontal velocity (which never changes). So the way to set it up is KEbefore=½m[(3.7)2+vy2], KEafterm(3.7)2, PEbefore=0, PEafter=(9.8)(0.70)m. If you put this all together, you will find that you need a (suprise!) vertical component of velocity of vy=3.7 m/s.

FOLLOWUP QUESTION: 
Alright I left a crucial part out of my last question about the vertical jumping velocity and the horizontal running velocity. The velocity that the man jumps vertically with 3.7 m/s allows him to reach up to the 3.05 m rim. The second man runs horizontally - grabs a rope (pendulum) - and this should take him up to the same height given his same reach. The first problem you can do like you said with the potential and kinetic energies or constant acceleration formulas. The second I can do as well. I'm ok on the math, i just don't know how that could be? The first man is directing all of his energy in one direction while the second man is grabbing a rope which would take him both vertically AND horizontally. How can the same velocity get that man to travel the same vertical height but in the second case - a further distance?

ANSWER:
The method is still the same. But now he starts with a velocity of 3.7 m/s and ends (at the top of the swing) with zero velocity. So, applying (PE+KE)before=(PE+KE)after, (0+½m(3.7)2)=(9.8mh+0). Note that this looks just like your first problem—it starts out with all kinetic and ends up with all potential energy. The shape of the path is irrelevant, i.e., if it starts at the ground with some speed and ends up somewhere else at rest, then it makes no difference how it got there. Don't worry about the "further distance" because no energy is lost or gained by the horizontal motion.


QUESTION: 
You have a long rope. You tide it into a tree and go walking around the world. When you come back to the tree from the other side. You push the rope tight and untide it from the tree and and tide the ropes together. Is this possible, or will it float.

ANSWER:
Sorry, this could not possibly work with a real rope. For one thing, the rope would sag under its own weight before you got to the horizon from the tree. But, just suppose you had this rope in a circle a few meters above the ground and all the way around the world. Look at a small piece of the rope and see if it is in equilibrium. There is its own weight, straight down, and the tension in the rope pulling left and down and right and down. No forces up, so how could that little piece of the rope be in equilibrium?


QUESTION: 
If you stopped an Earth satellite dead in its tracks, it would simply crash into Earth. Why, then, don't the communications satellites that "hover motionless" above the same spot on Earth crash into Earth?

ANSWER:
Because they only appear to hover motionless. Don't forget, the earth is spinning around once every 24 hours. The period of a communication satellite is 24 hours, it is not standing still.


QUESTION: 
it's a question the answer to which I wish to use as an analog when I make (always free) talks to citizen (and other) groups regarding homelessness; and specifically in response to the complaint by some in the audience that the homeless need to just pick themselves up by their own bootstraps and stop being a burden on society. I keep trying to explain to them that once one has fallen all the way down (as opposed to just tipping over a little, or even falling to one's knees; and especially once they've slipped so through certain kinds of society's cracks), it actually takes more effort to get back up again than it took to knock the person down. (And, trust me, it does.) [The Physicist: The questioner wishes to compare the energy necessary to tip over a cylinder of radius R whose center of gravity is a distance h above the floor to the energy required to lift it back up.] What is the amount of energy needed to tip it over from vertical to horizontal compared with (versus) the amount of energy needed to pick/tip it back up and make it vertical again? I'm looking for a ratio.

ANSWER:
To tip it over, you have to move the center of gravity (COG) so it is above the point on the floor where the cylinder touches the floor; to do this you must raise the COG a distance h[√(1+(R/h)2)-1]. The work necessary to do this is Wfall=mgh[√(1+(R/h)2)-1]. If R is much smaller than h, this may be approximated as Wfall≈½mgh(R/h)2. The work necessary to lift it back up is Wlift=mg(h-R). Again, if R is much smaller than h, Wliftmgh. So, the ratio is Wlift/Wfall≈2(h/R)2. For example, if h=5R, Wlift/Wfall≈50; it takes 50 times the work to lift as to push over!


QUESTION: 
Does an object immediately start to lose speed once an unbalanced force stops acting on it? For example, does a baseball achieve its highest speed at the exact moment it leaves the pitcher's hand and then begin to slow down due to air resistance or is there a period of time after it leaves the pitcher's hand when it might actually pick up speed before beginning to slow?

ANSWER:
You have to look at all the forces in a given problem. The mass will always respond instantaneously to whatever forces are on it at the moment. After a pitched ball leaves the pitcher's hand, the hand has nothing to do with what happens thereafter. The important forces on it are its own weight (gravity) straight down, and whatever air drag does to it. Air drag is pretty significant on a baseball—a 95 mph fastball loses about 10-15 mph by the time (about ½ second) it gets to the plate. So the instant it leaves the pitcher's hand, it starts speeding up in the down direction (due to weight) and slowing down in the direction it is moving. The air drag always slows it down, the gravity can either speed it up or slow it down depending on whether it is moving up (slows it) or down (speeds it). The net effect is usually to slow it because the pitcher has to loft the ball a bit for it to reach the strike zone high enough, so there is little net change in the vertical component of the velocity. Of course, the preceeding does not apply if you are interested in a breaking (curve, slider, etc.) ball. In that case there will be air friction forces not strictly opposite the direction of the velocity; I believe that a curve ball will sometime "break" at the last moment due to a downward force due to the spin on the ball which adds to gravity and makes the ball fall faster than it would just because of gravity.


QUESTION: 
Why does a rugby ball dropped vertically from the ground with it's long axis at an angle to the ground bounce backwards? What provides the force in the horizontal direction and in which way does it act?Plus if the Contact force on the ball is resolved and it does not pass through the ball's centre of mass in which way will the centre of mass of the ball move and how does this motion relate to the line of action of the contact force ?

ANSWER:
You are certainly right, if there were no force with a horizontal component, the ball would have to rebound straight back up although with some spin about the center of mass. However, what you refer to as the "contact force" has two components, normally called the normal force and the frictional force. If the surface were frictionless, the ball would come straight back up. But there is friction, a horizontal force, and the direction it points in is the direction which the ball will accelerate while there is contact. If there were no friction (f in the picture), the normal force (N) would exert a torque about the center of mass which would make the ball rotate clockwise. So, the ball would want to slide to the left during the collision and so the friction will point to the right and that is the horizontal direction the ball will move after bouncing. The center of mass will accelerate in the direction of the sum of all three forces on it (including the weight W). The torque about the center of mass will determine how the ball rotates after bouncing:

  • if N+f passes through the center of mass, it will not rotate;

  • if N+f passes above the center of mass, it will rotate clockwise;

  • if N+f passes below the center of mass, it will rotate counterclockwise.

When I bounce an American football (I don't have a rugby ball), I find it goes clockwise.


QUESTION: 
How much force does it take to move a 3600 lb object?

ANSWER:
Again we have that question "how much force does it take…" when there is no answer. Suppose the object were on a very smooth level surface (virtually firctionless). Then it would take maybe 1/10 oz of force to get it moving 1 ft/s if you pushed for a second. But suppose the object were on a rough surface. It might take a million pounds of force just to break it away from the friction. Suppose that the object were welded to the floor. Then maybe it would take a trillion tons of force to get it moving. The only sensible question you can ask without giving more information (and a definition of what is meant by "to move") is how much force does it take to lift a 3600 lb object? That would be 3600 lb.


QUESTION: 
I know that the full Law of Conservation of Energy can be shown as: "KE{gained} + PE{gained} = KE{lost} + PE{lost} + (F{friction} * {distance})" The "F{f}d" part is equal to the energy lost/gained by thermal energy due to friction. This would suggest that thermal energy due to friction is equal to the force of friction times the distance. However, I cannot find anyone that says this, or even backs this up! Most of the answers I find (of which their are very few) are differential calculus equations, none of which agree with each other. What is the actual formula, and why isn't it "F{f}d?"

ANSWER:
First, that is not at all how I would write the energy equation. I prefer E2=E1+W where E2 is the energy at the end, E1 is the energy at the beginning, and W is the work done by all external agents. When I write E this means the sum of kinetic energy and potential energy. All external agents mean all forces for which we have not introduced a potential energy function. I tell my students this is easy to remember because it says "what you end up with is what you started with plus what you added". If the external agent is friction, the work it does is negative because it takes energy away. This is not what you should call "Law of Conservation of Energy" because energy is not conserved but changes by W. So, the work done by friction is W=-Fd. The law of energy conservation states that the energy of an isolated system never changes. So, where did the energy which friction "took away" go if we imagine doing the whole experiment in some enclosed box to isolate it from the rest of the universe? Friction must be converting some of the energy we started with into some other form of energy (not showing up as obvious kinetic or potential energy), as you correctly feel. Most of the energy will end up as thermal energy, everything gets hotter. Also, a sliding object makes noise and sound waves carry energy. Thermal energy and sound waves are just microscopically another form of kinetic energy.


QUESTION: 
Consider a ball with mass M attached to a very long cord with linear density λ. If you calculate the ball's maximum height when you throw the ball up with initial velocity v, using F = dp / dt, F = - (M + λy) g, and p = (M + λy) dy / dt, you will obtain something different of what you obtain trying to calculate the maximum height using conservation of energy. How energy is dissipated in this case? The cord is straight below M and it is longer than the maximum height. In the beginning of the movement, only the ball moves, and the cord is on the ground. As the ball goes up, it pulls the cord. [I have changed your notation slightly.]

MY FIRST (INCORRECT!) ANSWER:
Let's first do energy conservation. Energy is conserved because the only force on the ball + cord is gravity. (There is also the force of the table on the stationary part of the string, but it does no work. E1=½Mv2, E2=Mgh+λhgh)=Mghλgh2 where h is the maximum value of y. Setting E1=E2, you must solve the quadratic equation ½λgh2+Mgh-½Mv2=0. You get two solutions, one for which h<0 which is discarded. The final solution is h=(M/λ)(-1+√{1+[λv2/(Mg)]}). As a check, if λ=0 this may be shown to be h=v2/(2g), the well-known answer for no hanging cord. Now, you say you solved it by solving F=dp/dt? I was unable to do so. I can get the differential equation for Newton's second law, but it is a nonlinear, second-order differential equation: 0=(M+λy)(g+d2y/dt2)+λ(dy/dt)2. Maybe I could solve this if I brushed up on differential equations, but I am not willing to do that!

MY SECOND ANSWER:
The questioner, Emilio Matos of Brazil, sent me his solution solving the F=dp/dt equation. He was much more clever than I and achieved a solution of h=(M/λ)(-1+(3√{1+[3λv2/(2Mg)]}). A copy of his solution (in Portuguese) is attached; note his slightly different notation: h for y, m for λ, v0 for v, hmax for h. The graph shown above compares the correct solution with my incorrect solution assuming energy conservation; the correct solution has a much lower maximum height indicating substantial loss of energy. My statement that "the only force on the ball + cord is gravity" is incorrect because the mass is increasing. In order for the "next piece" of cord to be moved from rest to moving up, a force must be exerted on it; that means that the next piece of cord exerts a downward force on the end of the moving cord which does negative work on the system. This is most easily seen by imagining a mass m sliding on a horizontal frictionless surface (avoiding the complication of gravity) with speed v and picking up a mass dm at rest. An impulse of Fdt=dp=(m+dm)(v+dv)-mv=vdm+mdv must be delivered to dm to speed it up, and so m experiences an impulse which slows it down. Of course, this is nothing more than  F=dp/dt, but with a focus on the nature of F. In the case of interest, there is the additional force of gravity; the important thing to realize is that this force is not the only force because if it were the acceleration would simply be -g which it clearly is not. Hence, energy is not conserved. Incidentally, the correct solution for h also reduces to h=v2/(2g) for a weightless cord showing that, just because an answer reduces to a known result in a particular limit, the answer is not necessarily correct! Now that we understand the problem, I am not going through the calculations necessary to use  energy to solve this problem (by calculating the work done by F) because it would simply go over the same ground already covered by Mr. Matos' solution.


QUESTION: 
if a bullet was travelling at 823meters per sec and hit an object that stopped it dead how much force per square cm would be exerted onto the target?

ANSWER:
Here is the question which I get in one form or another which indicates how poorly understood the concept of force is! First, you want to ask what the force exerted by the bullet is, not the force per cm2. If you were able to get the force, you could divide that by the cross sectional area of the bullet to find the effective force per unit area; but you cannot get the force because it depends on how quickly the bullet stops. If you mean by "stopped it dead" that it stops instantaneously, then the force would be infinite. The average force is the change in momentum (mass times velocity) divided by the time to stop. So, you need also the mass of the bullet. Suppose the bullet had a mass 0.02 kg, then the change in momentum (0.02x823) is about 16 kg m/s. If it stops in 0.01 s the average force is 1600 N=360 lb, if it stops in 0.001 s the average force is 16,000 N=3600 lb.


QUESTION: 
My teacher did a demonstration in class where there was a mass of 1 kg suspended by a string above it and then he attached a string below the mass. For the first demonstration, he pulled quickly to show that a rapid acceleration on the string below the mass would break that string. For the second demonstration, he showed that if you gradually accelerate the pulling of the string below, the string holding the mass would break instead. Could you please explain why this is the case, including the interaction between the two tensions between the strings? I understand that a greater acceleration means a greater force, but I have difficulty understanding why the gradual pull breaks the upper string and what "breaking the tension of the string" means.

ANSWER:
If you pull up gradually with a force F and the mass M is at rest, then the forces on the mass are its weight, -Mg, the lower string, -Tlower, and the upper string Tupper=F. Newton's first law states that 0=F-Mg-Tlower. So, the tension in the upper string is greater than the tension in the lower, F=Mg+Tlower. So, as you gradually increase F, the tension in the upper string will reach the breaking point first. I assume that in the second case, the lower string starts out slack and the mass is pulled upward by the upper string. When the lower string suddenly gets taught, there is suddenly a very large acceleration in the downward direction which can only be provided by the lower string. Therefore, there is suddenly a large tension in the lower string which exceeds the breaking strength and breaks it.


QUESTION: 
I had a question regarding relativity. If I'm not mistaken, the principle of relativity states that different reference frames are essentially equivalent in that you can't discern a "stationary" reference frame from a "moving" one (that basically there is no such thing as absolute position). So, i was considering a collision of two objects, one of which is much more massive than the other (say, a train and a person). From an observer relative to the person, the train is approaching at a certain speed (say, 5 mph). if the train were to crash into the person, the result would be very bloody and cause severe damage to the person. If, however, this were viewed by an observer moving alongside the train, it would appear as if the person simply walked into the train at 5 mph, and would emerge damaged, but not severely harmed. How is it that this doesn't contradict the principle of relativity? An event's result seems to depend on a "stationary" reference frame (that either the person or the train is moving), when according to relativity the result should be the same regardless of the observer's frame of reference.

ANSWER:
I'm sorry, but your assumption that the 5 mph train will be a violent accident is simply wrong. If you are hit by a 5 mph train you will suffer the same injuries as if you run into a train at a speed of 5 mph. Think about it—as the train approached you could simply grab on to the front and start moving with it. It would hurt a little bit, but no more than if you ran real hard into the train at rest. If the train were going 100 mph, it would kill you, but if you were launched at 100 mph into the train, you would die with similar injuries.


QUESTION: 
If all inertial frames of reference are equivalent, and there is no way to identify which of two objects which are becoming further apart is moving, how is it possible to determine which of two objects is accelerating? For example, how do we know that a train is accelerating away from a platform, and not vice-versa, if it is impossible to determine which is moving when the train is traveling at a constant velocity? If all inertial frames of reference are equivalent, and there is no way to identify which of two objects which are becoming further apart is moving, how is it possible to determine which of two objects is accelerating? For example, how do we know that a train is accelerating away from a platform, and not vice-versa, if it is impossible to determine which is moving when the train is traveling at a constant velocity?

ANSWER:
One way you can identify an inertial frame is if Newton's first law is true (an object at rest or moving with constant velocity has zero net force on it). For example, sitting at the train station there is a pendulum bob hanging straight down; the earth is pulling straight down with a force (called the weight) and the string is pulling straight up with a force which may be measured to have the same magnitude as the weight. Now, if somebody is accelerating in a train going by and also has a pendulum, it will not hang straight down but at an angle. But there are still only two forces on that pendulum bob, the weight straight down and the string not straight up. So, if you are in the train's frame of reference, you conclude that Newton's first law is not true.


QUESTION: 
why do paratroopers roll on landing

ANSWER:
When you hit the ground, the ground exerts a force up on you to stop you. The force which it exerts depends on how long the collision lasts. This is a result of Newton's second law which says that F=mΔvt where m is your mass, Δv is your change in speed, and Δt is the time to stop. You have no control over your mass or the speed you hit the ground, but you can make Δt longer and rolling is one trick for doing that. The longer you can make your collision with the ground last, the less it will hurt. Also, just jumping off a table, for example, it is more painful if you hold your legs straight than if you bend them while stopping.


QUESTION: 
In a projectile motion, which has a farther range (neglecting air resistance, or in a vacuum) fired from a catapult with constant force and same angle of flight, a paper(crumpled) or a rock? Knowing that the paper's mass is less than the rock, I thought that the paper would have a farther range since mass is inversely proportional to acceleration; but my teacher says otherwise, that they would have the same range. Who is correct?

ANSWER:
The problem is not specified completely, so either of you could be right. Think about it: if the (constant) force acted on the rock for 5 minutes but on the paper for 1 second, the rock would go way farther. You can go one of two ways—specify the time the force acts on each or the distance over which the force acts on each.

  • If the forces exerted on both are the same and the time which the force acts for each is the same, then you are right and your reasoning is right. Your reasoning is right because the acceleration of the less massive paper is larger and if the times of acceleration are the same, it ends up going faster; the ratio of the velocities will be given by v1/v2=m2/m1. If the rock is 4 times more massive than the paper, it will have ¼ the initial velocity of the paper.

  • However, if the catapult exerts equal forces over the same distance (the distance of the arc of the arm, presumably), the paper will still have a larger initial velocity than the rock. But it will not be because of acceleration being different since the two will spend different times experiencing the force. Here each acquires the same kinetic energy because equal work was done on them. In this case, the paper still goes faster but the ratio of velocities is v1/v2=√(m2/m1). If the rock is 4 times more massive than the paper, it will have ½ the initial velocity of the paper.

The second scenario is likeliest to be a description of a catapult. A very light stone will get launched in a much shorter time than a very heavy stone and both will experience the force over the same distance. In all fairness to your teacher, he was trying to illustrate that, without air drag, two projectiles with identical initial conditions will go over identical paths. This requires stipulating equal initial velocities, though, not equal forces.


QUESTION: 
My son has a question that I can't seem to find an answer to researching on the web. WHAT FORCE IS BEHIND A GOLF BALL WHEN HIT ON THE MOON? I appreciate your time on answering this for my son. I think it has something to do with Velocity, but not sure!

ANSWER:
I think there is some confusion about what force is here. Here are all the forces a golf ball experiences here on earth:

  1. The club, traveling with some speed, hits the ball and exerts a contact force on it for a very short time but it is a very big force and it results in the ball acquiring a very large velocity. As soon as the ball leaves the club, there is no force "keeping it moving". If there were no other forces, it would keep going forever with the speed with which it left the club.

  2. Once it is started, gravity pulls down on it which is the force which eventually does bring the ball back to the ground.

  3. As it flies through the air, it experiences air drag which can be very complicated. Essentially, it is a force trying to slow it down and the bigger the speed is the bigger this force is.

  4. If it happens to be spinning, the air drag can act asymmetrically so that the ball curves. This is what is called a hook or a slice in golf (depending whether it curves left or right, respectively, for a right-handed golfer).

  5. Of course, when it hits the ground, it experiences forces from the ground which ultimately bring it to rest.

What is different on the moon?

  1. If the club is the same club with the same speed, there is no difference for this force. Therefore, the ball launches just the same as on earth.

  2. The moon is much smaller than the earth and the result is that the gravity on the moon is much weaker. Therefore, this force (trying to pull the ball back down) is much smaller and the ball will go a lot farther.

  3. Since there is no air on the moon, there is no drag and this also results in the ball going much farther.

  4. The ball will not curve on the moon, regardless of how much spin it has.

  5. When it hits the ground, things are about the same as on earth except that all the forces are smaller, again because of gravity being smaller, so the ball rolls farther before it stops (also because it is going much faster when it hits the ground than it would have been on earth).


QUESTION: 
Why was work done defined as Force x Displacement? Is there a sound reason or was it simply derived from the work energy theorem?

ANSWER:
Work was not defined as Fd, it arises naturally from Newton's second law (the source of essentially all classical mechanics). You start with Newton's second law, F=ma, rewrite it as F=m(dv/dt), use the chain rule F=m(dv/dx)(dx/dt)=mv(dv/dx), rearrange mvdv=Fdx. On the right of this equation is work done if m moves the small distance dx; on the left is what turns out to be how much the kinetic energy changes when that work is done. If you integrate this you get
Δ(½mv2)=FΔx=W.
Maybe this was too much detail for you.


QUESTION: 
If 225 pounds falls 24 inches straight down to the ground, propelled only by the force of gravity, what is the force of impact?

ALMOST SIMULTANEOUS QUESTION: 
what is the striking force of an object that weighs 300lb falling from a height of 3 feet

ANSWER:
My God, will I never stop getting this question? It depends entirely on the details of the collision. In essence, what you need to know is how long did it take for the weight to stop? If you had checked the FAQ page you would have found an answer. There is no such thing as "striking force".


QUESTION: 
You and a child half your height lean out over the edge of a pool at the same angle. If you both let go simultaneously, who will tip over faster and hit the water first? Or will they hit at the same time?

ANSWER:
The child will hit the water first. To understand why, model a person as a stick of height L and mass M. Newton's second law says Iα=(MgLsinθ)/2 where I is the moment of inertia about the end of the stick, α is the angular acceleration, θ is the angle the person makes with the vertical, and g is the acceleration due to gravity. The right side is the torque exerted on the falling object and angular acceleration measures how quickly the rate of falling is increasing. So, whoever has the biggest α hits first. The moment of inertia needed is I=ML2/3, so solving for α, α=[(MgLsinθ)/2]/[ML2/3]=3gsinθ/(2L). So, mass does not matter and the larger L is the smaller α is. To convince yourself that this is reasonable, imagine a race to fall over between a matchstick and a 100 ft high chimney.


QUESTION: 
I fly alots of transatlantic flights and a friend of mine in discussion have a disagreement. Here's the problem- If I am flying in a plane at 60,000 feet and he is flying in a plane at 20,000 feet and our airspeeds are 400 mph. Who covers more ground faster the higher aircraft or the lower aircraft? I contend the higher aircraft will have a higher ground speed than the lower aircraft even though they are both travelling at the same speed of 400 mph. I believe the further you are away from the earth the distance you travel in a straight line is lengthened by the curvature of a sphere. Whose right?

ANSWER:
I think it is a good idea to define "ground speed". Let's say it is the speed of the plane's shadow as measured by somebody on the ground. Then if you have a speed v and are a distance h above the earth's surface, the ground speed vg can be shown to be vg=v(R/(R+h)) where R is the radius of the earth. Alas, you are wrong! (By the way, I do not know what you mean "travel in a straight" line because if you fly at some altitude you are flying in a circular path.) To make this plausible, consider the following: two planes, each with speed 400 mph, one at altitude 1000 ft, the other at altitude 100,000 miles and both are keeping their altitudes constant; which will go all the way around the world first? Because the altitudes of your example are so small compared to the radius of the earth, there is practically no difference. I calculate the ground speeds to be 398.9 for the higher plane and 399.6 for the lower. [If you care, here is the proof. The ground speed and the plane both have the same angular velocity which is essentially the rate you are going around the circle, for example 1 revolution per month is the angular velocity of the moon around the earth, 1 revolution per day is the angular velocity of the earth on its axis, 1 revolution per year is the angular velocity of the earth around the sun. If angular velocity is measured in radians per second, then the angular velocity is v/r where r is the radius of the circle v is going. Then, vg/R=v/(R+h).]


QUESTION: 
My students hear all the time that there is ZERO G. I tell them that astronauts are free falling, like someone jumping out of an airplane. But I hear professionals say zero G all the time and the kids are confused. How do explain that gravity exists everywhere, and then they hear of zero g on the space station ?

ANSWER:  
First, let me say that your example (which I have not included here) is not a good one. Let me attempt to give an answer to the question regarding zero-g (sometimes referred to as weightlessness). Those terminologies are, strictly speaking, incorrect. What is your weight? It is the force which the earth exerts on you. It is not what is measured by a scale; a scale measures the force you exert on it, not the force the earth exerts on you. Of course, in every day life the force which you exert on the scale happens to equal your weight, but if you are in an elevator accelerating up the scale will read more than your weight. If you are in an elevator free falling down, the scale reads zero but your weight is still the same. If you say zero-g you imply that the gravitational field is zero but that is certainly not true anywhere near the earth (except at that point where the earth's and moon's gravity cancel). I presume that you have taught your students about centripetal acceleration. Anything which moves in a circle of radius R with speed v has an acceleration toward the center of the circle of v2/R. A satellite in a circular orbit has just the right speed such that this acceleration is equal to g so, as you correctly state, it is the same as the free falling elevator—you feel like there is no gravity. All this is standard physics.

Here is another perspective: Einstein's principle of equivalence states that there is no experiment you can do to distinguish whether you are in a gravitational field or in an accelerated frame of reference (that is, the freely falling elevator and having zero gravity are physically indistinguishable from inside). So, if it looks like a duck and quacks like a duck, we might as well call it a duck!

(One final note: Note that I have ignored the tiny correction that if you are orbiting with your feet "down" your feet have a slightly different acceleration from your head.)


QUESTION: 
if an object is falling at a fixed rate of 500 feet per minute, what g-force will that object experience upon impact on the earth's surface. If possible include the formula so other rates could be used, since I would also like to calculate the g-force for the forward motion at different velocities. FYI: this is an attempt at calculating the best combination of conditions for an off-airport landing in un-inviting terrain by an aircraft experiencing complete power loss.

QUESTION: 
I fell down the stairs two years ago and am still wondering what effect the impact might have had on my brain. I fell head-first from the top to the bottom, and hit the wall where it meets the small landing at the bottom - with my head. My body kind of crumpled to my left. I fell 11 steps of normal height, with my body turning head first, without touching the wall or railing. The landing at the bottom is about 3 feet from the stairs to a plaster wall. The impact was at the top of my head. Please let me know the fall's velocity and force of impact of my head - and if you can, how my brain would have moved inside my skull after the impact.

ANSWER:  
Both of these questions are unanswerable because the force is proportional to the acceleration, that is the time rate of change of velocity. So knowing the velocity when an object hits and the fact that it is at rest afterward gives you the change in velocity but you cannot compute the rate of change of velocity without a time. So, if an object changes its speed by 500 ft/min =2.54 m/s, its mass is 100 kg, and it stops in 0.5 s, the average acceleration is 2.54/0.5=5.08 m/s2 and the average force experienced is ma=100x5.08=508 N=114 lb. Since the weight of 100 kg is about 220 lb, the force of the ground must be 114+220=334 lb. So the force you would feel is larger than your weight by 334/220=1.5, so you would feel a 1.5 g-force. This is an example, but if either of these questioners really wants an answer, more data are needed.


QUESTION: 
I don't understand Newton's Third Law. If it is true then surely, for example, it is impossible to move your hand through a table since the reaction will always equal the weight.

ANSWER:  
Newton's third law (N3) says that if one object exerts a force on another, the other exerts an equal and opposite force on the one. Many students misunderstand this law as you demonstrate in your example. Instead of talking about a hand, let us assume there is a book on the table. Are there any forces on the book? Yes, there is its own weight straight down (let's call that force W) and maybe the table, which touches the book, also exerts a force on the book (let's call that force T). Since the book is not accelerating, the total force on it must be zero (that is Newton's first law, N1) and so T must be a force straight up which is of the same magnitude as W. These forces are equal and opposite because of N3, right? WRONG, WRONG, WRONG! These forces are equal and opposite because of N1 and they have absolutely nothing to do with N3. They cannot be a N3 pair because both are on the same body (book) and N3 addresses forces on different bodies. So, what is the N3 (reaction) force which pairs with the force T? Since T is the force the table exerts on the book, N3 tells us that the book exerts a force down on the table which has the same magnitude as T. And, what is the N3 (reaction) force which pairs with the force W? Since W is the force the earth exerts on the book, N3 tells us that the book exerts a force up on the earth which has the same magnitude as W. That's right, the book exerts a force on the whole earth. N3 can never cause something not to move because the relevant forces are on different objects.


QUESTION: 
An acquaintance and I are having a heated discussion relating to the 1960 jump from 103,000 ft from a gondola by Joe Kittinger. According to several reports, Kittinger reached speeds over 600 MPH after he jumped. Since I can't prove that he did, I'm no physicist, he believes he must be correct. How can I determine the speeds that were reached in this jump?

ANSWER:  
In principle, this is a simple free fall problem. In practice, we need to worry about air resistance since that becomes important in real life at high speeds. However, there is very little air above about 60,000 feet, so let's assume that there is no air resistance and see how far he has to fall to reach a speed of 600 mi/hr and if it is less than about 40,000 feet he probably achieved that speed. The acceleration due to gravity is about 21.8 mi/hr/s; that is a freely falling object will gain about 21.8 miles/hour as each second clicks by. One pertinent physics equation is v=at where v is the speed (assuming we start from rest), a is the acceleration, and t is the time. So, putting 600 in for v and 21.8 in for a we can solve for t: t=27.5 s; in other words, after about a half a minute the object will be going 600 mi/hr. The second pertinent equation is sat2 where s is the distance traveled in time t. Solving for s I find s=12,000 ft, that is he will have a speed of 600 mi/hr when he is at about 90,000 ft, still far above where there is signficant air. (Incidentally, in the second calculation I used a=32 ft/s/s so the units would come out right, viz. feet.) So, I would say that yes, he must have gone at least 600 mi/hr. I did a little research and saw 714 mi/hr quoted as the highest speed he achieved. Once he starts encountering significant amounts of air he will begin slowing down.

FOLLOWUP QUESTION: 
This is a follow-up, and didn't know if I should post it online or not, since you've already answered it. But, the person with whom I'm having this discussion still insists you're answer is wrong. He fancies himself smarter than a nuclear-physicist, I guess, and, by his calculations , the top speed that Kittinger could have reached is 350 mph. Here is his argument and his calculations, referring to your response.

"That's just a repeat of what the other professor said, and in both cases they conveniently ignore drag. If you're going to ignore drag then ignore it and the guy keeps falling at increasing speed. Why stop accelerating at the point that corresponds to what the claims are? When you plug the drag variables into NASA's algorithm Kittinger doesn't get to 614mph. The professors don't bother to verify that the air is too thin to have any effect. Tell the college professor's to go to the Chemical Engineer's Handbook and look up Fluid and Particle Dynamics. In there is a table that describes the activity of bodies in free fall through a fluid. When they're going slow, they are stable. As they increase in speed they first start to tumble erratically, then they start spinning about their axis of least inertia. The tumbling starts somewhere around Mach 0.4 and the spinning around Mach 0.6 That's the flat spin the story eludes to, and the college professors ignore. If he fell at 614mph he's at Mach 0.9 and is in a flat spin. The big problem is all the contradictory statements that are attributed to Kittinger. He the first supersonic skydiver. He gets to an estimated 614 mph or 714. He has no sense of speed, yet he knows he keeps accelerating after the drag chute opens. That chute opens at 13 secs, or 16 secs or at 96000 feet. First of all drag doesn't work like a break. A body sitting still has no drag. As it speeds up drag increases and keeps increasing until the force propeling the object and the force of drag are equal. Then the object stops accelerating and moves at constant speed as long as force and drag stay the same. Drag is related to Velocity(speed) by the drag equation. D = 0.5 x Cp x p x A x V^2 So as long as you keep the Cp,p, & A the same, there is one value for drag for every value of speed (V) I'm not going into all the other crap, but the force pulling Kittinger down and causing him to speed up is gravitiy working on his weight. At roughly 300 lbs it takes 300 lbs of drag to stop him from accelerating. He's the shape of a brick, roughly half as wide as long and 1/3 thick as long. The Cp of a brick is 2.1. We use that to compute air flows through our tunnel kilns at the brickyard. The density of the air is about 0.00004 slugs/cuft. He is exposing about 15 sqft of area to the "wind". So if we plug in all those numbers in the Drag equation and solve for V(speed), he stops accelerating at 690 fps or 470mph. I'd say the air is thick enough to make a difference. But that doesn't take into account the drag on the small parachute he deployed to keep himself from going into that flat spin. It's 6' in diameter that's 28.26 sqft of area. The Cp for a round chute is 1.5. So as long as the drag on the chute and the drag on Kittinger doesn't total 300 lbs or more he is still accelerating. At 13 secs the total drag is 254lbs and his speed is 283 mph. He'll accelerate for less than a second more and get to about 290 mph. At 16 seconds the total drag is 384 lbs, so he'll decelerate from 347 mph with a pretty good jerk. I have no idea where or why the 96000 ft comes from, so I'm going to ignore it. He supposedly used a timer to deploy the chute. The speed of sound is around 660mph and tumbling would start somewhere around Mach 0.4 which is 0.4 x 660 = 264mph. That's damn close to the 13 sec mark. We'll never know based on the info available, but I don't think he got over 300-350mph."

Here are the facts, as presented in an article at http://www.centennialofflight.gov/essay/Explorers_Record_Setters_and_Daredevils/Kittinger/EX31.htm
1. Kittinger jumped from 102,800 ft.
2. His weight is approximately 300 lbs.
3. He was falling in a backward orientation
4. After falling for 13 secs, a small chute, 6 ft. in diameter, opened.
5. He feel for 4 min. and 36 secs more bringing him to 17,500 ft.

ANSWER:  
Your friend certainly makes some quite good points, although he is maybe a bit overemotional and maybe a little hostile to us college professors. It is true that we often simplify problems to get to the core of a problem. So let me be a little more careful and go over the calculation of your friend the way I would do it since some of the numbers he quotes are undocumented and some of them (in particular slugs/ft3) are completely incomprehensible to a modern physicist! Your friend must be an engineer. His equation is quite correct, that is the terminal velocity is given by v=[(2mg)/(rACp)]1/2 . Now it is easy to see how disputes can arise because the answer, of course, depends the choice of constants some of which are not easy to estimate (for example, I would say approximating the man as a brick is what some nutty college professor might do). The density of the air at 100,000 ft is about 1/100 the density at sea level and, since the density at sea level is about 1.3 kg/m3, I will take r=1.3x10-2 kg/m3. I searched the web for tables of drag coefficients and found that a parachutist has Cd=1-1.4 (not 2.1 as your friend assumed); I will use 1.2. The cross sectional area also requires a rough estimate since it depends on whether he is falling, for example, feet first or is falling "belly flop" orientation. I will assume the latter so as to get as small a speed as possible and I will estimate the area as about A=2 m x 0.5 m=1 m2 (your friend uses 15 ft2, about 1.4 m2). Using m=136 kg (300 lb) and g =9.8 m/s2, we are ready to estimate v; I get v=413 m/s= 924 mi/hr.

But, there is important information which you did not tell me the first time through--the opening of the small chute. So now the cross sectional area is about 2.5 m2 and the drag coefficient is about 1.42 (I model it as an open hemisphere into the wind). So now I get a reduced terminal velocity of v=240 m/s= 537 mi/hr.

Now I have the terminal velocities, what happens in our specific example? The space is too limited here to put in all the details, but I have assumed a constant density for the first 4000 m (about 12,000) of fall. I went back to an intermediate mechanics book to find the dynamic analysis of the falling body with quadratic velocity dependent drag force and I could apply (knowing the terminal velocities from above) the analysis to this specific problem. In the first 13 seconds I find that he falls about 1000 m and ends up with a speed of 130 m/s (291 mi/hr). Then, after he has fallen 3000 m more he will have a speed of 208 m/s (465 mi/hr). But he is still accelerating but now the air gets denser so his acceleration decreases even more; nevertheless, as he falls, since he is still fairly far below terminal velocity (240 m/s) he will end up going faster than 465 mi/hr.

So who is right here? Well, your friend is right in that we will never know based on information we have. I can easily imagine that I have made a factor of two error in the density, the area, or the drag coefficient (and so could your friend); increasing all by a factor of two would reduce the terminal velocity by almost 2/3 which is the difference between 600 mi/hr and 200 mi/hr. The results are too sensitive to modest changes in the parameters.


QUESTION: 
Let's say I have a metal rod about a half an inch thick and 300,000 kilometers long. Then say I give one end of said rod a mighty whack with a hammer, propelling it forward by one inch in a mere fraction of a second. My questions is, wouldn't the impact of my hammer cause the other end of the rod to move forward one inch just as rapidly as the end where I whacked it? And would this violate Einstein's law that states that nothing can move faster than "C"? Or would the far end of the rod have to wait one second after my whacking my end before moving forward by one inch?

ANSWER:  
Have you thought about the implications of your question? I figure the mass of the rod would be about 1010 kg. Suppose that you exert a constant force such that after 0.1 s it is moving with a speed of about 0.5 m/s; it would have moved about an inch in this time. The force is the change in momentum divided by the elapsed time so, roughly speaking, the required force is about 1012 N. Where are you going to get such a force? Anyhow, to the meat of your question: no, the other end would not start moving instantaneously. It could not begin moving until at least one second later than your end started moving for the reason you state: no information can travel faster than c. In reality, it would be much longer than one second because your "mighty whack" will compress the rod and this compression will move with the speed of sound in the metal and this compression is what travels to the other end to move it.


QUESTION: 
Hi I'm a 52 year old high school teacher and this is a problem I could not solve in the new curriculum. Here it is, word for word:

A red ball is stationary on a billiard table OABC. It is then struck by a white ball of equal mass and equal radius with velocity u( -2i + 11j ) where i and j are unit vectors along OA and OC respectively. After impact the red and white balls have velocities parallel to vectors -3i + 4j, 2i + 4j respectively. Prove that the coefficient of restitution between the two balls is 1/2.

ANSWER:  
First, allow me a little rant! It is utterly ridiculous that this problem is part of a high school curriculum. Coefficient of restitution (COR) is one of the least important concepts in classical mechanics. Furthermore, it is nearly always defined in terms of a one-dimensional collision which the collision in your problem is not, so it looks like the problem writer is trying to confuse the reader (which I consider to be poor educational method). Furthermore, I find that I do not get 1/2 for the coefficient of restitution when I work the problem. I will outline the solution to the problem and give my results. You can reconstruct the solution and see if I have made any errors. First, the COR e is related to the energy loss Q in the collision by Qmv2(1-e2) where m=m1m2/(m1+m2) (reduced mass) and v is the incident velocity (if one of the two particles is at rest as it is here). It makes no difference what the actual masses are since they are equal, so I shall choose m1=m2=1 kg such that Q=31.25(1-e2) J. (I have used v2=125 m2/s2 as given in the problem.) Now, just calculate Q to get e. The information given about the recoiling velocities is their directions, not their magnitudes; to get the speeds you must do momentum conservation. The red ball moves at an angle of 53.10 above the negative x axis and the white ball moves at an angle of 63.40 above the positive x axis. Conserving momentum in x and y directions I now find the speeds of the red and white balls: vr=7.5 m/s and vw=5.59 m/s. Hence the energy after the collision is 43.75 J and before the collision 62.5 J, so Q=18.75 J. Solving now for COR: 18.75=31.25(1-e2), e=0.63, not ½.

There is actually another way you can do it: if you work in the center of mass system it essentially looks like a one-dimensional collision since the two particles after the collision move colinearly apart with speeds of 3.54 m/s each so that the speed of separation is 7.08 m/s and the speed of approach before the collision is 11.18 m/s. The COR is defined as the ratio of the speed of separation over the speed of approach which works out to, you guessed it, e=0.63! Now I have more confidence in my solution.


QUESTION: 
Real life question: Tire pressure on and off the car: I am getting a new tire for my car. While it's on the rack, they check the pressure and it's a perfect 32 psi. They put the tire on my car, then lower the 3000 pound car back down on it, and say see ya' later. I say,"Shouldn't you check the pressure while the weight of the 3000 pound car is down on it?" "Nah", they say, it doesn't change. That doesn't make sense to me. I actually asked this question to my brother who is a ultra-high vacuum physicist at Sandia Labs, and he didn't know. I also asked this question at the famous Cartalk.com forum and got laughed out of it.

ANSWER:  
Here is the basic physics, the ideal gas law: PV=NRT where P is pressure, V is volume, T is absolute temperature, N is the amount of gas, and R is a constant of nature. Let's assume that T stays the same when the car is lowered off the rack. Now, presumably the volume of the gas in the tire decreases a little bit; therefore, the pressure must increase a little bit to keep the product PV equal to the constant NRT. However, the volume changes by a very small amount compared to the total volume of the tire, so for all intents and purposes (but not exactly) "it doesn't change".


QUESTION: 
I am a librarian assisting a library patron. The patron says at one time he had a book that gave him a formula to compute the weight of an object. If you put an object, such as a car, on a tire or ball or something that is pressurized, and you know the PSI, you can measure the size of the point of contact with the ground (the flat surface of the tire on the ground) and calculate the weight of the object.

ANSWER:  
Consider a piston of cross sectional area A, vertical, which has a pressure P under it and a weight W sitting on it and everything is in equilbrium; for simplicity, neglect the weight of the piston itself or imagine it to have been absorbed into W. We must not forget that there is an atmospheric pressure Pa pushing down on the cylinder. Then Newton's first law specifies that the sum of all the forces must add to zero, and so PA-W-PaA=0 (pressure time area equals force) so W=(P-Pa)A. But (P-Pa) is what is called the guage pressure, it is the pressure which most pressure guages read, the amount over (or under) atmospheric pressure. So 30 psi means, usually, 30+14.7 psi since Pa=14.7 psi. This seems to me to be equivalent to your question. Let's check it for reasonableness: suppose a car has each of its four tires in contact with the ground by an area of 6"x4" and the tire (guage) pressure is 30 psi. Then the weight of that car would be 6x4x4x30=2880 lb which is about what cars weigh.


QUESTION: 
A metal weighing 10kg and wood also weighing 10kg are both dropped from a building which would hit the ground first? If they are of the same shape,which would hit the ground first?

ANSWER:
If air drag is negligible, they would hit the ground simultaneously. If air drag is a factor, then the one with the larger density will hit first. A good approximation to air drag force is F=
¼Av2 where A is the cross sectional area presented to the wind and v is the speed; this equation is only true if SI units are used. The one having the larger density will have the smaller volume and therefore the smaller A; hence, at any given v the denser will have the smaller air drag force. The question, then, is how high does the building have to be in order for air drag to be important. I assume the two are spheres, the density of the metal is 8000 kg/m3 (near iron), and the density of the wood is 800 kg/m3 (near oak). You can find the terminal velocity of each by finding the speed at which the drag force equals the weight. Without any details, I find the terminal velocities of the metal and wood balls are 167 and 78 m/s respectively. The plot at the right compares the velocities of the oak and iron balls after they have fallen a given distance and also shows how each would move if there were no air drag. As you can see, even dropping from 100 m neither ball is anywhere near terminal velocity and neither velocity is much reduced from free fall. If you drop the balls from less than about 30 m, they will, for all intents and purposes, hit simultaneously.


QUESTION: 
There are numerous problems connected with air drag acting on the falling object. What if it acts on the object thrown up with a given initial speed v(0) and has a given terminal velocity of v(term)? What height would an object rise to?

ANSWER:
This is a standard problem, and if the object is such that we can approximate a drag force of the form F=-cv2, the solution is h=(vt2/(2g))ln(1+(v0/vt)2). The graph to the right shows heights achieved for various initial velocities if the terminal velocity is 10 m/s compared to the case of no air drag.


QUESTION: 
Here is the scenario: a pool ball is traveling at certain pace with an enormous amount of spin on it. Once it hits the rail at about a 45 degree angle is it possible for it to shoot off of the rail and travel faster than it was going before when it was on it's way to the rail?

ANSWER:
It does not even need to have "an enormous amount of spin", a modest amount will probably do it. The ball, coming into the rail, has translational kinetic energy and rotational kinetic energy. The friction between the ball and the rail can transfer some of the rotational energy to kinetic energy, so after the collision it will have a higher speed and a smaller spin.


QUESTION: 
Why do I go faster than my lighter weight friends when on a bike? I have noticed this downhill, but also on flats, I'm pedaling less. Assuming we both have top of the line bikes (we do). I thought things fell at the same rate regardless of mass. I think momentum and friction have something to do with this, but can you remind me of the principles involved here?

ANSWER:
If it were not for air drag, everybody should be pretty much equivalent. However, the air drag F depends mainly on the cross sectional area A presented to the wind and the speed v; a useful approximation I often use in these answers is F
≈¼Av2. (This works only if you use SI units.) So, assuming that you are not a whole lot fatter than your friends, you all experience the same amount of drag force. But, wait a minute, the acceleration is given by Newton's second law, a=F/m so, if you have a mass bigger than somebody else, your acceleration will be smaller, that is you will slow down more slowly. What it boils down to is that you have more inertia.


QUESTION: 
I have just started a new job and have been asked to do a presentation on the ABS (Avalanche Bag system). Upon reading about the product they kept quoting that the system works on the "Law of Inverse Particles." I figure ok, it's a Law, it should be fairly simple to research, but when I started typing it into search engines I am getting no feedback on this Law at all. So my question is does this Law even exits? If so is there maybe another name that it goes by? If it does exist who created it? Any help that you can off me on this topic will be very much appreciated, in the mean time I'll keep on checking books, sites and asking old teachers!

ANSWER:
I have never heard of the law of inverse particles. However, just looking over how this thing is supposed to work, it just looks like Archimedes' Principle to me. Archimedes' principle says that when an object is placed in a fluid it experiences an upward force, called the buoyant force, which is equal to the magnitude of the weight of the displaced fluid. Hence, if the object has a density less than the density of the fluid, it will float, if it is greater it will sink. I am guessing that the snow in an avalanche acts much like a fluid and, if you can significantly decrease your overall density, you will "float" on the surface, much like a cork floats on water. The way you decrease your density is to increase your volume with the inflated airbags. Why they invent this screwy name for this escapes me, must be an effective marketing technique.


QUESTION: 
As a gas bubble rises in a column of liquid, it expands due to the decrease in pressure from the liquid above it. Intuition tells me that the diameter of the bubble will increase at a decreasing rate; but is that true? Does the bubble's diameter increase at a decreasing rate over time, does it increase linearly with time, or does it expand at an increasing rate?

ANSWER:
Assuming temperature stays constant, volume is inversely proportional to the pressure. Since the pressure increases linearly with depth, the volume of the bubble will decrease linearly with depth. But the volume is proportional to the cube of the radius, so the radius does not decrease linearly with the depth. I have derived the following equation to approximate how the radius R varies with the depth d: R/R0=3
√[PA/(PA+ρgd)]≈3√[1/(1+0.1d)] where R0 is the radius of the bubble at the surface, ρ is the density of water, and PA is atmospheric pressure. The graph at the right shows the variation to a depth of 25 m.


QUESTION: 
I need you to settle an argument between my wife and I. We are traveling down the road at 60 mph. My wife throws a piece of balled up gum out the window. When it hits the ground, my wife thinks it will roll backwards. This is not possible and I know it will roll forward at a speed less than 60 mph due to friction. Who is right?

ANSWER:
You are
…unless, of course, it bounces back from some obstruction like a big stone in the road or something.


QUESTION: 
Since kinetic energy = 1/2 mv^2, it follows that, for a given mass, a doubling of velocity requires a quadrupling of energy. It seems to be certain then, that 2 joules are required to accelerate a 1 kg mass from zero to one meter per second in one second (ie accel=1m/s2)

Change in kinetic energy is equal to work. W1=½x1x12-½x1x02=½ J, not 2 J.

Now, according to the KE formula, to further accelerate the mass to 2 meters per second requires an additional 6 joules... because at that velocity, the energy now possessed by the mass is 8 joules - 4 times higher than when it was moving at 1m/s.

W2=½x1x22-½x1x12=3/2 J, not 6 J.

Here is my question: (in the form of a thought experiment) Scenario: I am in space Let's suppose I use 2 joules of energy to accelerate the mass to 1m/sec, relative to me. A spaceship, passing by with velocity 1m/sec and going in the same direction as the mass, has a second person onboard who intercepts that mass. (Because it's and the ship's speed are equal, the mass is effectively at rest with respect to the ship. Now let that person - on the moving train - apply 2 joules of energy to again accelerate the mass to 1 m/sec relative to HIM. So the actual velocity of the mass is now 2m/sec, relative to the first observer (me) (Assume same direction...) So, between us, only 4 joules of energy have been required to get the mass to 2 m/sec instead of the 8 implied by the KE equation. I have no difficulty in grasping that energy goes up as the square of velocity, but for the life of me, I cannot understand why both observers cannot use 2 joules of energy to effect an increase of 1 m/sec each Surely, what is true for one must be true for the other... ? (ie. that it only requires 2 joules to effect an increase in velocity of 1m/sec, relative to that frame) I have posted this on a Physics Forum and the only responses I have gotten allude to the fact I am getting it all messed up by changing the frame of reference and hence, the numbers won't add up. They say KE is not frame invariant and I can understand that - but no-one can explain where my logic is wrong... Can you possibly enlighten me ?

ANSWER:
You have made some errors in your preliminary calculations above which I have corrected. Those who told you that kinetic energy is different in different frames are certainly on the mark. But, I will show that the work done in one frame is also not the same, that is, if you do W Joules of work, somebody moving relative to you will see you do a different amount of work, that is, the energy expended is not something everybody will agree on. I am going to set up a scenario which I hope will show you where you have gone wrong. The ship goes by with a speed of 1 m/s and the guy on the ship exerts a 2 N force on a 1 kg mass, initially at rest in the ship, for 1 s. The acceleration is therefore F/m=2 m/s2. The distance (in the ship) the mass travels is x=
½at2=½x2x12=1 m and the speed it ends up with is v=at=2x1=2 m/s. The work done is therefore W=Fx=2x1=2 J. Now, view this whole thing from your perspective. You still see a force of 2 N being applied, the mass at the start has a velocity of v0'=1 m/s, the acceleration is still a'=F/m=2 m/s, and the time the force acts is still 1 s. So now, the distance (in your frame) that the force acts over is x'=v0't+½a't2=1x1+½x2x12=2 m and the speed at the end is v'=v0'+a't=1+2x1=3 m/s. The work done is therefore W'=Fx'=2x2=4 J. Note that for both observers, the work done equals the change in kinetic energy: W=2 J=½x1x22-½x1x02=2 J and W'=4 J=½x1x32-½x1x12=4 J. You were willing to accept that kinetic energy was different in different frames but failing to see that energy expended is different in different frames.


QUESTION: 
I noticed in a textbook that 'g' varies directly with distance inside the Earth's radius but is inversely proportional to it outside. I understand the reason for the latter but would like to know more about the former - thanks! Please keep it simple.

ANSWER:
Here is as simple as I can make it: First, the assumption must made that the mass of the earth is uniformly distributed, that is, the density is constant everywhere. This is roughly true on a large scale. Second, you must know that Newton's universal law of gravitation says that the force of gravity from a spherical object is proportional to the mass M of the object you are being attracted to and inversely proportional to r2 where r is the distance to the center of the sphere. [You erred when you said g is inversely proportional to r, it is r2.] Third, there is one more thing you need to know, called Gauss's law which states that only the mass inside where you are attracts you, the mass outside of r exerts no gravitational force on you. So, as long as you are outside, F~Mtotal/r2; but, if you are inside, only part of Mtotal, Mpartial, exerts the force on you, F~Mpartial/r2. But, think about it for a minute: Mpartial is proportional to the volume of the smaller interior sphere and that is proportional to r3 (V=(4/3)
πr3), so F~r3/r2~r. [~ means proportional to here.]


QUESTION: 
When you face a wall, place your toes against it, and try to rise up on the balls of your feet, why can't you?

ANSWER:
When you stand up straight, your center of gravity is above a point between your heel and toes and you do not tip over because there is no net torque which would tip you over. When you go up on your toes and are unable to lean forward, your center of gravity now exerts a torque about your toes which rotates your body backward. Try standing on your toes when you are not against the wall; in order to remain standing you will find that you must lean forward so that your center of gravity will be directly over your toes.


QUESTION: 
I was doing some water sports and was wearing a wet suit and a 50N (50 Newtowns) rated buoyancy aid. I weigh 93Kg. Cliff jumping into the sea, after resurfacing I bobbed gently on the surface with no swimming effort. Later wearing same equipment on fresh water river, fell of a kayak and promptly sank! To get to and stay on surface I had to swim or actively tread water. The question is what value N rated buoyancy would I need to "bob gently" in fresh water?

ANSWER:
You do not weigh 93 kg, your mass is 93 kg and your weight is 93x9.8=911 N. Suppose that you just float with your head above water in salt water. Then the buoyant force the water is exerting up is 911-50=861 N. Archimedes' principle says that this is the weight of the displaced water, and let's say salt water has a density of 1030 kg/m3; so, the volume of your body less head must be 861/(1030x9.8)=0.085 m3. So, the buoyancy you can expect in fresh water if just your head is up is 0.085x1000x9.8=836 N because the density of fresh water is 1000 kg/m3. So, to have identical buoyancy to the salt water situation, you need an additional 25 N (861-836) buoyancy.


QUESTION: 
If a man was to be proportionately increased in size, would it be more difficult for him to perform a pull-up? All his body percents are exactly the same including his muscle, fat, ability, ect. The pull-up bar is also high enough that his feet would not touch the ground in either situation.

ANSWER:
Scaling of biological organisms has been a topic of interest for a long time. I think Leonardo de Vinci was the first to think about it. Anyhow, I think the answer to your question is that it would be harder to do a pullup. Here is why: The weight of something is proportional to the cube of its dimensions. So, if you doubled all the dimensions of a man, his weight would increase eightfold. But, the strength of bones or muscles is dependent on their cross sectional areas, a bone or muscle doubled in size would have about four times the strength.


QUESTION
I've been told that when a gas is cooled, it takes up less volume, and therefore its density increases. I've also been told that a gas always fills its volume (i.e. a container). These two ideas, however, seem to contradict one another. On the one hand, cooling a gas causes it to contract and decrease in volume, which would increase its density. However, on the other hand, a gas will always fill its volume, which means its volume doesn't change when cooled. Therefore, if the volume doesn't change when cooled since a gas will always fill its volume, then its density would not increase. In other words, if a gas will always fill its volume, then cooling it would not increase its density since the gas always fills its volume and therefore remains at the same volume. In a closed container, if a gas expands to fill the volume of the container, then how could cooling the gas increase its density? How could a gas fill its container/volume on the one hand, but then on the other hand, when its cooled, decrease in volume and therefore increase in density? If a gas fills its container/volume, it wouldn't increase in density when cooled since it FILLS its volume. Therefore, since it fills its volume, the volume of the gas would remain constant. Therefore, since the volume of the gas would remain constant, its density would not increase when cooled. So my question is, how could a gas increase in density when cooled if that same gas always fills the volume of the container it's in?

ANSWER:
T
hat's a pretty long question which requires only a short answer. If the amount of gas remains constant, the relation between pressure, volume, and temperature is PV/T=constant. The situation you describe has a constraint that volume remains a constant, so P/T=constant which means that as the temperature goes down the pressure goes down; and, the volume stays the same. So, if you cool a gas in a rigid container, the density remains the same.


QUESTION: 
how much time will it take to travel down a 1000 ft , 30 degree ilcline with a 225 lb payload on a 200 lb soap box race car and if pay load was lighter would the time be shorter with lighter load?

ANSWER:
I have dealt with this kind of question before, but maybe it is time to revisit the whole thing with one answer. First, consider the ideal situation where there is no friction of any kind. Physicists do not like English units, so I am going to convert everything to SI units: 1000 ft=305 m, 425 lb=193 kg. I assume that you do not want all the details of my calculations, just the pertinent results. The time does not depend at all on what the mass is (if friction plays no role). There is an acceleration down the incline which is a=
½g. I find that the time to the bottom is about 11.2 s and the speed at the bottom is about 54.9 m/s=123 mph. I assume you are not crazy enough to be in a soap box car going that speed, so friction must play a role. There are two kinds of friction you have to consider:

  • Friction due to the moving parts, like bearings, like the wheels rolling on the ground, like wheels moving on axels, etc. Empirically, we find that this kind of friction increases proportionally with how hard the moving parts are pressed together and this, of course, is proportional to the total weight of car plus rider. But, the force impelling the car down the hill is also proportional to the total weight, so once again, the speed at the bottom is independent of the load. There is something called the coefficient of friction which tells you how much friction force there is for a given force pressing the surfaces together. For example, if this coefficient is 0.3 for a 100 lb box moving on a horizontal floor, you would have to push with a 30 lb force to keep it moving with a constant speed. If I take 0.3 to be your coefficient of friction, your speed at the bottom would be about 26.8 m/s=60 mph. Since the acceleration is about half what it was, the time is about twice as long, 22 s.

  • For objects which are going faster than a few mph, air drag becomes important. Air friction depends on two things, the shape of the object and the speed it is going. It does not depend on the mass of the object. Actually a pretty good approximation to the magnitude of the force of air friction is F≈¼Av2 where A is the area presented to the wind (this works only if F, v, and A are in SI units). So, this force, which points up the incline just like the moving parts friction, causes the car to slow down more. But, Newton's second law says that a=F/m so that if F does not depend on m, the acceleration (amount of slowing down due to this force) gets smaller as m gets bigger. It gets a little complicated to actually calculate the time and speed for this case, but the important part is that this is the only place I can find where the weight of the whole car makes a significant difference.

So, the bottom line is that if you go fast enough for air drag to be important (and I suspect you do), the heavier of two otherwise identical cars should win.


QUESTION:
Just like a ball when hits a wall it bounces back, why dont people bounce back in the same way, when opposite force is exerted on them by the wall (as in case of ball)??

ANSWER:
It is because of elasticity. When two objects collide, how they move after the collision is determined by what happens to energy after the collision. If you drop a superball on the floor it rebounds almost as high as from where it was dropped; if you drop a putty ball, it does not. When the superball hits the floor, it compresses but it compresses like a spring which "stores" the energy. So the energy of the moving ball is moved into the energy of the spring. The spring now decompresses and gives the energy back to the motion of the ball and it flies back into the air. It obviously did not give all the ball's energy back since the ball did not quite rebound to where it was dropped from; the little bit of energy lost shows up as heat and sound. The putty ball also compresses when it hits but it bears no resemblance to a spring because after you compress it (which takes energy), it does not decompress and give energy back. Here, the energy of the ball's motion is used to compress the putty ball and that energy shows up as heat and sound entirely. The superball (neglecting the little amount of energy lost) is referred to as an elastic collision and the putty ball is a perfectly inelastic collision.


QUESTION:
I am falling from a plane at terminal velocity. I have a ball in my hand. I throw it towards the ground. What happens? Does the ball accelerate away from me, or does it simply fall beside me? What about if it was a very heavy ball such as a cannon ball compared say to a baseball or a plastic 'WalMart' pool ball?

ANSWER:
Terminal velocity is determined by the geometry of the object, the density of the fluid (air), and the mass of the object. The terminal velocity in air for an object of mass m and cross sectional area A can be roughly approximated by vt
≈√(4mg/A) (only in SI units). You and the ball have some terminal velocity together. When you depart from the ball you both have different terminal velocities since the masses and geometries have changed. Suppose it was a cannon ball. Then your net mass gets smaller and your geometry does not change much, so you will have a lower terminal velocity and slow down; the ball has a considerably smaller cross sectional area than it did when it was "part of" you and therefore a larger terminal velocity, so it will accelerate down until it reaches its new terminal velocity. Suppose it was a styrofoam ball. Then neither your net mass nor your geometry change much, so you will have about the same terminal velocity and continue falling at about the same rate; the ball has a considerably smaller mass than it did and therefore a smaller terminal velocity, so it will experience an upward acceleration and slow down until it reaches its new terminal velocity, you will overtake it and it will appear to go up (but is actually just going down more slowly).


QUESTION:
My question has to do with traction and the movement of a wheel(a wheel alone). Traction is essential for its movement both linear and circular. But if we throw a wheel forward it rolls some meters and then it stops(and falls). Which force is responsible for the decrease in its velocity? Cause if traction is parallel to the ground facing backwards then linear movement 's negative accelleration is explained but not angular negative accelleration. If traction is parallel to the ground facing forward then angular negative accelleration is explained but not linear. If traction is zero then which force decreases both velocities linear and angular?

ANSWER:
One of the reasons I love doing Ask the Physicist is because I often learn things I did not know or had never thought about. You would think that a guy who has been teaching introductory physics courses for nearly 50 years would find this question simple. But, indeed I was puzzled by it because, as I have found by thinking about it and talking to some friends, I wasn't thinking beyond the friction force (which questioner calls traction) being simply the only force in the horizontal direction and obviously stopping the forward motion after some distance. I never addressed the angular acceleration of the wheel before. This answer will be long-winded because that is what I do when I have learned something which pleases me! What frictional forces are important to understand the rolling of a wheel? Most introductory physics classes talk only about the contact forces of static friction and kinetic friction. Kinetic friction is not applicable to this problem because the wheel is not slipping on the ground, and static friction might be important, but not necessarily. If we have a round wheel rolling on a flat horizontal surface (don't look at the figure yet!), there are three possible forces
—the weight which must be vertical, pass through the center of mass, and (assuming it is a uniform wheel) pass through the point of contact; the friction, which must be parallel to the surface and pass through the point of contact; and the normal force which must be perpendicular to the surface and pass through the point of contact. If you now sum torques about the point of contact (as noted by the questioner), there are none! So, there can be no angular acceleration; if we have stipulated that the wheel does not slip, then there can be no linear acceleration either and the wheel will roll forever and no friction is required. But we all know better! A real wheel will eventually slow down. The key is that there is no such thing as a perfectly round wheel or a perfectly flat surface, one or both must be deformed. In that case, we have to think about a new kind of friction called rolling friction, the friction the wheel has because of the rolling. This is different from the static friction, and static friction may still be present still to keep the wheel from slipping. A perfectly round wheel cannot have rolling friction as I showed above, it must deform which means that there is no longer a "point (or line) of contact" but now an area of contact. Since the normal force is only constrained to act somewhere where the two are in contact, it is now possible (in fact inevitable) that this force will not act through the center of mass of the wheel. That is the whole key to answering this question. So, finally, the answer: refer to the figure where I have drawn the forces mg, N, and f. The weight is still constrained to be vertically down and pass through the center of mass (blue cross). The normal force is constrained to be vertical and act somewhere where the wheel and ground are in contact, drawn a distance d to the left. The frictional force (which now includes both static and rolling friction) is constrained to act at the surface and parallel to it. I choose a coordinate system with x to the left and y up; the axis (red cross) about which I will sum torques is at the ground directly under the center of mass and positive torque results in an angular acceleration which is positive when acceleration of the center of mass is positive (counterclockwise around the axis). All is now straightforward: ΣFx=-f=ma, ΣFy=N-mg=0, Στx=-Nd=Iα=Ia/L where I is the moment of inertia about x and L is the distance from x to x. Finally, N=mg, a=-f/m, and d=fI/(Lm2g).

Finally, a couple of real-world provisos. Of course, N is really distributed over the whole area, but the dynamics can be done by assuming it effectively acts all at one point just like we assume the weight acts all at the center of mass. And, the rolling friction might not really act at the surface of contact since it arises from the deformation of the wheel and it might not be purely horizontal since it is not directly a force due to the contact with the floor. So, there are still some idealizations in my analysis, but there are always idealizations when dealing with friction. And, the problem could have been equally well done assuming the ground, not the wheel was being deformed. One could also have done the analysis by summing torques about the center of mass and using the parallel axis theorem Icm=I-mL2.

I would like to acknowledge a very useful discussion over pizza with friends and colleagues Edwards, Love, Meltzer, and Anderson.


QUESTION:
A fly is hovering in a car. the car is going 45 MPH with the windows up. the car hits a solid object and suddenly stops. Does the fly hit the front window, the back window, or nothing at all?

ANSWER:
This is tricky. The stock answer would be that the fly would smash into the front windshield just like you would if you weren't wearing your seat belt. But, the fly is actually hovering with respect to the air, and so whatever happens to the air will be what happens to the fly. The air does not all smash into the windshield but more or less stays just where it is relative to the car. The reason for this is that air is a collection of molecules most of which are already going much faster relative to the car than the car is relative to the ground. An average air molecule is going about 1200 mph. Think of half of them going (relative to the road) toward the back with speed like 1155 mph and half going toward the front with speed like 1245 mph, so their average is still 1200. When the car stops, the fly sees no significant change and just continues hovering where he is. He has such a small mass that he does not have enough inertia to overcome the air drag he would be experiencing and hit the windshield.


QUESTION:
How many Newtons are exerted when a 300 lb. man falls 3 ft.?

ANSWER:
Read faq page. This question has no answer.

QUESTION:
We are working to produce a safety harness and the strap material we are using has a maximum Newton rating - we were trying to get an idea of what Newton rating would be needed to support a 300 lb. man if he fell 3 ft. Being hunters (tree stand safety harness) - perhaps we are wording the question incorrectly. Can you clarify your response?

ANSWER:
What matters is how long it takes the falling guy to stop. The mass of a 300 lb guy is about 130 kg, the acceleration of gravity is 10 m/s2, and so the weight of the guy is about 1300 N. You need that strong a strap just to hang him there at rest. If he falls 3 ft (about 1 m) he will be going about 4.5 m/s. So, let's call F the force needed to stop him and t the time it takes him to stop; I reckon that F
≈130(10+(4.5/t)). For example, if he takes ¼ s to stop, F≈3600 N to stop him. The straps are probably pretty unstretchy, so your best bet would be to make the harness out of a stretchy material because, don't forget, the bigger F is the more it is going to hurt during the stop.


QUESTION:
what is the upward force in pounds on the bottom of an empty basin caused by a groundwater depth of 8 feet above the tank bottom? The basin has a base of 300 square feet. What is the formula? This is in regards to not emptying a tank fully for repair as the upward force may displace it from its resting place.

ANSWER:
The key here is to find the pressure at a depth d (8 ft in your case) below the surface of a liquid of density
ρ. This is given by P=ρgd where g is the acceleration due to gravity; when you know the pressure, you can find the force F on the bottom by multiplying the pressure times the area A (300 ft2 in your case), F=PA=ρgdA . So, there is your formula. You are probably wanting to do this in English units, but physicists hate to work in English units. I did it in SI units where ρ=103 kg/m3, g=9.8 m/s2, d=8 ft=2.44 m; so I found P=2.39x104 N/m2 and converted to P=499 lb/ft2. So, F=499x300=149,700 lb. Actually, I found that it is pretty easy to do this in English units: the weight density of water is ρg=62.4 lb/ft3 and so F=62.4x8x300=149,760 lb.


QUESTION: 
Please can you explain what happens to the energy released by the shattering of a glass on a hard surface? We are told that the energy on Earth has remained constant since the formation of the planet so what is the fate of the energy produced by this event?

ANSWER:  
What makes you think energy is released? Why does a piece of glass not just spontaneously break? The fact is, you must put energy into the glass to make it break. If you drop it, it has kinetic energy when it hits and then the surface does work on it by exerting forces on it. So the question should be what happened to the energy which got put into the glass to break it. It takes work (energy) to break molecular bonds which were holding the glass together before it broke; there goes some of the input energy. It makes a big crash; there goes some more of the energy (sound). It will heat up a little bit; there goes some more of the energy.


QUESTION:  
I am revising some poorly-written curriculum materials for my Integrated Technology class, and have stumbled upon this paragraph: "Energy can not be seen since, unlike matter, it has no mass. However, it can be measured, sensed, or otherwise detected. For example, we cannot actually see the energy of a light bulb traveling through air, but we can see its illumination reflected from this laboratory manual or the walls of this room." While there is some correct information here, the statement that energy can't be seen strikes me as wrong. It seems to me the issues are: 1) We can only see energy in the visible portion of the spectrum and 2) it must be directed toward our eye for us to see it. The fact that light is not matter doesn't, in fact, matter. In fact, I would say that we can't see matter without (visible light) energy. Is my assessment correct (the explanation, as written, is wrong or at least misleading), or can you illuminate me on this subject?

 ANSWER:
It is most unsettling to a physicist to read garbage like this! Energy is, simply stated, the ability to do work. It comes in all forms, not just electromagnetic (light). A moving car has energy due to its motion (kinetic). A falling skydiver has energy by virtue of his altitude (potential). A cup of hot soup has energy by virtue of its temperature (heat). A bottle of gasoline has chemical energy stored which can, as you know, be exploited. To speak about "seeing" energy is ludicrous; can you see a moving car, a falling skydiver, a cup of soup, a beam of light? It is also inaccurate to say that energy does not have mass; as we know from the theory of relativity, mass is energy—you know, E=mc2!


QUESTION:  
A sky-diver is falling toward earth. A tunnel has been previously excavated completly through the earth at exactly the location of the skydiver's landing. He continues his dive through the tunnel without touching the sides of the tunnel. I believe that Newton would have had him stop at the earth's core. Where would Einstein have him stop?

 ANSWER:
First, since this is clearly an idealized problem, let us neglect air friction (which is, of course, not negligible because the skydiver has a terminal speed before he hits the ground). Until he enters the tunnel he is accelerating with a constant acceleration down. When he enters the tunnel, he experiences less and less force as he goes deeper because there is less and less of the earth pulling on him (all of the earth outside him exerts no force) until finally at the center he has zero force on him but he has his highest velocity of the whole trip since he has been speeding up the whole time. Now as he moves away from the center he slows down. When he reemerges at the other end of the tunnel he has exactly the same speed as he had when he entered it. He continues until he reaches the altitude from which he originally jumped at which point he turns around and begins the process all over again. Newton an Einstein would both agree on this. If air friction were included, he would not go as far and if he happened to stop at the center of the earth, he would stay there forever. If the air friction were included the general solution to the problem would be that he would oscillate back and forth going less far each time until he finally stopped in the center. There is an interesting aspect of this problem: when the skydiver is inside the tunnel, he moves exactly like he were a mass on an ideal spring.


QUESTION:  
if their is a truck full of birds and the birds are sitting at the bottom of a trailer ,does it weigh the same as if all the birds are flying?

 ANSWER:
There is more than one answer to this question. Let us assume that the birds are hovering or moving with constant velocities. In that case, each bird stays in flight because the air exerts a force up on him equal to the bird's weight; but Newton's third law requires that the bird therefore exerts an equal downward force on the air. Therefore the net weight of the whole truck is unchanged. All forces internal to a system cancel out in the calculation of the net force on that system because of Newton's third law. Another possibility would be if the birds have an acceleration with a vertical component; the simplest example is that all the birds are in freefall inside (probably not what you had in mind by "flying") in which case the birds would not contribute to the weight (neglecting any air friction or buoyancy). A similar question was answered earlier which might interest you.


QUESTION:  
If I shoot a modern arrow straight up into the air with a bow that exerts 75 lbs of thrust does the arrow travel to the top of it's flight in the same amount of time that it takes to fall back to earth?

 ANSWER:
A freshman physics class would say yes but that is if you neglect air friction. On the way up the arrow has two forces on it: its weight down and the air friction also down. On the way down it also has two forces on it: its weight down and the air friction up. The problem is complicated somewhat by the fact that the air friction force depends on velocity, so it is not possible, without stipulations on the initial speed and the details of the air friction to determine which will be quicker, up or down, but since the forces on them are definitely different if there is any air friction at all, the times will definitely differ.


QUESTION:  
Is it possible for a helicopter to hover above the Earth and have the Earth move under as it rotates so that the helicopter would "travel" around the Earth in 24 hours?

 ANSWER:
Inasmuch as the earth drags its atmosphere with it, the helicopter would have to travel at a speed much larger than helicopters go relative to the air to remain in the same position in space as you stipulate. So, no it is not possible.


QUESTION:  
Does a hockey puck reach it's maximum velocity at the last instant of contact with the hockey stick, or is there such a thing as (for lack of the proper terminology) 'residual acceleration' I have considered this at great length.. >From what I know in physics if there is no longer force being applied to the puck there should not be any accelertion after the contact.... Almost a logical deduction.. however seeing replays and it's almost an instinct telling me that I might be wrong.. It doesn't feel like the stick is travelling at a the same velocity as the puck is.. 100mph shots and all.. Perception maybe different than reality.. anyway, if I am wrong and there the puck continues to accelerate for a short distance after leaving the stick,. it's because of the flex in the stick, the energy transfered into the puck has not fully caused it to accelerate.. possibly like the flex in a golf ball..

 ANSWER:
A short answer to your long question: The puck cannot continue to accelerate (gain speed) after it leaves the stick. In the physics sense, it does accelerate but with a negative acceleration, that is it slows down.


QUESTION:  
My question regards a discrepancy between answers I get when I work a problem using two different approaches. This problem was given to me several years ago, and although I got the right answer then, I was told that I could have also used a different approach. I have tried several times since to do this problem the other way, but have had no luck. The problem is this: if a pole of length L is stood on end and released, if the initial velocity is neglegible, what is the final velocity of a point at the end of the pole when it hits the ground. I believe I originally solved this problem using the concept of the potential energy being converted to kinetic energy (mgL=(1/2)mv2), getting an answer of v=(2(g)(L))^(1/2). I am now trying to solve it using calculus by integrating the equation v2=2ad over all da from 0 to 9.81m/s2 (where d is the arc length of the falling pole), and taking the square root of both sides. When I do this I end up with an answer of v=(( PI)(g)(L))^(1/2). What am I doing wrong?

 ANSWER: 
Neither of your methods is correct. The second method, using v2=2ad is wrong because this equation is valid only if the point has uniform acceleration which it does not. The first method  is incorrect because you treat the stick as a point mass at the end of the stick falling straight down which is not what is happening. The only way to solve this problem in closed form is to use energy conservation as you have tried to do; however, you must use rotational dynamics to do this. So the potential energy at the beginning is mgL/2 because the center of mass of the body is located halfway up the stick; the kinetic energy when the stick hits the ground is I
w2/2 where I is the moment of inertia of a the stick about its end (which is ML2/3 for a uniform stick of mass M and length L) and w is the angular velocity of the stick (which is v/L if v is the speed of the point at the end). If you now equate the energies before and after and do the algebra, you will find v=(3gL)1/2.


QUESTION:  
I am teaching physics at a Kansas high school, and my students and I were debating over a problem. The problem is as follows: Two 20-N boxes are placed on both ends of an ideal rope. The rope is then passed over two pulleys fixed on each of the two edges of a fixed table, so that one 20-N box hangs on one pulley, and the other 20-N box hangs on the other pulley (the rope being continuous over the two pulleys). We have been debating over the value of the tension in the rope, whether it's 20 N or 40 N. Half of the class agrees with me that the tension is going to be 20 N, but I cannot convince the rest of the class that the tension is not 40 N, but rather, 20 N. (Or is it me who's wrong?)

 ANSWER: 
Breathe easy--you are right! The tension in this rope is 20 N. The tension at a point in a rope is defined as the force which it exerts on what it is attached to. So you need to focus on one box or the other. You now say, what are all the forces on that box? One is its own weight which we know to be 20 N vertically down. Another is the force which the rope exerts on it; technically we know neither the magnitude nor direction of this force, but it is called the tension in the rope at the point of attachment. There are no other forces on the box. Since it is in equlibrium, all (both) forces must add to zero, so the tension is 20 N straight up. And, for a rope of negligible mass the tension must be the same everywhere on the rope; otherwise if you inserted a tiny mass in the rope and the tension were different on one side than the other, the tiny mass would not be in equilibrium.


QUESTION:  
If you were to drop an ice cube and a shot glass from 4 feet up, which would land and break first? Why?

QUESTION:  
My question to you involves the feather and bowling ball freefall experiement in a vacuum. Now I understand that by removing the factors of air resistance that you free up the feather to fall much faster than it normally would. But the part I'm having trouble with is that they do indeed fall, gravity is obviously still involved and the ball still outweighs the feather many times over. In fact the term "weight" is defined by the amount of force gravity exerts on an object's mass. (not totally clueless, I just look like I am) So by that term should not the ball fall much more quickly since gravity has that much influence on it?


 ANSWER: 
Since these two questions are so similar, I will answer them with one answer. Ignoring or removing air friction results in all falling objects as having the same vertical acceleration, 9.8 m/s/s; that is the speed of any freely falling object increases by 9.8 m/s as each second ticks by. Hence, the ice cube and the shot glass and the feather and the bowling ball will all hit simultaneously. To understand the "why" you need to know two things:

  • Newton's second law states that the acceleration of an object is proportional to the force it experiences and inversely proportional to its mass; if we define a unit of force to be that which causes one kilogram to have an acceleration of 1 m/s/s, then a=F/m if F is in Newtons.
  • An object's weight, the force which the earth exerts on it, is proportional to its mass. So we can write that W=mg where, for now, g is just some constant; if we measure the weight in Newtons and the mass in kilograms, then we find (by simply measuring) that g=9.8 m/s/s.

Putting these two things together for the case where the force in Newton's second law is the weight of an object, then a=W/m=mg/m=g. Note that m cancels out; this is the key to understanding. Although the weight is proportional to the mass, the inertia (resistance to acceleration) is also proportional to the mass, so all objects have the same acceleration.


QUESTION:  
I am writing in the interest of hopefully resolving a question which had arisen in my workplace. One gentleman poses the hypothetical situation of a motionless tank sitting on solid ice which he describes as "very slick and smooth - so much so that if one were to toss a penny across the surface then it would glide on endlessly." He posits that the tank is then started and attempts to move forward. His position is that the tank will not be able to move as the treads would simply spin on the ice. His detractor posits that the treads are moved by the wheels inside the treads and that this would be able to propel the tank forward. So, would this tank be able to move forward or not? If so, what propertys of physics would make it be able to move and, if not, why would this tank not be able to move forward? The gentleman's scenario also posits that there is no friction between the tank treads and the ice. Is it realistic, physically speaking, to posit these two surfaces touching and no friction existing between them?

 ANSWER: 
How genteel you are! The gentleman who says that the tank will not move forward if the ice is perfectly frictionless is correct. It is the force of friction which accelerates the tank forward, not the force which the wheels exert on the treads; if the wheels exert a force on the treads, then Newton's third law says the treads exert an equal and opposite force on the wheels so the two cancell each other out. No it is not possible to have a perfectly frictionless surface; it is possible to get a good enough approximation, however, to do an experiment which should convince the second gentleman.


QUESTION:  
I recently heard that if you shoot a bullet straight out with one hand and drop a bullet to the ground from the same height with the other hand then both bullets will hit the ground at the same time. Is this true and why. It would seem that the bullet being shot would have much more energy behind it abd take longer to expend that energy and hit the ground.

 ANSWER: 
This is what happens assuming that there is no air friction (which is actually a pretty poor approximation for something going as fast as a bullet). But, if there were no air, they would hit simultaneously. The reason is that they both move identically in the vertical direction, that is they both have the same acceleration vertically whereas the one shot moves with constant speed horizontally. So, imagine having a horizontal spotlight which cast a shadow of the shot bullet on a screen, a way of looking separately at its vertically motion. The shadow would move in exactly the same way as the dropped bullet.


QUESTION:  
If you were to drop two equal boxes of equal dimensions built with the same material and you filled one with ping pong balls and the other with bricks. You than proceded to drop them both simaltaneously out of an airplane at 2000 feet. Which one would hit the ground first and why?

 ANSWER: 
In an elementary physics course you are taught that all objects have the same acceleration, so from that perspective the motion of the two boxes would be identical and they would therefore hit the ground simultaneously. However, in the real world there is air friction which is normally neglected when you first learn about free fall. However, air friction is not negligible particularly if speeds get very large. So, as you stipulate, the geometries of the two objects are the same, they will both have equal air friction for a given speed. However, when the air friction force equals the weight of the object, it will stop accelerating and fall at constant speed; therefore the lighter box will stop accelerating sooner and therefore hit the ground later.


QUESTION:  
I always wondered about this question and was hoping you can shed some light on it for me. I noticed on Star Trek, they have gravity on their spaceships. I know this is a sci-fi program and realize its all fictional based on some scientific facts. But, I always wondered , why can't we put gravity on the spacestation we have orbiting our planet now? I know we have simulated 0 gravity on earth. Why can't gravity be simulated on a spacestation in space?

 ANSWER: 
The only way we know to "simulate gravity" is to be in an accelerating frame of reference. If you are in a frame accelerating down with an acceleration of 9.8 m/s2 near the earth (e.g. a freely falling elevator) you will feel weightless. That is why you are weightless in the space station because it is in free fall, just like the elevator but it is a free falling projectile. To create "gravity" you would have to have a local acceleration of 9.8 m/s2; the way to do this is to have a very large donut shaped space station (say maybe a few hundred meters across) which is rotating with the right speed so that you could walk around the inside of the outer surface feeling as if there were gravity.


QUESTION:  
The discussion is if there was such a thing as a indoor wakeboard training facility would the hydro dynamics be such that if water speed was equal in the following scenarios could equal amounts of 'big air' be obtained.

  1. A 200# wake boarder is being towed behind a boat at a distance of 60' at 20 mph. The wake height is 24" high and has a 45 degree angle on the outside of the wake. The wake boarder has sufficient skill to cut in from outside the wake and obtain a height of 36" and travel 20'. Rope mounting height is 36" above water level.
  2. The same 200# wake boarder is in a wave pool capable of creating the same water conditions as being towed behind a boat at a distance of 60' at 20 mph. Same wake characteristics and same skill level from the same wake boarder. Will the Wake boarder be able to obtain the same height and distance as he would behind a boat? Same rope, rope height same everything.

Some of us say yes some say no. Some are of the opinion that forward momentum generated by the boat would enable greater distances to be obtained. Others say that the height and distance would be equal because it is the speed of the water going by and the ability to generate speed perpendicular to the direction of the water that provides the speed to obtain height and distance.

 ANSWER: 
You are going to have the water going by in #2 with the same speed as the boat in #1--good. You are going to have a wake with the same characteristics (same height and motion relative to the wakeboarder)--good, although I am not convinced that it would be easy to actually do this since the wake shape has to be essentially at rest relative to the wakeboarder whereas the water per se is not. What you do not talk about is the air. In #1 there is a significant wind in the face of the wakeboarder so you better have a big fan for #2. I believe that air friction will play a very important role after he lifts off, that is he will essentially fly like a ski jumper. I think that if you don't forget the air, it will be a pretty good simulation. Incidentally, the boat has nothing to do  with the mechanics of the wakeboarder because they are not in contact so the boat cannot exert any forces on him. Whatever happens to the wakeboarder depends only on his interactions with water, air, and rope (and, of course gravity).


QUESTION:  
if you shot a bullet directly up, when it came back down would its speed be the same as when it left the gun or does terminal velocity prevent that.

 ANSWER: 
If there were no air friction, the speed would be the same. But there is significant air friction which takes kinetic energy away from the bullet, so it is going more slowly when it hits the ground. Incidentally, because the earth is rotating, the bullet would not hit precisely at the point from where it was fired straight up.


QUESTION:  
Usually when I get bored, I spin my chair. First I take a spin with my legs in. When I strech my legs out, the speed decreases, which is abvious because of the laws of the centrifugal force. The weird think, though, is when I retract my legs again, the speed INcreases after it has decreased. I asked my scince teacher about this conundrum, and her suggestion was that perhaps I retracted my feet fast enough to give my spin energy enough to spin faster. To me that just doesn't seem right.

 ANSWER: 
First of all, there is no such thing as "the laws of the centrifugal force", so though it may be obvious to you that you slow down, it should not be! What you are experiencing here is conservation of angular momentum. Angular momentum is the product of the angular velocity (for example, the number of revolutions per second) and the moment of inertia the object has relative to the axis about which it is rotating. Moment of inertia is essentially a measure of how far away, on average, the mass of the object is from the axis of rotation. So, when your legs are in your moment of inertia is smaller than when your legs are out. The angular momentum of an object remains constant (is conserved) if there are no external torques on the system. You and the chair spinning can be approximated as having no torque, although there is some due to friction which will eventually, as you know, cause your angular momentum to vanish (because you stop spinning). But, over the short term, your angular momentum stays about the same, so if you put your legs out your angular velocity must decrease and if you pull them in, it must increase. This is the same principle dancers and skaters use to increase their spin rate by pulling in their arms.


QUESTION:  
I was just wondering why when you rub two items together (like your hand on carpet for example) heat is generated? Where does that energy come from?

ANSWER: 
It comes from you. When you cause the two surfaces to move against each other you do work (a force acting over a distance) and work adds energy to a system, in this case heat energy.


QUESTION:  
A recently published book on baseball pitching claims that an arm/hand assembly travelling at 31 mph will allow for a ball to be released at 95 mph. The author's contention is that, because of the law of conservation of momentum, the larger mass of the arm travelling at 31 mph will transfer momentum to the smaller mass of the ball, thus causing it to accelerate to 95 mph in order to satisfy that law. It is my contention that there can be no transfer of momentum unless there is a sudden decrease in velocity of the carrier due to an external force being applied, thus allowing the carrier to "give up" it's momentum to the unconstrained object (the ball). I proposed this question. What event has occurred from a point 1 millionth of a second prior to ball release and a second point 1 millionth of a second after release that causes momentum to be transferred? The masses of the arm and ball have not changed. Therefore, the only variable in the momentum equation to change is velocity. If the velocity is reduced by internal means, muscular relaxation, the momentum will be not be transferred to the ball but dissipated within the body. There is no external force, like a brick wall to stop the arm, therefore there can be no transfer of momentum to the ball. Am I correct here? Any help would be greatly appreciated.

ANSWER: 
I feel quite certain that the speed of the part of the "arm/hand assembly" (as you call it) which is in contact with the ball is moving with the same speed as the ball when it is released. It is totally fallacious to talk about the speed of the "arm/hand assembly" because each part is moving with a different speed:

  • the shoulder has a small speed due to motion forward of the body
  • the elbow has a speed which is the vector sum of the velocity of the shoulder and the velocity of the elbow due to its rotation about the shoulder
  • the wrist has a speed which is the vector sum of the velocity of the elbow and the velocity of the wrist due to its rotation about the elbow
  • the hand (and ball)  has a speed which is the vector sum of the velocity of the wrist and the velocity of the hand due to its rotation about the wrist

So, you see, it is not so simple.


QUESTION:  
What would the effect on the perceived gravitational pull of our planet if the rotation (or spin) of the planet was increased by a factor of 4? I would assume that the increased momentum at the outer edges would cause items on the surface to experience a lower gravitational pull. Unfortunately, I can not locate any equations to summarize this. Can you lend any assistance?

ANSWER: 
The answer to your question depends on where you are. The effect would be zero at the poles and maximized at the equator. I will work it out for the equator. The first thing you need to do is calculate your acceleration. Since you are moving in a circle, your acceleration is v2/R=(2pR/T)2/R=4pR/T2=4x3.14x6.4x106/(3600x24/4)=0.17 m/s2; here v is the speed of something on the equator, R is the radius of the earth, and T is the period ("length of day" which you stipulate to be 24/4=6 hours). This is the centripetal acceleration and the direction is toward the center of the earth. The sum of all forces on you must add up to your mass (m) times this acceleration. The forces on you are your own weight mg=9.8m (down) and the force that the ground exerts up on you, call it N. So we have N-9.8m=-0.17m, or N=9.63m. The force N is the force by which you perceive how heavy you feel, so you feel about 1.7% lighter than if the earth were not rotating; if you were standing on a scale, which measures the force you exert on it, not your weight, it would read 1.7% smaller than your actual weight mg.


QUESTION:  
Why do keepers of the famous clock Big Ben in London add or remove coins from the bottom of the pendulum to speed up or slow down the clock if the mass of pendulum has nothing to do with the pendulum's period?

ANSWER: 
Because Big Ben's pendulum is not a simple pendulum, that is it is not a point mass attached to a massless string. It is what is often referred to as a "physical pendulum" and its period is determined by its moment of inertia and the moment of inertia depends on how much mass there is and where it is.


QUESTION:  
Hi, quick question, a few friends and I had a little debate the other day. Basically the premise of the debate was that if humans somehow started to dump all our garbage into space the mass of the earth would be affected enough to somehow affect earth's orbit. Their contention was that everything is produced by the earth as in what humans make from it therefore if we started jettisoning it into space it would affect earth's mass to alter its orbit however miniscule. I saw something really wrong with this, in my view there are many variables not being taken into account here that affect earth's mass and its relation to the sun and its orbit and that garbage is too small a variable or another counteracting variable would negate any effect it could have if it was sent into space. Yes a little bit of a weird topic lol, but any insight would be appreciated.

ANSWER: 
If the earth's mass suddenly decreased by half, the period of the orbit (one year) would be totally unaffected. The period of the orbit of a satellite is independent of its mass if the sun's mass is very large compared to the earth's mass (2x1030 kg vs. 6x1024 kg, a ratio of about 3/1,000,000). To be exact, it is the reduced mass of the system which matters; that would be msunmearth/(msun +mearth
), but, since msun>>mearth, the reduced mass is incredibly close to mearth, and the dependence of the orbit on the mass is completely negligible. Technically, however, your friends are right that a change of mass will affect the orbit; it is just that for even a huge change in mass (a large fraction of the earth's total mass) the effect would be so small as to be unmeasurable.

Another problem to consider is the possible disruption of the earth's orbit because of how the mass was ejected. For example, in the extreme case where you jettison half the earth's mass into space with some velocity the remaining half would recoil with the same velocity which would certainly seriously alter the orbit. But if you jettisoned a million tons, the earth's recoil would be negligible.


QUESTION:  
what would happen if a small hole was drilled right through Earth, that is from one side to another going through the Earth's core, and I so happen to drop an object into that hole, would the object fall right through, fall towards the middle and stay there, or result in a ping pong effect? (Given that the posibillty of creating such a hole is possible, without a catastrophic event happening)

ANSWER: 
This is a classic introductory physics problem. The object will speed up until it gets to the center and slow down until it gets to the hole opening at the other side, will then start speeding up back toward the center, etc. If the earth is modeled as having uniform mass density (which is only roughly accurate), the motion of the object will be what we call simple harmonic motion, that is it will behave exactly as if it were attached to a giant spring.


QUESTION:  
In physics class we performed an experiment called projectiles launshed at an angle (to the horizontal). Now we made a setup where we let a marble shoot away with the help of an elastic band. We then measured the distance the marble travelled when letting it shoot from different angles, 20, 30, 40,45, 50, 60, 70 degrees. I would appretiate some theory concerning this experiment because I thaught that the marble would reach the longest distance from the angle 45 but the distance of the angles 40, 45 and 50 were very similar....would you please explain to me a little about what a "fair test" is  and what factors do actually affect the range of this elastic band.

ANSWER: 
Well, an elastic band is not a terribly good energy source because it tends to get "fatigued" with repeated use so it will not produce reproducible results.  But, even if you were using a better launcher, say a good steel spring, your results would not be surprising.  The expression for the range of a projectile is R=v2sin(2
q)/g where v is the speed at launch, g is the acceleration due to gravity, and q is the angle of launch.  To the right I have plotted the range as a function of the angle for an experiment where the maximum range is 5 m.  The maximum is at 450.  But the thing to note is how slowly the range is changing when you are at angles close to 450; the right-most graph is the same as the other graph except that it is plotted only over the range 40-500.  As you can see, any angle in this range will give about the same value for the range--5 m.


QUESTION:  
If action & reaction are always equal in magitude & opposite in direction, why don't they always cancel one another & leave no net force to accelerating a body?

ANSWER: 
Newton's third law states that if one object exerts a force on a second, the second exerts an equal and opposite force on the first.  Therefore, the "action/reaction" forces are never exerted on one body.  If you select a body to study, its motion is determined only by the forces exerted on it, not by forces exerted by it.  Students often make mistakes with this "action/reaction" thing because they tend to identify any pair of equal and opposite forces as being an "action/reaction" pair.  For example, a 1 lb book sitting on a horizontal table has two forces on it, its 1 lb. weight pointing down and a force of 1 lb which the table exerts up on it (usually called the normal force); these have nothing to do with Newton's third law but are equal and opposite because the book is in equilibrium and the force the table exerts is therefore required to be 1 lb up.  If we now look at the table, the book exerts a 1 lb force down on it because of Newton's third law; the "action/reaction" pair is the force the table exerts on the book and the force which the book exerts on the table.  Lots of novice physics students want to say that the weight of the book is the 1 lb force down on the table--this is totally false since this is a force on the book, not the table.


QUESTION:  
I am having trouble understanding torque.  What is it?  A force? A tendency?  Something else? Also, how does it differ from inertia?  Does it not resist circular motion? I am very mathematical, so maybe an explanation and/or derivation of the formula t=Fd could help me.

ANSWER: 
It is difficult to come up with an answer since you ask a question, "...how does it differ from inertia?" which indicates that you know very little about rotational physics!  I will try, but you must first understand translational physics, where everything may be treated as a point mass.  In this arena, you might very well ask "...what is force and how does it differ from inertia?"  

  • Inertia, which is called mass (m), is the property which measures how resistant something is to being accelerated (a) if it is pushed or pulled; the amount of push or pull, called force (F), measures how effective the push or pull is at causing acceleration.  All this is stated, of course, by Newton's second law, F=ma.  

In rotational physics the whole scenario sounds very much the same:

  • Inertia, which is called moment of inertia (I), is the property which measures how resistant something is to having an angular acceleration (a) about some axis; that "thing" which causes the angular acceleration, called torque (t), measures how effective the "thing" is at causing acceleration.  All this is stated, of course, by Newton's second law in rotational form, t=Ia.  

For translational physics, the only thing which determines inertia is how much stuff (mass) there is; for rotational physics, inertia is determined by how much stuff there is but also how it is distributed: an object with most of its mass near its axel is much easier to get spinning than is an object with equal mass but much of it far from its axel.  For translational physics, the force is determined only by hard we push; for rotational physics, torque is determined by how hard we push but also where we push and the direction in which we push.  For example, suppose you want to open a door and you push as hard as you can but you push at the hinged end of the door; the door doesn't open, does it?  This is because the torque increases as you increase the "moment arm", the distance from the rotation axis to where the force is applied (your t=Fd equation).  Actually, your equation is a bit too simplified to completely understand torque.  If you have an open door and push at the edge of the door which is opposite the hinges but you push straight toward the hinges, you do not close the door even though F and d are both big.  It is really only the component of the force Fp which is perpendicular to d which contributes to the torque, so t=Fpd.


QUESTION:  
Okay, I have a stick of mass M and length L held so that it makes and angle theta with respect to the floor.  The stick is not hinged on the floor, and the contact between the end of the stick and the floor is frictionless.  I release the stick and it falls to the floor.  How do I find the horizontal distance the left end (the end on the floor) travels during the fall?  And with what speed does the right end (the end initially in the air) hit the floor?

ANSWER: 
This is easier than you think.  Since there are no forces which have horizontal components, the center of mass of the stick must fall straight down.  The distance the end will move is (L/2)(1-cos
q).  The second question (not quite so easy!) should be addressed by energy conservation.  At the time the stick is released, it has only potential energy E1=Mg(L/2)sinq .  Just before it hits, the speed of the end will be v and the speed of the center of mass will be v/2.  The energy, purely kinetic, will be the kinetic energy of the center of mass plus the kinetic energy about the center of mass: E2=(1/2)M(v/2)2+(1/2)Iw2=(1/8)Mv2+(1/2)(ML2/12)[(v/2)/(L/2)]2=Mv2/6.  Equating E1 and E2 now gives v=[3gLsinq ]1/2


QUESTION:  
I have been curious about this problem for some time.  If a solid ball of mass m and radius r is resting on a block of mass M and a horizontal force is applied to the block, how would I find the maximum value of the force F that moves the block without causing the ball to slide.  This assumes there is friction between the block and the ball, but not between the ground and the block.  I think F acting on the ball has to equal the force due to friction between the two objects, but I'm not sure if, when finding this frictional force, you consider the mass to be m+M or just m.  And I feel like I'm leaving something out. Thanks!

ANSWER: 
This is a tricky problem.  We know the ball has a frictional force forward on it given by f=
mmg if the ball is about to slip.  Therefore, the center of mass of the ball has a forward acceleration of am=mg.  However, this is not the forward acceleration of the block.  So now consider the sum of torques about the center of mass of the ball.  This must equal the angular acceleration of the ball about its center of mass times the moment of inertia of the ball about its center of mass: t=fr=mmgr=(2mr(2mr2/5)(at/r) where at is the acceleration of the rim of the ball with respect to the center of mass of the ball.  Therefore at=5mg/2.  Now, since the center of mass itself has an acceleration, then the acceleration of the point of contact between the ball and the block as measured in the laboratory is at-am=aM=3mg/2.;  I have called this aM because it is also the acceleration of M since the ball is not slipping.  So, finally, F-f=F-mmg=M(3mg/2), so F=mg(m+3M/2).  Note that the ball exerts a force on the block of magnitude f backwards because of Newton's third law.

Note that I have used a theorem of classical mechanics: The sum of torques about the center of mass of a rigid body is equal to Ia even if the center of mass is accelerating. 


QUESTION:  
The extremes of "perfectly elastic" and "perfectly inelastic" collisions are somewhat confusing as most books show something between the extremes.  For instance, a billard ball that hits the billard table's side bumper and moves away with the same constant velocity that it approached with would have 0 change in the ball's momentum, because: mass of ball * (+velocity) = (mass of ball) *(-velocity), hence net = 0, assuming only x component direction. Kinetic energy is conserved, but momentum is not?  s the way to think about this collision, or think of a small object (the ball) colliding with a much larger object (the table). Just seems "fuzzy"?

ANSWER: 
Well, perfectly elastic simply means that kinetic energy is conserved.  Perfectly inelastic is a little more subtle since it is defined as all the kinetic energy in the center of mass system as disappearing in the collision; it is easy to recognize, though, since it corresponds to the interacting particles being stuck together afterwards.

The billiard ball collision you cite you have incorrectly analyzed.  There is not zero change in linear momentum of the ball because linear momentum is a vector.  The change is a vector away from the bumper with magnitude 2mv.  The momentum of the ball is not conserved because there is an external agent which delivers an impulse (force x time) to it; that agent, of course, is the bumper.  If you want to see momentum and energy both conserved, you must look at some isolated system: imagine the pool table to rest on a frictionless floor so that the system is the ball plus (the much more massive) table; then the table would recoil a tiny amount and the ball would have slightly less speed after the collision such that energy and momentum would both be conserved.  In the limit that the mass of the table approaches infinity, the ball would have the same speed after the collision as before but the momentum of table would be nonzero after the collision even though it would be at rest!  This is because zero x infinity need not be either zero or infinity.  This assures momentum is conserved.  Of course, there is no such thing as infinite mass and the real world has the table essentially at rest after the collision.


QUESTION:  
My question is rather simple, but has created much controversey among my friends and I. When I am at an amusement park, I often wonder... Which part of the rollercoaster goes faster? The front, or the back? I immediately come to the conclusion that the rollercoaster MUST be going the same speed, in the front and the back. My friends disagree with me, and say that the back is going faster. I'm pretty sure that this is untrue. Could you explain the reason our confusion, and give a brief answer.

ANSWER
For a roller coaster, every car moves with the same speed.  The reason is that all the cars are constrained to move on the track.  If they were not constrained to move on the track, however, one part of the train could easily move with a different speed from another.  Imagine, for example, that the train rotated around the front car which was standing still (except for its rotational motion); then each car, as you moved toward the other end, would be moving faster than the one before.  When I was a kid we used to play "crack the whip" when ice skating: a long line of kids would skate across a pond and the kid at one end would essentially stop and the kids near the other end would get a very exciting and fast ride!


QUESTION:  
At science centers, you can usually find the activity with the bicycle wheel and a spinning chair? Can you explain the gyroscopic precession principles involved with that activity and what makes you spin when you change the angle of the bicycle wheel? Also, can you please explain that using simple language...most websites I have visited about this question are difficult for a layman to understand.

ANSWER: 
Precession is not an issue here; rather, conservation of angular momentum must be appreciated.  Imagine that you are holding a spinning wheel with its axel pointing vertically and that you are standing on a turntable which can rotate without significant friction.  Although there are external forces on you (your weight, the weight of the wheel, and the force the table exerts up), none of these forces exerts any torque on you and the wheel. The total angular momentum of a system which has no external torques on it must never change which is what it means to say that the angular momentum is conserved.  Angular momentum is a vector quantity.  If you look down on the spinning wheel and it is spinning counterclockwise, it has an angular momentum vector which points up and has a magnitude which I will call L.  You are not spinning and have an angular momentum, therefore, of zero.  Therefore the total angular momentum of you + the wheel is L upward and this is what it must always be.  Now you flip the wheel over so that the angular momentum of the wheel is L downward, so the angular momentum of the wheel has changed by an amount of 2L downward.  But, the total angular momentum cannot change so an angular momentum of 2L upward must pop up somewhere; the only part of the system which can get angular momentum is you, so you must start spinning such that your angular momentum is 2L upward which means, since up means counterclockwise as seen from above, you will spin counterclockwise as seen from above.  Here is a movie.

Incidentally, if you care, conservation of angular momentum is simply a consequence of Newton's first law in its rotational form.  In translational form, Newton's first law says that, if there are no external forces on a system, the linear momentum will be conserved; in rotational form it is that, if there are no external torques on a system, the angular momentum will be conserved.


QUESTION: 
I want to ask a few questions about "coefficient of friction"

1.Apart from experimentally determine the coefficient of friction between two surfaces, is there any theotical way to do it? For example, if we know the structure of the two materials, can we then proceed to calculate the coefficent of friction, and if we can, how? What factors (temperature, charge on the sufaces, etc.) are involved in the calculation?

2.If we know the coefficient of friction between, say, surface A and surface B and name if x, that between surface B and surface C be y, can we then say that the coefficient of friction between A and C would be at maximum x+y? Similarly can we say it be at least the minimum of (x,y,abolute(x-y))?

3.In my textbook it tells me that coefficient of static friction between two clean steel surfaces is approximately 0.6 and the coresponding value of kinetic friction is still 0.6, which means the value differ very little, the same is true for telflon, which in my book it says both value are 0.04, however for lubricated steel surfaces, the values for static and kinetic friction are 0.09 and 0.05, which is nearly a double, the coefficient of static and kinetic friction between two glass surfaces is 0.9-1.0 and 0.4, which is once again nearly a double. What makes the difference between them? Is there any pair of surface that has a great difference between coefficient of static and kinetic friction?

Lasly is there any reference you would recommend me about this topic? 

Thanks very much for answering my questions.

ANSWER: 
Ah, such an innocent-sounding question!  Every freshman physics student learns about the very simple idea of a coeffiecient of friction, the proportionality constant between the magnitude of a frictional force and the normal force (how hard the surfaces are pressed together).  And that there are two types, static and kinetic, and that they are so easy to understand.  But here is the catch: f=
mN is a statement of an approximate experimental fact, not a law of physics.  You cannot derive it from known laws, you cannot deduce it from first principles.  Surprisingly, friction is fairly poorly understood.  I have gone to the website for the Tribology (study of interacting surfaces) Laboratory at the University of Florida and find the following statement: "When two bodies are in contact and relative motion, a finite force is required to maintain this motion, the friction force . A coefficient of friction is calculated by dividing the friction force by the normal force. To date, despite considerable efforts, there is no model capable of predicting friction coefficients from first principles. Thus, careful and proven experimental techniques represent the most sophisticated and reliable technique for investigating, designing, and assessing the tribological worthiness of new materials."  The most famous example of a large difference between static and kinetic friction is the violin bow; here the instrument is excited by the bow repeatedly slipping/sticking on the string.  The two links I have given here would be good places to start your research.  Do a google search on "theory of friction"; much of what you find will be quite technical.


QUESTION:
About that "moving molecules" question--let's say we had a glass of water. The water molecules would have the "tendency" to form hydrogen bonds with each other. Wouldn't we need a constant input of energy to break these bonds and keep the water in a liquid state? If so, where does this energy come from? Is it, for the most part, radiation?

ANSWER:
The answer is pretty much the same as the one just below.  Yes, the water molecules want to bond to other water molecules and that is exactly what happens when the water freezes to ice.  But they do not bond to each other if you have liquid water above 00C because the kinetic energy which an average molecule has is greater than the energy which is necessary to break the bond.  It is kind of like saying: suppose we have two heavy balls which each are moving very fast in opposite directions and we decide to tie them together with a little rubber band; the rubber band would just break. However, if you take the kinetic energy away from the water molecules (i.e. cool the water down) then eventually they will bond together.  Just as below, there is no need to "keep" everything moving; Newton's first law takes care of that.  Incidentally, radiation has absolutely nothing to do with anything here.


QUESTION:
Can you explain to me what exactly KEEPS molecules moving? With no energy being added, they should just eventually stop, shouldn't they? Where does this energy that keeps them moving come from? In the end, does it all come down to radiation from the sun? 

ANSWER:
You have fallen into one of the most common traps regarding misunderstanding how the universe works.  Newton's first law states that an object which experiences no net force will continue to move with constant speed in a straight line.  What this means is that if something is moving and nothing is pushing or pulling on it, then you do not have to do anything to keep it moving.  In terms of energy, if something has a certain amount of energy, then it will retain that energy until some external agent changes it; this is called conservation of energy.  I am not sure what you have in mind with your question, but probably the molecules moving around in a gas.  As you probably know, the temperature of a gas is a measure of the average kinetic energy per molecule.  If the gas is in thermal equilibrium with the walls, then when a molecule hits the wall it rebounds (on the average) with the same kinetic energy it had beforehand.  You don't have to do anything to keep it moving.  Incidentally, if Newton's first law were not true we would never have sent probes to the distant planets like Saturn and Jupiter or even the close ones like Mars and Venus.  The reason is that if we had to KEEP the probe moving by burning an engine the whole way we could never carry enough fuel.  What actually happens is that we burn up almost all the fuel escaping the earth and acquiring a high speed and then we just turn off the engines and coast the rest of the way.


QUESTION:
I heard speeding up (gradually stepping more on gas pedal) while driving on a long, sharp curve was better (safer) than slowing down (gently braking). Is this true?

ANSWER:
Let us assume that the main safety concern in turning a curve is the possibility of skidding.  My answer will address that concern.  There are lots of other issues that could affect "safety" such as whether you have better control while accelerating a front-wheel or rear-wheel drive car, or whether you have a four-wheel drive car, or the fact that braking occurs on all four wheels but acceleration uses (usually) only two, etc.  These are engineering issues which I don't want to get into.  In the safe use of a car, the force of static friction between the road and the wheels provides the force to brake the car, accelerate the car, or move a car around a curve.  The fact that it is static friction confuses some people since the car is obviously not static.  However, if all is well, the tires are at all times at rest where they touch the road.  An important feature of static friction is that there is a limit to how much you can get; for example, if a heavy box is sitting on the floor, you can push gently and it won't move, push harder and it won't move, push harder still and it won't move...but eventually it will "break away" and start sliding across the room.  This is what happens when you slam on your brakes and start to skid, "peel out" when accelerating, or break into a skid when turning a curve too fast.  When you go around a curve, the static friction from your tires must provide a force equal to mv2/R where m and v are the mass and speed of the car (this could be smaller for a banked curve) and R is the radius of the curve.  Let us assume that you are going through with the fastest speed possible so that if you went any faster you would skid.  This force is labeled 'turn' in the two pictures above which show the forces on the "footprint" of a tire on the road; this 'turn' force is pointing toward the center of the circle about which the car is turning.  The two pictures show what happens if you either accelerate or brake: an additional force is required and the sum of the two, labeled 'net', is bigger than the static friction can give and you will go into a skid.  Therefore, if you fear you might be approaching a curve going too fast, you should do your slowing down before you start the curve, don't try to do it while you are negotiating the curve.  


QUESTION:
If there is a ball that is traveling on a vertical axis at 2m/s and hits a 45 degree angle, it will then travel at 2m/s to the left. Now lets say that before the collision the ball and the angle were traveling to the left at 3m/s relative to the planet Venus (or whatever). After the ball collides with the angle it's total velocity, relative to venus will be 5m/s to the left. Before the collision, relative to Venus, the ball was traveling at 3.8m (I used the pythagorean theorm with 2m/s and 3m/s). How is it that the ball has increased velocity relative to Venus without any energy being added to the system. The angle should only redirect the velocity. If the ball hit any kind of angle that was fixed in space relative to Venus there would be no way to change the velocity from 3.8 to 5. How is this possible?

ANSWER:
I can answer your question, but you have made it far harder than it need be.  Furthermore, I am not sure that you have gotten the velocity of the ball after the collision right for the second scenario, but it doesn't really matter since we can look at a much easier example (easier because it is a one-dimensional collision, whereas yours is two-dimensional) and learn where this mysterious added energy comes from.  Suppose we have a ball which collides elastically with a stationary wall; if the ball comes in with speed v to the right, then it rebounds (to the left) with speed v also--energy is conserved.  Now, if the wall is moving toward the ball (to the left) with speed u, the velocity of the ball after the collision will be v'=v+2u to the left and the velocity of the wall will still be u'=u to the leftHow did I get  these speeds after the collision?  Well, these are the speeds only if the wall has an infinite mass.  We get the results by conserving linear momentum and kinetic energy for a collision of a ball of mass m and a wall of mass M and then let M approach infinity.  So they are approximate solutions to the real problem if M is enormously bigger than m which is what you are presumably assuming.  But these solutions cannot be exactly true because no wall is really infinitely massive.  But, if it were, it would have an infinite amount of energy before the collision if it were moving; it then loses the amount of energy which the ball gains, but if you subtract a finite number from infinity you still get infinity.  So it sort of looks like energy materialized from nowhere but it really comes from the infinite energy which the wall has!  So, basically, any energy which the ball gains must come from the wall.

Incidentally, the general solutions to the problem if M is not infinite are
     v'
=[(m-M)v-2Mu]/(m+M)
     u'
=[(m-M)u+2mv]/(m+M)
which you can show are approximately v'=-(v+2u) and u'=-u for very large M, that is M>>m. (The minus signs denote that they are both traveling to the left.)


QUESTION:
I have a question regarding Kinetic Frictional Force. The magnitude of the kinetic frictional force is equal to the coefficient of kinetic friction multiplied by the value of the normal force; therefore, there is no dependence on surface area. Based on that, I was wondering if it were fair to say that tire companies that advertise that wider tires "grip the road" better than average size tires (made of identicle material), are deceiving consumers? Since wider tires cost considerably more, I was just wondering if they were misleading the public or if there were other forces/laws that apply to this scenario that do involve surface area. 

ANSWER:
This is not an easy question you have asked!  I did a little research and here a few particularly enlightening answers to this question which I have found (stolen!) on the web:

  • "In general, most of the responses were speculative, and centered on factors such as the nature of the tire-road surface interface, heat dissipation, tire deformation and elasticity, stability, and a myriad of other environmental and engineering constraints. The 'best' answer suggested that the tire problem did not fall under the standard Coulomb (dry) friction parameters, thus using tires as an example was comparing apples to oranges. 

    "Indeed, further digging supports the latter statement. According to Engineering Mechanics: Vol. 1, Statics (2nd ed), JL Meriam and LG Kraige, Wiley and Sons: New York, 1986, the coefficient of ROLLING RESISTANCE, while analogous to the coefficient of static or kinetic friction, is really an entirely different beast. It would be most difficult to describe fully without a free body diagram, but is a function of many factors, including, but not limited to: road and tire deformation and the resultant pressure over the area of contact, elastic and plastic properties of the mating materials, wheel radius, speed of travel, and roughness of the surfaces. Meriam and Kraige state, '... depends on many factors which are difficult to quantify, so that a comprehensive theory of rolling resistance is not available.'"  (Plagiarized from Biomch-L Newsgroup)

  • "This is a good question and one which is commonly asked by students when friction is discussed. It is true that wider tires commonly have better traction. The main reason why this is so does not relate to contact patch, however, but to composition. Soft compound tires are required to be wider in order for the side-wall to support the weight of the car. softer tires have a larger coefficient of friction, therefore better traction. A narrow, soft tire would not be strong enough, nor would it last very long. Wear in a tire is related to contact patch. Harder compound tires wear much longer, and can be narrower. They do, however have a lower coefficient of friction, therefore less traction. Among tires of the same type and composition, here is no appreciable difference in 'traction' with different widths. Wider tires, assuming all other factors are equal, commonly have stiffer side-walls and experience less roll. This gives better cornering performance."  (Plagiarized from physlink.com)

  • "Now let's look at what happens at the tire-to-road interface. In a perfect physic's laboratory world, one usually asserts that the friction force (i.e., the braking force) between two objects is a function of two things: the coefficient of friction at the interface, and the normal force. (Muddying the waters more than a little bit is the fact that this is only true for two "smooth" surfaces sliding against each other. It doesn't take into account that rubber is relatively malleable, and tends to fill the small nooks and crannies in a roadway, thereby changing the nature of the "braking" problem into one containing both frictional and shearing-type forces. This turns out to be a VERY difficult complication. Worse, the vulcanized forms of rubber used in most automotive tires has rather weird coefficient properties that don't behave nice and linearly. But I digress...)"  (Plagiarized from datsuns.com)

You can find a lot more by going to Google and searching on friction AND "surface area" AND tires.


QUESTION:
My physics students and I are debating which principle better explains why airplanes and helicopters are able to fly - is it Newton's 3rd law or Bernoulli's principle? Also, if a helicopter is stationary and you have a scale with enough capacity, can you measure its weight while it levitates above the scale? Can an airplane do this if it were able to levitate?

ANSWER:
Indeed, both are important in the design of a wing or a helicopter propeller, but Bernoulli's principle is the more important one (that is, normally the pressure difference due to the difference in velocity of air over the top and bottom of a wing provides most of the lift).  Newton's third law (N-3) also plays a role for the following reason: Because of the shape of the wing, the air which came straight in before it encounters the wing leaves the wing with a downward component of its velocity; the wing must therefore have exerted a downward force on the air so, because of N-3, the air must have exerted an upward force on the wing.  The balance between the two types of lift can be altered by changing the "angle of attack", the angle which the wing makes with the direction of flight.  If the angle of attack becomes bigger, the air is thrust more in the downward direction than otherwise, so N-3 becomes relatively more important.  (Although, changing the angle of attack also increases the Bernoulli effect because the speed of the air over the top will be greater relative to over the bottom when the wing is tilted.)  I would guess that sustained level flight with the airplane upside down (which is possible) would depend primarily on N-3 and therefore require a rather large angle of attack.  Incidentally, there is a limit to how much lift you can gain by increasing angle of attack; eventually, air on the upper of the surface of the wing will be unable to flow smoothly, will become turbulent, and all the Bernoulli lift will be lost.  This is called a stall.  I have had some flying lessons where stalls were purposely caused and I can assure you, the plane begins falling like a rock which shows that N-3 alone will not keep the plane in the air!

The answer to your second question is no.  If you are standing under a helicopter which is hovering, you experience the downward force of the air which is thrust down.  However, this force is nowhere near the downward force of the whole weight of the helicopter.  If you were able to "gather" all this downward wind and if the wind lost no energy coming down from the helicopter, the net force down would still be much smaller than the weight of the helicopter.  This again emphasizes that the role played by N-3 is relatively small.


QUESTION:
We recently had an unfortunate incident in Chicago where scaffolding fell from the John Hancock Building and killed three people who were in their vehicles. This led to a barroom discussion on what could kill a pedestrian if it was dropped off the Empire State Building. Some are of the opinion that something as small as a penny could do the job. Others say it would have to be something the size of a rock.

ANSWER:
This is a great question, but it does not (which is the case for most "real-world" questions) have a simple answer.  I think that there are essentially four things you need to understand in order to discuss the problem--force, pressure, acceleration due to gravity, and terminal velocity.  

  • You probably know all about acceleration due to gravity.  If there were no air friction then all objects would have a constant acceleration down of approximately 10 m/s/s.  What this means is that when you drop something from rest, it will have a speed down of 10 m/s after 1 second, 20 m/s after 2 seconds, 30 m/s after 3 seconds, etc.  I will not burden you with any equations but will assume that you will trust me to calculate for myself and tell you the answer of how fast something will be going after having fallen some distance or some time!  Obviously, this will be necessary to us later.
  • In the real world, there is air friction and this can seriously affect the acceleration due to gravity.  Because the air friction depends on how fast something is moving (for example, imagine the force of the wind on your hand sticking out the window of a car going 20 and a car going 80), it turns out that instead of accelerating forever, a falling object eventually reaches some terminal velocity and then falls with constant speed after that.  How big that terminal velocity is depends on lots of things, but very much on the size and shape of the object.  Some examples: a parachute causes the terminal velocity to be so low (but only when open!) that you can safely jump out of an airplane with it; an ant could jump off the empire state building and experience no injury; in fact, cats have fallen out of apartment windows 40 stories up and survived; a dust particle has a terminal velocity so close to zero that it will stay in the air for years because of drafts pushing it back up every time it starts to fall.
  • Of course, you know what force is qualitatively.  Quantitatively force is mass times acceleration.  So, for example, if you want to stop a 1000 kg (about 2000 lb) car going 10 m/s (about 20 mi/hr) in one second, its acceleration would have to be 10 m/s/s; the force you would have to exert is 1000x10=10,000 newtons (about 2500 lb).

I'll come back to pressure later.  So, let's drop a penny off the empire state building and neglect air friction.  I would guess that the height is something like 100 stories x 3 m/story=300 m.  The speed it would acquire would be about 80 m/s.  Now, it hits your head which must stop it before it enters the brain, maybe 5 mm inside.  So, I figure that it must stop in about .0001 seconds which means that it must have an acceleration of 800,000 m/s/s!  If we estimate the mass of the penny to be 1 gram=1/1000 kg, them mass times acceleration is about 800 newtons (200 lb).  So your head must exert an upward force of 200 lb on the penny and the penny, therefore, will exert a force of 200 lb down on your head!  So the question is, could your head withstand such a force?  Yes and no.  Imagine a penny laying on your head and a 200 lb man standing on it.  This would quite likely not cause your skull to be crushed.  On the other hand, imagine a penny standing on its edge and a 200 lb man standing on it.  In all likelihood, his weight would drive the penny through your skull.  So, if the penny hit you flat side, you would probably survive.  If it hit you edgewise, you might not.  If you now factor in air friction, the penny would be going slower, maybe only half as fast; so in the final analysis, I would say you have a fair chance of surviving a penny strike.  Something much more massive than a penny, say baseball, would almost certainly do you in since the force is proportional to the mass and the final speed would be about the same.

  • Finally, why is the penny more dangerous on its edge than on its side?  Because in the former case the whole force is spread over a much smaller area of your head than in the latter.  Force per unit area is called pressure, and pressure on your skull is more indicative of strength against breaking than force is.  A sharp dagger dropped from the second story of a building would have no problem getting through your skull because the force is all applied over essentially a point so the pressure is almost infinite.

QUESTION:
We know that when a ball is thrown it is being affected by air resistance, lighter balls will then have a short range (as a ballon does), heavier balls, due to its mass, will have smaller initial velocity and thus have a short range also. So I'm going to investigate which texture (density) of ball will have the longest range. However, I have encountered a few problem.

1. Should I assume my hand exert constant force or constant power?

2. In deducing the equation for the range, I came across with the following equation:
1/R * LN(1-(bR/mvCOS(z)))=bTAN(z)/mg - (b^2)/(v(m^2)COS(z))
where R is the range, m is the mass of the ball, z is the angle of projection, b is the resisive parameter (I assume air resistance to be bv), and g the gravitational acceleration.  But I'm not able to solve for R, can you help me solve that?

ANSWER:
You have several serious misconceptions here which you should surely straighten out by doing more reading and research.  The trajectory of a projectile is determined solely by its initial velocity, the constant you refer to as b, and, if b is not zero, the mass.  Your statement that "heavier balls, due to its mass, will have smaller initial velocity" is wrong.  Initial velocity is whatever you give it.  So you make a mistake to think of your hand as having anything at all to do with the problem.  I am not saying that mass has nothing to do with the problem, but it enters only in a subtle way as you will see below; if there were no air friction, the trajectories of all objects with identical initial conditions are identical.  But mass is not the only thing which determines what the effects of air resistance will be.  For example, a parachute surely has much more mass than a BB but encounters enormously more air resistance.  If you want to do an experiment comparing trajectories of different things, you will need to figure out how to give those things all the same initial conditions (particularly speed and direction at the beginning).  Once your hand or whatever is no longer in contact with the thing, your hand or whatever is irrelevant.

So, you found a formula and you want me to solve it!  First of all, there is an error in your formula (you are missing one minus sign).  The correct formula is ln(1-x)/x=-[1+(bv/(mg cos(z)))] where x=(bR)/(mv cos(z)).  I should tell you that this is a well studied problem and is standard in almost all intermediate level classical mechanics textbooks, so you will find this formula in just about any classical mechanics textbook.  It is, alas, what is called a transcendental equation, that is one which cannot be solved in closed form.  That doesn't mean you can't solve it for a specific case (specific b, m, z, v), you just have to do it approximately or numerically.  Since R is the only thing you are varying for a specific case you could, for example, program a computer to keep computing the left side of the equation until it was close enough (say within 1%) to the right side of the equation.  Or, you can solve it graphically.  Since your formula is of the form 

           ln(1-x)=-Cx

where C is just some constant and x is proportional to R.  So, if you plot both ln(1-x) and -Cx vs. x, where they cross will be the x corresponding to the R you seek.  An example, for C=2 is graphed to the left.  

Finally, there is an elegant way to approach the problem if x is very small.  By small I mean small compared to one; notice that x cannot be bigger than or equal to 1 because ln(0) is negative infinity and logs of negative numbers do not exist (for your purposes).  In that case you may approximate ln(1-x) as the first few terms of a power series expansion, -x+x2/2-x3/3.  Using this you can show that you may approximate R=(2v2 sin(z) cos(z)/g)[1-(4bv sin(z))/(3mg)].  This is a formula which you can evaluate directly, but don't forget that it is approximately correct only for (bR)/(mv cos(z)) small.

An excellent reference for this topic is Mechanics by Keith Symon published by Addison-Wesley but many other texts will cover it also.


QUESTION:
I have recently being investigating, damping of a rigid pendulum?  I used the equation y1=y0e-kt (the -kt is to the power of e, y0 is the initial amplitude and y1 is the final amplitude). I damped the pendulum using different areas of card and timed how long 5 oscillations took. I would like to know how I would justify using 5 swngs and not 10 for example?

I had to draw a graph of Area against k. I found that T(period) is independent of area. However, I would like to find out if I could expand this investigation to obtain more results. Also, is it true that a simple pendulum is an approximation of simple harmonic motion? and does s.h.m only occur when there is a small amplitude?
QUESTION TO THE QUESTIONER:
Please clarify what you mean by area.
REPLY:
We used card to dampen the ruler, by place it at right angles to the plane of oscillation. To vary the damping, I varied the area(centimetres squared) of card.

ANSWER:
OK, so the picture I have is that you used a ruler as a pendulum and you attached cards of varying areas to the end and perpendicular to the direction of swing to induce varying air friction.  Let me first answer your last question: as you state correctly, a pendulum is not an example of simple harmonic motion, one characteristic of which is that the period is independent of the amplitude.  However, to a very good approximation it is SHM if the amplitude is not too large.  For a pendulum, amplitude is the maximum angle through which it swings.  How large is too large?  That depends on how accurately you measure things.  My guess is that you are using a stopwatch or maybe just the second hand of a clock to measure times; you would probably find constant periods for your pendulum up to maybe as large as 400.  In fact it would be an interesting extension of your study to measure period as a function of amplitude: measure the period for amplitudes of say 100 to as big as you can get (if your ruler is mounted so that you can have amplitudes bigger than 900 you could go all the way up to almost 1800.)  If you then plot your data you should find a straight horizontal line up to some maximum angle and then it should go up.  
Now, we come to the crux of your experiment which, as I see it, is to observe the effect of damping on the motion.  So you measured the time of five oscillations and found that you got the same answer for all dampings.  This is a good start.  What you have shown is that different small oscillation pendula with different damping (that is different k because that is what you vary by changing the card area) all have the same periods.  In fact, that is what you are supposed to find if you study the theory of the damped oscillator which predicts that you get exactly the same oscillatory motion except that the amplitude decreases exponentially as a function of time.  The graph to the left shows what theory predicts.  You see that the damped oscillator has exactly the same period as the undamped one.  So you have verified that aspect of the theory.  However, what you need to do now are some measurements of the amplitude as a function of time and see if it really is exponential.  The way you would do that is to measure the angle of the pendulum each time it reaches the amplitude (when you release it, one half period later, one half period later, etc).  This is a little trickier measurement since you would have to tape an angle scale to the wall or something like that.  What would be really cool is if you could take a video of it and then make the measurements on the tv or computer screen.  If the theory is right, then you should get a series of points which can be plotted vs. time (one half period per point) which fall off like the blue lines above. And, the data for different damping should now look different with the larger cards falling faster than the smaller cards.  Now, finally, if you are good at math, you should be able to extract the constants k for each card so you can see how k depends on area.  The easiest way to do this is to plot your data on semilog graph paper (or plot the logarithm of amplitude vs. time); if the data are truly exponential, you should get a straight line and k is essentially the slope of that line.

Incidentally, when measuring these amplitudes it would be a good idea to use a very long pendulum so that the period becomes much bigger because it would be easier to measure because it would go slowly longer near the end of each swing; and you could tape a more accurate (that is bigger) angle scale to the wall which would be easier to read.


QUESTION:
Hi!
I'm involved in a "science problem of the week" contest. The idea is to research each question, making contact with anyone who might be of help. Perhaps you could assist me with this latest question?  Thanks!

Consider the following scenario... 

In this situation, a person sits in a seat and pulls down on a rope. As a result, the seat and the person in it rise up.  If the seat and a man together weigh 180 pounds, with how much force must the man pull down on the rope in order to raise himself and the seat up with a constant speed? 

ANSWER:
This is a classic problem often appearing in textbooks.  Shown in the figure above is the picture from Serway's text Physics for Scientists and Engineers (I'm probably violating some copyright by reproducing it!)  So I will refer to the boy and swing instead of your man and seat since that is the picture I found.  Since the boy and swing are to move up with constant velocity, this is a Newton's first law problem, i.e. the sum of all forces on anything you choose to look at must add up to zero.  The easiest thing to look at is the boy+swing.  The forces on the boy and swing are its 180 lb. weight (down), the tension of the rope tied to the swing (up), and the tension of the rope the boy is pulling on (up).  But the ropes tied to the swing and on which the boy is pulling are the same rope so they have the same tension.  Therefore, twice the tension in the rope must equal 180 lb, so the tension in the rope is 90 lb.  Technically, the problem has not yet been solved since what is asked for is how hard the boy must pull down on the rope and we have found how hard the rope pulls up on the boy; however, Newton's third law tells us that these two forces must be equal and opposite, so the final answer is 90 lb down.


QUESTION:
When placing an hourglass on a scale, will its mass increase, decrease, or stay the same as the sand is falling?

ANSWER:
Here is a question with just layers and layers of answers!  I will address all the possibilities I can think of:

  1. First, let's just think of the problem purely classically (relativity is an issue as you will see below).  Let us assume that the hourglass has nothing in it except sand (in particular, there is no air).  During the time the sand is transferring from the upper to the lower chamber there is, at any time, a fraction of the sand falling and therefore not exerting any downward force on anything.  Therefore, during the time the sand is falling there will be some fraction, say 1%, of the sand not being weighed, so the scale will read less at this time.
  2. Next let there be air in the hourglass.  During the time when sand is falling the air will exert two upward forces on the sand, a frictional force and a buoyant force.  If the air exerts an upward force on the sand, then Newton's third law stipulates that the sand exerts an equal but opposite force on the air which is part of the system being weighed, so the amount by which the weight is less will be smaller than if there is no air in the hourglass.
  3. A possibility related to part 2 of the answer is that, if the forces upward on the sand are not small compared with the weight of each sand particle (for example think of the particles of sand as being a very fine powder or think of the hourglass being filled with molasses), then the falling particles will quickly acquire a terminal velocity where the upward forces are exactly equal to the weight of the particles; in this case the hourglass system will always have the same weight as the sand falls.
  4. Another small effect is that the earth's gravitational field is not really uniform, that is the weight, which is just the force which the earth exerts on the sand, gets bigger like 1/r2 as r decreases where r is the distance to the center of the earth; it is weight, not mass, which the hourglass measures.  Therefore the total weight when all the sand is in the bottom will be larger than when it was all in the top.  This is a very small effect since the radius of the earth is about 6,400,000 m and the distance the sand falls is only something like 10 cm, so this effect will cause the weight to increase by an amount (I calculate) of about 3x10-6%!
  5. Now we come to relativistic effects.  Einstein says E=mc2 where E is energy, m is mass, and c is the speed of light (about 3x108 m/s).  Let's think about the energy of one grain of sand.  At the top it has, relative to the bottom, a gravitational potential energy of mgh (where m is the mass of the grain of sand, h the distance it falls, and g the acceleration due to gravity) and no kinetic energy.  Let us ignore effects like 2-4 above for purposes of clarity, so imagine an evacuated hourglass in a uniform gravitational field.  As the sand falls its potential energy gets smaller and its kinetic energy gets larger, all the while keeping the total energy the same. When it hits the bottom, all of its energy is kinetic and equal in magnitude to mgh; but now it stops so this energy disappears.  Or does it?  No, energy is still conserved because the kinetic energy will be converted into heat and the sand+hourglass will warm up a bit.  But, if we suppose that the whole system will then cool back down to the same temperature as its surroundings eventually, there has been a net loss of energy Mgh where M is the mass of all the sand.  But a loss of total energy means a loss of total mass because E=mc2, so the change in mass is DM=Mgh/c2; if we again take the size of the hourglass (h) to be about 10 cm, I find DM/M=10-17 or about 10-15% change in mass!

In summary, 

  • #1 decreases the mass measured by something on the order of 1% while the sand is actually falling;
  • #2 and #3 cause the effect in #1 to be smaller or zero;
  • #4 increases the weight measured (but not the mass) by something on the order of 10-6% after the sand has fallen; and
  • #5 decreases the mass by something on the order of 10-15% after the sand has fallen and cooled down

This is probably way more than you wanted, but the answer you wanted should be in there somewhere!  And, I had fun thinking about it!