Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a cube of mass M resting on a rough surface such that the coefficient of friction between the cube and the surface is K. So in order to just slide the cube I need to apply a minimum force of KMg, g= acceleration due to gravity. Now IF I apply a force, F which is very small as compared to KMg in magnitude then the body will not slide because the surface will exert an equal amount of friction force on the body. Now consider a sphere of same mass resting on the same surface having the same coefficient of friction. Now when I apply a force very very small in magnitude as compared to KMg, then an equal amount of frictional force will be exerted by the surface on the sphere at the point of contact. This frictional force will cause the sphere to rotate as it exerts a torque about the center of the sphere. Now as the sphere rotates the point of contact slides and hence now frictional force of magnitude KMg acts on it, pushes it forward and slowly sets the sphere on rolling motion. But This certainly constructs a perpetual motion machine and hence violates conservation of energy. May I know where was I wrong, where is the flaw in this so called perpetual motion machine?

share|improve this question
1  
"Now as the sphere rotates the point of contact slides" It does? Why? I've never seen a billiard ball do that. –  Mike Dunlavey Oct 16 '11 at 22:26
    
@Mike: he means the point of contact changes places on the sphere. This is an inane question. It is obviously the case that the friction force does not set the sphere moving faster than it already is moving. –  Ron Maimon Oct 17 '11 at 4:14

4 Answers 4

Coefficient of friction applicability

Saying the frictional force is $kMg$, opposite the direction of motion is a particular physical model. This model is not fundamental, and is certainly not 100% true, although it works pretty well for lots of practical situations. It applies for sliding between two surfaces, and it does not apply for a rolling ball.

Wheel friction

This is a topic that many struggle with. The two surfaces in question, the ground, and the wheel or ball, are not moving relative to each other. The surface of the ball touches the ground, but it does not slide. You can try, but there is no good way to construe this situation in such a way that the coefficient of friction model is valid, although there may be an appropriate analog for rolling motion.

Perpetual motion

Whatever frictional forces do occur in a rolling situation will be dissipative, as frictional forces always are. This applies for rotation as well as linear motion, although there are interesting (transient) exceptions. For instance, when a bowling ball is released with no spin, friction converts some linear momentum into angular momentum, but still leads to a more entropic state, creating heat in the process.

Wheel friction both:

  • Retards linear momentum
  • Retards angular momentum.

Given the stipulation that the rotation is in-phase (a characteristic of rolling). Hope that helps.

share|improve this answer

Zassounotsukushi's answer is good, but I want to emphasize that there is a mistake here:

when I apply a force very very small in magnitude as compared to KMg, then an equal amount of frictional force will be exerted by the surface on the sphere at the point of contact.

The surface does not exert an equal force. The way friction works is that it exerts enough force to stop all relative motion between the part of the ball that contacts the table and the table itself. That force may or may not be equal in magnitude to your very small push.

Let's start with a cube. Say you have a cube sitting on the table, and you push on it with 1N of force, and that's smaller than the maximum friction force. Then, since we know the cube isn't going anywhere, the friction with the table must be pushing back with 1N of force. This creates a net torque. If your original push is through the center of the cube, so it creates no torque about the center, then the friction force does create a torque. It would seem that for any push, no matter how small, the cube rotates up off the table.

What stops this from happening is that the normal force also exerts a torque, and that torque cancels the torque from friction. If you put little springs on the bottom of the cube, some in front and some in back, you would see the cube tilt forward, compressing the front springs more than the back ones. The distribution of the weight of the cube becomes uneven. The higher normal force near the front of the cube produces a torque. The weight will continue moving up towards the front until this new torque is exactly strong enough to counter the one from friction.

With a sphere, this redistribution of normal force cannot occur because there is only one contact point. Instead, the sphere begins accelerating in the direction of the push. It does this in such a way that the bottom of the sphere remains motionless with respect to the table.

If we push on the sphere with 1N of force, the bigger the friction force is, the smaller the acceleration of the ball. However, the bigger the friction force, the more torque friction exerts. There's exactly one friction force for which the torque from friction and the net force from pushing minus friction produce the same motion of the ball. In this case it happens to be 2/7 N if you push the sphere horizontally on a line through the center.

share|improve this answer
    
The retarding torque for a rolling object is a pretty fun concept. Consider that the force from the ground must be almost perfectly normal to the surface, and certainly closer to normal than the vector from the center of contact pressure and the sphere center. You could actually compute friction coefficients given the material elastic modulus, unlike the normal coefficient of friction model. Good that you mentioned small force and initial change from being at rest. I was lost there. –  Alan Rominger Oct 17 '11 at 4:58

I think that you do have perpetual motion here - an ideal pushed ball with no friction will keep rolling forever (same as Newton's Law). Although I don't quite get the part about the contact point sliding, (I think that the Zass has the right take on that - remember that car tires squeal when sliding over the road, which hopefully is not too often in the cars you drive in).

I think to have 'real' perpetual motion (whatever that would be) you would need to create a machine that not only moved forever, but was able to supply some excess energy, even if only in the form of heat.

share|improve this answer
    
Well, from a completely theoretical point of view, an operating system with 100% efficiency is entirely allowable, and I don't think there's any reason to talk down that possibility. A good example I wouldn't mind asking about myself is that of a quantum system. How much energy does the electron orbiting a nucleus in its ground state dissipate? Well, none. Although this might be disputable, I still think it counts as moving, in addition to the fact that several forms of energy are involved. –  Alan Rominger Oct 17 '11 at 1:59
2  
Agreed that there is lots of stuff that moves 'forever' but the search for 'perpetual motion' usually means the search for some sort of over unity or new physics - like magnetic motors and other such mostly nonsense stuff. –  Tom Andersen Oct 17 '11 at 2:25

Understand the difference between sliding and rolling.

When you run, on a nice dry surface, are your feet sliding? Of course not.

Suppose you are an octopus, and have a sneaker on every foot, and go for a run. Are your sneakers sliding? or are you rolling?

Just think of a rolling wheel as having an infinite number of sneakers, end-to-end. Not one of them is sliding.

Friction only comes into play when things are sliding.

Fun Fact: When gear teeth mesh, like in a car's transmission, would you think they are sliding in and out of contact? Nope, they're rolling.

share|improve this answer
    
"Friction only comes into play when things are sliding." No, that's not right. If something is rolling and accelerating, there's probably friction forces working. That's the case here. –  Mark Eichenlaub Oct 17 '11 at 15:09
    
@Mark: You know what I'm trying to say. Until the force at the interface exceeds the friction force, you can treat the objects as locked together. –  Mike Dunlavey Oct 17 '11 at 17:44
    
No, I don't know what you're saying. Your comment has actually obscured it for me. What is "the force at the interface"? How is a rolling sphere "locked together"? –  Mark Eichenlaub Oct 17 '11 at 23:37
    
@Mark: Assume the pavement's not moving, OK? Something with friction, like a rubber sneaker, is in contact with it. Suppose the downward weight on it is 100lb, and the coefficient of friction is 0.5. So if a sideways force (trying to slide it) is less than 50lb, the sneaker does not move (or accelerate). When things don't accelerate, it's because the sum of forces on them is zero, so the pavement is responding with equal and opposite forces. If the sneaker is more like a tire, rolling forward, the same thing applies. At each point in time, the rubber on pavement is not sliding. –  Mike Dunlavey Oct 18 '11 at 0:24
1  
Yes, I understand. But that does not mean that friction plays no role. The friction forces are there, even when there's no sliding. When there's no sliding, the friction forces simply aren't dissipating energy. –  Mark Eichenlaub Oct 18 '11 at 1:36

protected by Qmechanic Jan 27 '13 at 19:33

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.