# How does rotational kinetic energy increase in rolling downhill?

1. Nov 25, 2013

### exscape

1. The problem statement, all variables and given/known data

This is not a homework problem. Okay, it is -- but I have solved it correctly already, so the question is not there. I'm just not sure about a detail in one of many solutions.

We have a hollow cylinder with a uniform mass distribution rolling down an incline with some coefficient of static friction μ. It begins at rest. It has mass M and radius R, and so a moment of inertia M R^2. The incline makes an angle θ to the horizontal.

The goal of the problem is to find the linear acceleration of the center of mass, linear velocity when the object has moved a height h downwards (vertically downwards), and also the minimum coefficient of static friction required to avoid slipping.

2. Relevant equations

$\sum F = m a_{cm}$
$\sum \tau = I \alpha_{cm}$
$\alpha = a R$ for pure rolling

3. The attempt at a solution

To solve it, I first wrote a N2L equation for the center of mass, with a component of gravity downhill, and static friction uphill.
Combine that with an equation relating torque relative to the center, $\tau_C = F_{friction} R$ to the linear acceleration as above.

The answers I find are all correct, but I'm not satisfied. If we consider torque relative to the center (of mass), only static friction can provide any torque, since gravity acts though the center of mass. However, this seems to lead to a contradiction.

Does the torque caused by static friction do work on the cylinder, to increase its rotational kinetic energy?
If NO, how can the rotational kinetic energy increase without a torque that does work?
If YES, how can static friction do work here? The total kinetic energy is equal to the work done by gravity (M g h).

I suppose we can "resolve" this dilemma by instead calculating torque relative to the contact point, in which case gravity can now provide a torque. However, it seems to me that this analysis method should be just as valid, so how does one resolve this apparent contradiction?

2. Nov 25, 2013

### haruspex

Yes, the static friction does work in that reference point. The force is applied over the distance rolled. If you change the reference point to be the point of contact then the work is done by gravity instead.

3. Nov 25, 2013

### exscape

Huh, alright. So not only is static friction doing work, but it is doing positive work... I have to admit I didn't really realize that could happen (I tend to think of friction being a source of energy loss only, in the cases where it does work at all.)

The energy is clearly coming from the gravitational potential energy though, right? It just feels strange to me that a force other than gravity can convert gravitational potential energy into kinetic energy.

4. Nov 25, 2013

### haruspex

Analysis of rotation - angular momentum, torque.. - usually depends on the reference point you choose. The answers will be the same (subject to certain rules) but it will look different superficially. Here, which force is doing the work changes.

5. Nov 26, 2013

### willem2

But the static friction is decreasing the non-rotational kinetic energy at the same time, so the total work it does is still 0. If you let the cylinder slide down the slope without friction, it would move faster than it moves when rolling.

Static friction really can't do any work, because it pushes against a part of the cylinder that is not moving.

6. Nov 26, 2013

### exscape

Ah, right, thanks. So it is indeed valid to say that it does both positive and negative work, as long as it does no net work, even if it acts over zero distance?

There is just one thing still bothering me. If we do the analysis from the contact point/a point on the incline, friction cannot provide any torque... but it must still reduce the linear acceleration, or the final velocity would be a factor √2 higher. The linear acceleration (and therefore final linear kinetic energy) will be lower due to the static friction, so does it not do negative work for the linear motion, and zero work for the rotational, for a net negative?
How can the final linear kinetic energy be less with friction than without, if the friction does no work?

7. Nov 26, 2013

### Staff: Mentor

Just because a force acts to accelerate an object (via N2L) does not mean that that force did work on the system. Whether you choose to analyze torques about the center of mass or the contact point is up to you. Since the friction force acts on a part of the cylinder that is instantaneously at rest, no work is done. (The static friction is not an energy source.)

Note that the total mechanic energy is not less. Friction does no work, but it does enable some linear KE to be transformed into rotational KE.