# Rolling cylinder or slipping cylinder reaches bottom first?

1. Apr 3, 2017

### vcsharp2003

1. The problem statement, all variables and given/known data
Two identical cylinders are released from the top of two identical inclined planes. If one rolls without slipping and the other slips without rolling then which one will reach the bottom first? How will their speeds compare when they reach bottom of incline?

I am not sure if my attempt is correct, since this is a test question and I cannot verify my answer.

2. Relevant equations
Apply Work Energy Theorem to both situations
ΔKE + ΔPE = Worknet

3. The attempt at a solution
Let θ be the angle of incline with horizontal and h it's height. Let v be the velocity at bottom of incline.

for rolling, since force of friction does no work on rolling cylinder, ∴ Worknet = 0
∴ (0.5 x m x v2 - 0 ) + (0 - mgh) = 0
0.5 x m x v2 = mgh

for sliding, force of friction does work, ∴ Worknet ≠ 0
∴(0.5 x m x v2 - 0 ) + (0 - mgh) = - f x hsinθ
0.5 x m x v2 = mgh - fhsinθ

From above analysis, the velocity at bottom for rolling disk will be greater, which also means the rolling disk will reach bottom first.

2. Apr 3, 2017

### Staff: Mentor

For the rolling cylinder, where has the energy come from (or gone to) in order to make it spin? Think about the energy of rotation.

For the sliding cylinder that does not roll, can there be any friction at all?

3. Apr 3, 2017

### phyzguy

The one that is rolling gains angular momentum, and thus energy of angular motion. You need to include this in your analysis..

4. Apr 3, 2017

### vcsharp2003

That is a very nice point. Since cylinder slips, then friction should not be there. Friction is what causes cylinder to roll by supplying the necessary torque. Right?

In that case, the problem becomes simpler. The initial PE of cylinder in rolling case gets divided into final linear KE + final rotational KE, whereas, in sliding case the same PE goes wholly into it's final linear KE, and therefore the sliding case will have a higher velocity at bottom and also reach the bottom first.

Does above reasoning sound right to you?

5. Apr 4, 2017

### Staff: Mentor

Yes, that's it.

6. Apr 4, 2017

### kuruman

Perhaps a clarification is needed for OP's benefit that by "the speed of the cylinder" is tacitly meant "the speed of the center of mass". In the case of slipping, all points on the cylinder move at the same speed; in the case of rolling without slipping, some points on the cylinder move faster and some slower than the center of mass while the point of contact moves not at all (otherwise there would be slipping). Nevertheless, the argument in post #4 is valid when one considers that $PE=\frac{1}{2}m V^2_{CM}$ when there is only slipping and $PE=\frac{1}{2}m V^2_{CM}+ K_{rot.}$ when there is rolling without slipping.

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