# Energy lost in rolling, sliding and torsion

1. Apr 1, 2009

### qmul

Energy "lost" in rolling, sliding and torsion

Hey everybody,

I have a question concerning three states and the state of energy in them.

1. Rolling
2. Sliding
3. Torsion

For example:

If I have a steel cylinder with diameter d on an axis (no friction in the axis) rolling horizontally over a rubber surface, is the energy (work done), just the simple horizontal force in the axis times distance?
What happens, if I block the cylinder and slide it over the surface? Third question would be just to rotate the cylinder without horizontal movement at all.

Any ideas would be greatly appreciated ;-)

Phil

EDIT: defined the contacting surfaces. Surface cylinder = steel, 2nd Surface = rubber block.

Last edited: Apr 1, 2009
2. Apr 1, 2009

### Mapes

Re: Energy "lost" in rolling, sliding and torsion

Hi qmul, welcome to PF. The work done in the first and third cases is relatively low and often assumed to be zero; the reason is that there is essentially no resisting force under your assumption of no axis-cylinder friction.

The work done in the second case is the product of the distance and the frictional force between the cylinder and the surface. Does this answer your question?

3. Apr 1, 2009

### qmul

Re: Energy "lost" in rolling, sliding and torsion

To complicate maters - the surface is rubber and steel for the cylinder. So there is quite a bit of deformation done to the rubber while rolling / sliding / torsion. The force for rolling, measured in the axis, is significantly lower than in sliding. I guess these new factors will change the question, slightly, don't they?

4. Apr 1, 2009

### Mapes

Re: Energy "lost" in rolling, sliding and torsion

Sure, they'd imply a non-zero work for the first case, which is what you observed.

5. Apr 6, 2009

### qmul

Re: Energy "lost" in rolling, sliding and torsion

are there some general formulas for each of the cases - no matter if it is zero-or non-zero work?

6. Apr 6, 2009

### Mapes

Re: Energy "lost" in rolling, sliding and torsion

Yes:

$$P=Fv$$

$$W=Fd$$

$$F=\mu_k N$$

where P is power, F is the resisting frictional force, v is velocity, d is distance, $\mu_k$ is the coefficient of kinetic (moving) friction, and N is the weight of the moving object.