Energy lost in rolling, sliding and torsion

AI Thread Summary
The discussion centers on the energy dynamics involved in three states: rolling, sliding, and torsion of a steel cylinder on a rubber surface. It highlights that the work done in rolling and torsion is often minimal due to negligible resisting forces, while sliding generates significant work due to friction. The deformation of the rubber during these interactions complicates the energy calculations, suggesting that rolling may involve non-zero work. Participants provided formulas to quantify the energy involved, including power, work, and frictional force. Overall, the conversation emphasizes the importance of considering surface interactions when analyzing energy loss in different motion states.
qmul
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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.
 
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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?
 


Mapes said:
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?

Thank you for your fast reply Mapes!

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?
 


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


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


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.
 
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