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exscape
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Homework Statement
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.
Homework Equations
[itex]\sum F = m a_{cm}[/itex]
[itex]\sum \tau = I \alpha_{cm}[/itex]
[itex]\alpha = a R[/itex] for pure rolling
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, [itex]\tau_C = F_{friction} R[/itex] 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?