Rolling a cylinder down a slope.

[EthanB]

Homework Statement

A solid cylinder of radius R sits at the top of a slope of angle theta. When it rolls down, what is the minimum coefficient of friction (k) required to make the cylinder roll without slipping?

Homework Equations

Fx: mgsin(theta) - kmgcos(theta) = ma
Torque: kmgRcos(theta) = mR^2 * Alpha

The Attempt at a Solution

In order for the cylinder to roll without slipping, the tangential acceleration at the edge must be twice the acceleration of the center of mass (right?). Therefore, we solve for where Alpha = 2a/R.

From Torque:
kmgRcos(theta) = mR^2 * 2a/R
kgcos(theta) = 2a

Substituting for a from Fx:
kgcos(theta) = 2gsin(theta) - 2kgcos(theta)
k = (2/3)tan(theta)

The book's answer is (1/2)tan(theta).

 

[EthanB]

Oops... I was using the wrong moment of inertia. It won't let me delete the post. Thanks anyway.

 

[EthanB]

Actually, I was using the right moment of inertia. It's a hollow cylinder. Ok, so can anyone help me out?

 

[Doc Al]

In order for the cylinder to roll without slipping, the tangential acceleration at the edge must be twice the acceleration of the center of mass (right?).

That is incorrect. For rolling without slipping the instantaneous speed of the point of contact of the cylinder with the surface of the incline must be zero (with respect to the surface). Translate that into a mathematical statement so you can apply it.