# Cylinder on inclined plane

## Homework Statement

A block of a certain material begins to slide on an inclined plane when the plane is inclined to an angle of 11.86o. If a solid cyclinder is fashioned from the same material, what will be the maximum angle at which it will roll without slipping on the plane (in degrees)?

## The Attempt at a Solution

drew a free body diagram, and have the following equations

mgsin(theta) + friction = ma
torque = friction*R = I(angular acceleration) --> friction=(I*angular acceleration)/R
mgsin(theta) + (I*angular acceleration)/R = ma
a=R*(angular acceleration) and I = .5MR^2
... substitute those in, but then I get

theta = (inverse sin)[(.5*a)/g] -- how do i solve for a?

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## The Attempt at a Solution

drew a free body diagram, and have the following equations

mgsin(theta) + friction = ma
torque = friction*R = I(angular acceleration) --> friction=(I*angular acceleration)/R
mgsin(theta) + (I*angular acceleration)/R = ma
a=R*(angular acceleration) and I = .5MR^2
... substitute those in, but then I get

theta = (inverse sin)[(.5*a)/g] -- how do i solve for a?
Have you used the relation between $v$ and $\omega$?