# Mechanics of a rolling drum

I have no clue on how to approach this. The professor only discussed centripetal acceleration and angular velocity (As in $2πr\over T$ $= ωr$). Does the acceleration along the axis of the drum act in the same way that centripetal acceleration does? I understand that the angular velocity in this case is $ωR$, and the derivative of that is the angular acceleration, but how to I find $ω$? How does $a$ affect the period, and how does its position (In this case along the axis) affect that?

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You have been given a translational acceleration of $a$. You just need to relate this to the angular version $\alpha$. How can we relate the two? –  Will Jul 1 '13 at 15:10
Here's a hint: what is the relationship between $v$ and $\omega$? Think about where this comes from. Sure it acts on the entire drum, but the drum has symmetry so we can consider the gravitational acceleration through the center of mass. Anyway that's not the problem. You have been given an acceleration $a$, yes you know it's due to gravity but it doesn't matter where is comes from, you know its value. –  Will Jul 1 '13 at 15:20
Would you know what to do if you were given the acceleration of the edge of the drum with respect to the center? –  Will Jul 1 '13 at 15:37
Great, hopefully you understand where that relationship is coming from? So we want to know the acceleration of the edge. Now, if the cylinder's center has acceleration $a$ with respect to the slope, how is the bottom edge accelerating with respect to the center of the cylinder? –  Will Jul 1 '13 at 17:00
Right, so the edge is accelerating with magnitude $a$ with respect to the center and you can now find $\alpha$. –  Will Jul 1 '13 at 18:10

## 1 Answer

There is no oscillation or period on this problem. All you know is the slip condition which means that the center of the drum moves by $v = \omega R$.

Now differentiate this relationship to find $a$ as a function of $\alpha = \dot{\omega}$.

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