# Angular acceleration of a cylinder rotating around another

1. Dec 14, 2016

### Happiness

For Q11(b), what is the relation between the angular acceleration $\alpha$ of the bottom right cylinder and its horizontal acceleration $a_x$?

I get $\alpha=\frac{a_x}{\sqrt{3}R}$, which is half the given answer (7.84) below.

After the bottom right cylinder rolls around the top cylinder by an angle of rotation $\theta$ about its (the bottom right cylinder's) own center of mass, the cylinders will look as follows:

(The bottom left cylinder is omitted in the drawing.)

Let the axis of symmetry of the figure of 3 cylinders at $t=0$ be $x=0$. (In other words, the point of contact of the two bottom cylinders has an $x$ coordinate of 0).

Then the $x$ coordinate of the center of the bottom right cylinder, $x = 2R\sin\phi$, where $\phi=30^\circ+\theta$.

$\dot{x}=2R\cos\phi\dot{\phi}$

$\ddot{x}=2R(\cos\phi\ddot{\phi}-\sin\phi\dot{\phi}^2)=2R(\cos\phi\alpha-\sin\phi\omega^2)$, where $\alpha=\ddot{\theta}$ is the angular acceleration of the bottom right cylinder and $\omega=\dot{\theta}$ is its angular velocity.

At $t=0$, $\phi=30^\circ$ and $\omega=0$. Thus $\ddot{x}=2R(\cos 30^\circ\alpha)$. And we get $\alpha=\frac{a_x}{\sqrt{3}R}$.

I believe the given answer is wrong because I suspect there is an approximation involved when it uses an infinitesimal distance $d$ in deriving (7.84).

Last edited: Dec 14, 2016
2. Dec 15, 2016

### TSny

I don't believe that $\ddot{\theta}$ equals the angular acceleration $\alpha$ of the bottom right cylinder.