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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}##.

The given answer:

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).

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