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(B) The distance between the centers of disk $A$ and disk $B$ is $2R$. The velocity of the center of mass of disk $B$ is $v = \omega(2R)$.Since disk $

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$5$

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(B) The distance between the centers of disk $A$ and disk $B$ is $2R$. The velocity of the center of mass of disk $B$ is $v = \omega(2R)$.Since disk $B$ rolls without slipping on the circumference of disk $A$,the condition for no slipping at the point of contact is $v = \omega_0 R$,where $\omega_0$ is the angular velocity of disk $B$ about its own center.Substituting $v = 2\omega R$,we get $2\omega R = \omega_0 R$,which implies $\omega_0 = 2\omega$.The angular momentum of disk $B$ about the center of disk $A$ is given by $\vec{L} = \vec{r} 	imes \vec{p} + I_c \vec{\omega}_0$,where $\vec{r}$ is the position vector of the center of mass of $B$ relative to $A$,$\vec{p} = M\vec{v}$ is the linear momentum,and $I_c$ is the moment of inertia of disk $B$ about its center.$\vec{L} = M(2R)v + I_c \omega_0 = M(2R)(2\omega R) + (\frac{1}{2}MR^2)(2\omega) = 4MR^2\omega + MR^2\omega = 5MR^2\omega$.Comparing this with $n M \omega R^2$,we find $n = 5$.

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