Two cylindrical hollow drums of radii $R$ and $2R$ and of a common height $h$ are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise) respectively. Their axes,which are fixed,are parallel and in a horizontal plane separated by $3R$. They are now brought into contact.
$(a)$ Show the frictional forces just after contact.
$(b)$ Identify forces and torques external to the system just after contact.
$(c)$ What would be the ratio of final angular velocities when friction ceases?

(D) The figure shows the situation given in the question. At the point of contact,the velocity of the smaller drum is $v_1 = R\omega$ (upwards) and the velocity of the larger drum is $v_2 = 2R\omega$ (downwards). Due to this relative velocity,a frictional force $f$ acts on the smaller drum in the downward direction and on the larger drum in the upward direction.
$(b)$ The external forces acting on the system are the normal forces at the contact point and the reaction forces at the fixed axes. Let $F$ be the normal force at the contact point. The reaction forces at the axes are $F'$ and $F$. The net external force on the system is zero. The external torque about the center of the smaller drum is $\tau = F \times 3R$ (anti-clockwise).
$(c)$ Let $\omega_1$ and $\omega_2$ be the final angular velocities of the smaller and larger drum respectively. When friction ceases,there is no relative motion at the point of contact.
Hence,the tangential speeds must be equal: $R\omega_1 = 2R\omega_2$.
Therefore,the ratio of the final angular velocities is $\frac{\omega_1}{\omega_2} = 2$.