Oscillation of a cylinder hanging from a table A cylinder is tied to another object that is moving in circular motion on top of a table. Initially the object moves at an angular velocity that prevents the cylinder from moving. If the cylinder is pulled slightly downwards and starts oscillating, what is the period of the oscillation? (Given that the mass of the cylinder is M, the mass of the revolving object is m, and the initial radius of the revolution is R.)
 A: Viewing in the rotation frame, the mass $m$ has a centrifugal force pointing radically outward which is balance by the gravitational force of cylinder through the connection string. Under equilibrium, the system established a constant angular momentum $L_e$ (rotation radius $R$). When radius is drawn slightly smaller, the speed of rotation increases to maintain the constant angular momentum (due to the greater centripetal force during the pulling process in aspect of force). The greater speed renders a larger centrifugal force pulling the mass outward ignite the oscillation. We will analysis these several aspects.

*

*The balance
In the rotational frame, the centrifugal force
$$
   m R \omega_e^2 = M g.\\
    \omega_e = \sqrt{ \frac{M g}{m R} }.
$$
The angular momentum:
$$\tag{1}
   L_e = m R^2 \omega_e =  m R^2 \sqrt{ \frac{M g}{m R} } =\sqrt{m M g R^3}.
$$

*The force

After the equilibrium is established, the force is balanced rendering a constant angular momentum.
Therefore, for small change in the rotational radius $R \to R+ dR$ generates a restoring force $dF$:
$$\tag{2}
F = m R \omega^2 = \frac{L_e^2}{mR^3}\\
dF = -3 \frac{L_e^2}{mR^4} dR = -\left(3 \frac{ M g R^3}{R^4}\right) dR = - k dR.
$$
where $k$ is the restoring force constant.


*The oscillation

Eq.(2) provides the force constant
$$
 k = 3 \frac{ M g}{R}.\\
$$
The angular frequency of oscillation
$$
   \omega_{OSC} = \sqrt{\frac{k}{m} } = \sqrt{\frac{3 M g}{m R}}.
$$
