# Time Period of a Rolling Cylinder

In this post, we have discussed a problem based on the time period of a rolling cylinder. Try the problem yourself first, then read the solution.

The Problem: Time Period of a Rolling Cylinder

A solid uniform cylinder of radius (r) rolls without sliding along the inside the surface of a hollow cylinder of radius (R), performing small oscillations. Determine time period.

Solution:

Translational kinetic energy + rotational kinetic energy + potential energy= constant

$$\frac{1}{2}mv^2+{\frac{1}{2}I\omega^2+mg(R-r)(1-cos\theta)}=C$$
Now $$I=1/2mr^2$$
$$3/4m(dx/dt)^2+mg(R-r)\theta^2/2=C$$
Differentiating with respect to time,

$$\frac{3}{2}m(\frac{d{^2}x}{dt{^2}})^+mg(R-r)\theta\frac{d\theta}{dt}$$
Now, $$x=(R-r)\theta$$
$$\frac{3}{2} d^2x/dt^2(R-r)d\theta/dt+gxd\theta/dt=0$$
Cancelling (\frac{d\theta}{dt}) throughout

$$\frac{d^2x}{dt^2}+\frac{2}{3}\frac{gx}{R-r}=0$$
this is the equation for SHM, with
$$\omega^2=\frac{2}{3}\frac{g}{R-r}$$
$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{3(R-r)}{2g}}$$

### One comment on “Time Period of a Rolling Cylinder”

1. Arabinda Ghosh says:

Sir the terms moment of inertia, angular momentum should be mentioned about which point

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