$\rightarrow$ $Arc \ CP = Arc CP$

$R \ \theta = r \ \phi$

$\therefore$ $\phi = \frac{R \ \theta}{r}$

$\rightarrow$ Translatory displacement of center of cylinder = $(R – r) \theta$

$\rightarrow$ Total rotational displacement of cylinder = $\phi = 0$ = Relative displacement.

$\rightarrow$ $KE = (KE)_{Translational} + (KE)_{rotational}$

$\frac{1}{2} \ m\ [ (R-r) \theta]^2 + \frac{1}{2} I_G (\phi - \theta)^2$

$\rightarrow$ $PE = mg [ (R-r) – (R-r) \ cos \ \theta ] = mg \ (R-r) [ 1- cos \ \theta]$

$\frac{d}{dt} (KE + PE) = 0$

$\frac{d}{dt} [ \frac{1}{2} m [ (R-r) \theta]^2 + \frac{1}{2} I_G (\phi - \theta)^2 + mg (R-r) (1- cos \theta)] = 0$

$\frac{d}{dt} [ \frac{1}{2} (R-r)^2 \theta^2 + \frac{1}{2} I_G [ \frac{R\theta}{r} - \theta]^2 + mg (R-r) (1-cos \theta)] = 0$

$\frac{d}{dt} [ \frac{1}{2} m (R-r)^2 \theta^2 + \frac{1}{2} [\frac{1}{2} mr^2] . \frac{1}{r^2} [ R-r]^2 \theta^2 + mg (R-r) (1-cos \theta)] = 0$

$\frac{d}{dt} [\frac{1}{2} m (R-r)^2 \theta^2 + \frac{1}{4} m (R-r)^2 \theta^2 + mg (R-r) (1 –cos \theta)] = 0$

$\frac{d}{dt} [\frac{3}{4} m (R-r)^2 \theta^2 + mg (R-r) (1-cos \theta)] = 0$

$\frac{3}{4} m (R-r)^2 2 \theta \theta + mg (R-r) sin \theta . \theta = 0$

$\frac{3}{2}m (R-r)^2 \theta + mg (R-r) \theta = 0$

$\theta + [ \frac{mg (R-r)}{\frac{3}{2} m (R-r)^2}] \theta = 0$

$\theta + [ \frac{2g}{3 (R-r)}] \theta = 0$

$W_n = \sqrt{ \frac{2g}{2(R-r)}}$ rad/sec.