Determine the natural frequency of oscillation of a half solid cylinder of mass ‘m’ and radius ‘r’ when it is slightly displaced from mean position and released [USE ENERGY METHOD].

$\therefore$ $cos \theta = \frac{x}{\frac{4r}{3 \pi }}$

$x = \frac{4r}{3 \pi} cos \theta$ $\rightarrow$ $h = \frac{4r}{3 \pi } - \frac{4r}{3 \pi 1} cos \theta$

$\rightarrow$ Energy Method,

$K.E = \frac{1}{2} Ip \theta^2$

$PE = mgh = mg [\frac{4r}{3 \pi} - \frac{4r}{3 \pi } cos \theta ]$

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

$\frac{1}{2} Ip 20 \theta + mg (\frac{4r}{3 \pi}) sin \theta \theta = 0$

$Ip \theta + (\frac{4mgr}{3 \pi Ip}) \theta = 0$

$\theta + (\frac{4mgr}{3 \pi Ip}) \theta = 0$

$\therefore$ $W_n = \sqrt{ \frac{4mgr}{3 \pi Ip}} $ rad/sec

Where, Ip = mass moment of inertia of half solid cyclinder about which can be detrmined as follows:

$\rightarrow$ By parallel Axis theorem,

$Ip = Ig + m(Gp)^2$

But, $I_o = I_G + m (\frac{4r}{3 \pi })^2$

$I_G = \frac{mr^2}{2} – m (\frac{4r}{3 \pi })^2$

$\therefore$ $I_G = m [ \frac{r^2}{2} - \frac{16 r^2}{9\pi ^2}] $

$\therefore$ $I_p = m [ \frac{r^2}{2} - \frac{16r^2}{9\pi ^2}] + m [r - \frac{4r}{3 \pi }]^2$

$I_p = m [ \frac{r^2}{2} - \frac{16r^2}{9\pi ^2}] + m [r^2 + \frac{16r^2}{9\pi ^2} - \frac{8r^2}{3 \pi}]$

$= mr^2 [ \frac{1}{2} - \frac{16}{9 \pi ^2} + 1 + \frac{16}{9 \pi ^2} - \frac{8}{3 \pi }]$

$Ip = mr^2 [\frac{3}{2} - \frac{8}{3 \pi }] = mr^2 [ \frac{9 \pi – 16}{6 \pi }]$

$\therefore$ $W_n = \sqrt{ \frac{4mgr}{3 \pi \times mr^2 [\frac{9 \pi – 16}{6 \pi}]}}$

$W_n \sqrt{ \frac{8g}{r (9\pi – 16)_c}}$ rad/sec.