Q.3

Q.4

1] Newton's Method.

Assume, point of vibration at point 'P'

I T = e of all moments.

$+ I_p \theta = -kr \theta \times r$

$I_p \theta = -kr^2 \theta$

$I_p \theta + kr^2 \theta = 0$ . . . . . E O M

$\theta + \frac{kr^2}{I_p} \theta = 0$

To find IP, use parallel axis theorem.

$Ip - IG + m (dist)^2$

$= \frac{mr^2}{2} + mr^2$

$I_p = \frac{3mr^2}{2}$

$w_n = \sqrt{ \frac{kr^2}{Ip}} = \sqrt{ \frac{kr^2}{ \frac{3mr^2}{2}}} - \sqrt{ \frac{2kr^2}{3mr^2}} = \sqrt{ \frac{2k}{3m}} r/s$

Energy Method:

$KE = (KE)_m$ NOTE: Assume pure rolling motion.

$KE = (KE)rot$

$\frac{1}{2} I_p \theta^2$

$\frac{1}{2} [ \frac{mr^2}{2} + m(r)^2] \theta^2$

$= \frac{1}{2} [ \frac{3mr^2}{2} ] \theta^2$

keq.

PF = (PE) spring = $\frac{1}{2} k (r \theta)^2 = \frac{1}{2} \frac{kr^2 \theta^2}{keq}$

$w_n = \sqrt{ \frac{keq}{Ieq}} = \sqrt{ \frac{kr^2}{3mr^2}} = \sqrt{ \frac{2k}{3m}} r/s$

2]

$KE = (KE)_M + (KE)_disc$

$\frac{1 }{2} Mx^2 + \frac{1}{2} I \theta^2$

$= \frac{1}{2} Mx^2 + \frac{1}{2} I \frac{x^2}{r^2}$

$= \frac{1}{2} [ M + \frac{I}{r^2}] x^2$

$KE = \frac{1}{2} mew x^2$

$PE = (PE)k + (PE)2k$

$\frac{1}{2} kx^2 + \frac{1}{2} 2k (r \theta)^2$

$= \frac{1}{2} kx^2 + \frac{1}{2} 2k x^2$

$= \frac{1}{2} [\frac{[k + 2k]} {keq} x^2$

As F.O.M.

$Mx^2 + kx = 0$

So we can conclude that,

$[ m + \frac{I}{r^2}] x^2 + (3k) x = 0$