The center of a disk of radius r and mass m i | vedclass

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(A) Let the disk be displaced by an angle $	heta$ from the equilibrium position. The displacement of the center of the disk along the arc is $x = (R-

Vedclass · Accepted Answer

$\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}$

Vedclass · Answer

(A) Let the disk be displaced by an angle $	heta$ from the equilibrium position. The displacement of the center of the disk along the arc is $x = (R-r)	heta$.The total energy $E$ of the system is the sum of the potential energy of the spring,the gravitational potential energy of the disk,and the kinetic energy of the disk (translational + rotational).$E = \frac{1}{2} k x^2 + mg(R-r)(1 - \cos 	heta) + \frac{1}{2} m v^2 + \frac{1}{2} I \omega_{rot}^2$Since $x = (R-r)	heta$,$v = (R-r)\dot{	heta}$,and for rolling without slipping,$\omega_{rot} = \frac{v}{r} = \frac{(R-r)\dot{	heta}}{r}$. The moment of inertia of the disk about its center is $I = \frac{1}{2}mr^2$.$E = \frac{1}{2} k (R-r)^2 	heta^2 + mg(R-r) \frac{	heta^2}{2} + \frac{1}{2} m (R-r)^2 \dot{	heta}^2 + \frac{1}{2} (\frac{1}{2}mr^2) (\frac{(R-r)\dot{	heta}}{r})^2$$E = \frac{1}{2} [k(R-r)^2 + mg(R-r)] 	heta^2 + \frac{1}{2} [m(R-r)^2 + \frac{1}{2}m(R-r)^2] \dot{	heta}^2$$E = \frac{1}{2} [k(R-r)^2 + mg(R-r)] 	heta^2 + \frac{3}{4} m(R-r)^2 \dot{	heta}^2$Since the total energy is conserved,$\frac{dE}{dt} = 0$:$[k(R-r)^2 + mg(R-r)] 	heta \dot{	heta} + \frac{3}{2} m(R-r)^2 \dot{	heta} \ddot{	heta} = 0$Dividing by $\dot{	heta}$ (assuming $\dot{	heta} 
eq 0$):$\ddot{	heta} + \frac{k(R-r)^2 + mg(R-r)}{\frac{3}{2} m(R-r)^2} 	heta = 0$$\ddot{	heta} + \frac{2}{3} [\frac{k}{m} + \frac{g}{R-r}] 	heta = 0$Comparing with the standard $SHM$ equation $\ddot{	heta} + \omega^2 	heta = 0$,we get:$\omega = \sqrt{\frac{2}{3} [\frac{k}{m} + \frac{g}{R-r}]}$Thus,the correct option is $(A)$.

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