Question

A disc of moment of inertia ${}^{\prime }{I}_1^{\prime }$ is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed ${}^{\prime }{\omega }_1^{\prime }$. Another disc of moment of inertia ${}^{\prime }{I}_2^{\prime }$ having zero angular speed is placed coaxially on a rotating disc. Now both the discs are rotating with constant angular speed ${}^{\prime }{\omega }_2^{\prime }$. The energy lost by the initial rotating disc is

• A. $\frac{1}{2}\left[\frac{I_1+I_2}{I_1{I}_2}\right]{\omega }_1^2$
• B. $\frac{1}{2}\left[\frac{I_1{I}_2}{I_1-I_2}\right]{\omega }_1^2$
• C. $\frac{1}{2}\left[\frac{I_1-I_2}{I_1{I}_2}\right]{\omega }_1^2$
• D. $\frac{1}{2}\left[\frac{I_1{I}_2}{I_1+I_2}\right]{\omega }_1^2$

Answer 1

The correct option is D $\frac{1}{2}\left[\frac{I_1{I}_2}{I_1+I_2}\right]{\omega }_1^2$

Initial angular momentum of the system $L_i=I_1{w}_1+I_2{w}^{\prime }$ where $w^{\prime }=0$

$\therefore$ $L_i=I_1{w}_1+0=I_1{w}_1$

Thus initial kinetic energy of the system $K_i=\frac{1}{2}{I}_1{w}_1^2+\frac{1}{2}{I}_2(w^{\prime }{)}^2$

$⟹$ $K_i=\frac{1}{2}{I}_1{w}_1^2+0=\frac{1}{2}{I}_1{w}_1^2$

Final angular momentum of the system $L_f=(I_1+I_2)w_2$

Using conservation of angular momentum : $L_i=L_f$

$\therefore$ $I_1{w}_1=(I_1+I_2)w_2$

$⟹$ $w_2=\frac{I_1{w}_1}{I_1+I_2}$

Thus final kinetic energy of the system $K_f=\frac{1}{2}(I_1+I_2)w_2^2$

$⟹$ $K_f=\frac{1}{2}(I_1+I_2)\frac{I_1^2{w}_1^2}{(I_1+I_2{)}^2}=\frac{1}{2}\frac{I_1^2{w}_1^2}{I_1+I_2}$

Kinetic energy lost $\Delta K=K_i-K_f$

Or $\Delta K=\frac{1}{2}{I}_1{w}_1^2-\frac{I_1^2{w}_1^2}{2(I_1+I_2)}$

Or $\Delta K=\frac{1}{2}{I}_1{w}_1^2[1-\frac{I_1}{I_1+I_2}]$

$⟹$ $\Delta K=\frac{1}{2}{w}_1^2[\frac{I_1{I}_2}{I_1+I_2}]$