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 w ith constant angular speed $‘{\omega }_1^{\prime }$. Another disc of moment of inertia $‘I_2’$ having zero angular speed is placed coaxially on a rotating disc. Now both the. discs are rotating with constant angular speed $‘{\omega }_2’$. 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

Answer: (D)

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
$K_i=\frac{1}{2}{I}_1{w}_1^2+\frac{1}{2}{I}_2{\left(w^{\prime }\right)}^2$
$\Rightarrow K_i=\frac{1}{2}{I}_1{w}_1^2+0=\frac{1}{2}{I}_1{w}_1^2$
Final angular momentum of the system $L_f=\left(I_1+I_2\right)w_2$
Using conservation of angular momentum : $L_i=L_f$
$\therefore I_1{w}_1=\left(I_1+I_2\right)w_2$
$\Rightarrow w_2=\frac{I_1{w}_1}{I_1+I_2}$
Thus final kinetic energy of the system
$K_f=\frac{1}{2}\left(I_1+I_2\right)w_2^2$
$\Rightarrow K_f=\frac{1}{2}\left(I_1+I_2\right)\frac{I_1^2{w}_1^2}{{\left(I_1+I_2\right)}^2}=\frac{1}{2}\frac{I_1^2{w}_1^2}{I_1+I_2}$
Kinetic energy lost $\Delta K=K_i-K_f$
Or $\Delta K=\frac{1}{2}{I}_1{w}_1^2-\frac{I_1^2{w}_1^2}{2\left(I_1+I_2\right)}$
Or $\Delta K=\frac{1}{2}{I}_1{w}_1^2\left[1-\frac{I_1}{I_1+I_2}\right]$
$\Rightarrow \Delta K=\frac{1}{2}{w}_1^2\left[\frac{I_1{I}_2}{I_1+I_2}\right]$