Question

A disc of moment of inertia $'I_1'$ is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane w ith constant angular speed $‘\omega_1'$. 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 $L _{ i }= I _{1} w _{1}+ I _{2} w'$
where $w '=0$
$\therefore L _{ i }= I _{1} w _{1}+0= I _{1} w _{1}$
Thus initial kinetic energy of the system
$K_{i}=\frac{1}{2} I_{1} w_{1}^{2}+\frac{1}{2} I_{2}\left(w'\right)^{2}$
$\Rightarrow K _{ i }=\frac{1}{2} I _{1} w _{1}^{2}+0=\frac{1}{2} I _{1} w _{1}^{2}$
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 $\Delta K =\frac{1}{2} I _{1} w _{1}^{2}\left[1-\frac{ I _{1}}{ I _{1}+ I _{2}}\right]$
$\Rightarrow \Delta K =\frac{1}{2} w _{1}^{2}\left[\frac{ I _{1} I _{2}}{ I _{1}+ I _{2}}\right]$