Question

A circular disk of moment of inertia it is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_f$. The energy lost by the initially rotating disc to friction is

• A. $\frac{1}{2}\frac{I^2_b}{(I_t+I_b)}\omega^2_i$
• B. $\frac{1}{2}\frac{I^2_t}{2(I_t+I_b)}\omega^2_i$
• C. $\frac{I_b-I_t}{(I_t+I_b)}\omega^2_i$
• D. $\frac{1}{2}\frac{I_bI_t}{(I_t+I_b)}\omega^2_i$

Answer 1

The Correct Option is D

As no external torque is applied to the system, the angular momentum of the system remains conserved.
$\therefore L_i=L_f$
According to given problem,
$I_t\omega_i=(I_t+I_b)\omega_f$
or $\omega_f=\frac{I_t\omega_i}{(I_t+I_b)}\hspace25mm ...(i)$
Initial energy, $E_i=\frac{1}{2}I_t\omega^2_i\hspace25mm ...(ii)$
Final energy, $E_f=\frac{1}{2}(I_t+I_b)\omega^2_f\hspace25mm ...(iii)$
Substituting the value of $\omega_f$ from equation (i) in equation (iii), we get
Final energy, $E_f=\frac{1}{2}(I_t+I_b)\Big(\frac{I_i\omega_i}{I_t+I_b}\Big)^2$
$=\frac{1}{2}\frac{I^2_t\omega^2_i}{2(I_t+I_b)}\hspace15mm ...(iv)$
Loss of energy, $\Delta E=E_i-E_f$
$\frac{1}{2}I_t\omega^2_i-\frac{1}{2}\frac{I^2_t\omega^2_i}{I_t+I_b}$ (Using (ii) and (iv))
$\frac{\omega^2_i}{2}\Bigg(I_t-\frac{I^2_t}{(I_t+I_b)}\Bigg)=\frac{\omega^2_i}{2}\Bigg(\frac{I^2_t+I_b I^2_t-I^2_t}{(I_t+I_b)}\Bigg)$
$=\frac{1}{2}\frac{I_b I_t}{(I_t + I_b)}\omega^2_i$