Question

A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ${\omega }_i$. Another disk of moment of inertia $I_h$ 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. $\frac{1}{2}\frac{I_b^2}{(I_t+I_b)}{\omega }_i^2$
• B. $\frac{1}{2}\frac{I_t^2}{2(I_t+I_b)}{\omega }_i^2$
• C. $\frac{I_b-I_t}{(I_t+I_b)}{\omega }_i^2$
• D. $\frac{1}{2}\frac{I_b{I}_t}{(I_t+I_b)}{\omega }_i^2$

Answer 1

Answer: (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,
$I_t{\omega }_i=\left(I_t+I_b\right){\omega }_f$
or ${\omega }_f=\frac{I_t{\omega }_i}{\left(I_t+I_b\right)}\dots (i)$
Initial energy, $E_i=\frac{1}{2}{I}_t{\omega }_i^2\dots (ii)$
Final energy, $E_f=\frac{1}{2}\left(I_t+I_b\right){\omega }_f^2\dots (iii)$
Substituting the value of ${\omega }_f$ from equation (i) in equation (iii), we get Final energy, $E_f=\frac{1}{2}\left(I_t+I_b\right){\left(\frac{I_i{\omega }_i}{I_t+I_b}\right)}^2$ $=\frac{1}{2}\frac{I_t^2{\omega }_i^2}{2\left(I_t+I_b\right)}\dots (iv)$
Loss of energy, $\Delta E=E_i-E_f$
$\frac{1}{2}{I}_t{\omega }_i^2-\frac{1}{2}\frac{I_t^2{\omega }_i^2}{I_t+I_b}$ (Using (ii) and (iv))
$\frac{{\omega }_i^2}{2}\left(I_t-\frac{I_t^2}{\left(I_t+I_b\right)}\right)$
$=\frac{{\omega }_i^2}{2}\left(\frac{I_t^2+I_b{I}_t-I_t^2}{\left(I_t+I_b\right)}\right)$
$=\frac{1}{2}\frac{I_b{I}_t}{\left(I_t+I_b\right)}{\omega }_i^2$