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 $I_b$ is dropped coaxially on to 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. $\frac{1}{2}\frac{I_b^2}{(I_t+I_b)}{{\omega }_t}^2$
• B. $\frac{1}{2}\frac{I_t^2}{(I_t+I_b)}{{\omega }_i}^2$
• C. $\frac{I_b-I_t}{(I_t+I_b)}{{\omega }_t}^2$
• D. $\frac{1}{2}\frac{I_b{I}_t}{(I_t+I_b)}{{\omega }_t}^2$

Answer 1

Answer: (D)

$\frac{1}{2}\frac{I_b{I}_t}{(I_t+I_b)}{{\omega }_t}^2$

Loss of energy, $△E=\frac{1}{2}{I}_t{\omega }_i^2-\frac{1}{2}\frac{I_r^2{\omega }_i^2}{(I_t+I_b)}$

$=\frac{1}{2}\frac{I_b{I}_t{\omega }_i^2}{(I_t+I_b)}$

$\therefore △E=\frac{1}{2}\frac{I_b{I}_t}{(I_t+I_b)}{\omega }_i^2$

Hence, the answer is $\frac{1}{2}\frac{I_b{I}_t}{(I_t+I_b)}{\omega }_i^2.$