A circular disk of moment of inertia It is rotating in a horizontal… | NEET 2010 (AIPMT, Prelims)

Q86NEET 2010 · AIPMT · PRELIMSMedium

Angular Momentum & its Conservation

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 onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega$ . The energy lost by the initially rotating disc to friction is

Show answer and solution

Answer: (d)

Conservation of $L$ : $\omega = \frac{I_t\omega_i}{I_t+I_b}$ .
$K_i = \frac{1}{2}I_t\omega_i^2$ , $K_f = \frac{1}{2}(I_t+I_b)\omega^2 = \frac{1}{2}\frac{I_t^2\omega_i^2}{I_t+I_b}$ .
Loss $= \frac{1}{2}I_t\omega_i^2\left(1 - \frac{I_t}{I_t+I_b}\right) = \frac{1}{2}\frac{I_bI_t}{I_t+I_b}\omega_i^2$
So the answer is (d).

Report a mistake

Angular Momentum & its Conservation question from NEET 2010 (AIPMT, Prelims)

More questions on Angular Momentum & its Conservation