Two coaxial discs, having moments of inertia I1 and I1/2, are… | JEE Main 2019 (10 Apr, Shift 1)

Q62JEE Main 2019 · 10 APR · SHIFT 1Medium

Angular Momentum & its Conservation

Two coaxial discs, having moments of inertia I1 and I1/2, are rotating with respective angular velocities $\omega$ 1 and $\omega$ 1/2, about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef - Ei) is:

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Answer: (a)

Conservation of $L$ : $I_1\omega_1 + \frac{I_1}{2}\cdot\frac{\omega_1}{2} = \frac{3I_1}{2}\omega \Rightarrow \omega = \frac{5}{6}\omega_1$ .
$E_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}\cdot\frac{I_1}{2}\cdot\frac{\omega_1^2}{4} = \frac{9}{16}I_1\omega_1^2$
$E_f = \frac{1}{2}\cdot\frac{3I_1}{2}\cdot\frac{25}{36}\omega_1^2 = \frac{25}{48}I_1\omega_1^2$
$E_f - E_i = \frac{25-27}{48}I_1\omega_1^2 = -\frac{I_1\omega_1^2}{24}$ , i.e. energy lost $= \frac{I_1\omega_1^2}{24}$ .
So the answer is (a) $\frac{I_1\omega_1^2}{24}$ (in magnitude).

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