QJEEM24Phf13c: JEE Main 2024 Physics - Class 11 JEEM24Phf13c Solution: A thin circular disc of mass and radius is rotating in a…

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $L_{\text{initial}}$ and $L_{\text{final}}$ .

The initial angular momentum of the system is given by:

$L_{\text{initial}} = I_{\text{initial}} \cdot \omega$

Here, $I_{\text{initial}}$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

$I_{\text{initial}} = \frac{1}{2} M R^2$

Thus,

$L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega$

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $\omega'$ . The moment of inertia of the second disc is:

$I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2$

The combined moment of inertia of the system after placing the second disc is:

$I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2$

Thus, the final angular momentum of the system is:

$L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega'$

By the conservation of angular momentum:

$L_{\text{initial}} = L_{\text{final}}$

This simplifies to:

$\left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega'$

Solving for $\omega'$ :

$\omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega$

So, the new angular velocity of the system is:

$\omega' = \frac{2}{3} \omega$

Thus, the correct answer is:

Option D: $\frac{2}{3} \omega$