A thin circular disc of mass $\mathrm{M}$ and radius $\mathrm{R}$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. If another disc of same dimensions but of mass $\mathrm{M} / 2$ is placed gently on the first disc co-axially, then the new angular velocity of the system is :

<p>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 $L_{\text{final}}$.</p> <p>The initial angular momentum of the system is given by:</p> <p>$L_{\text{initial}} = I_{\text{initial}} \cdot \omega$</p> <p>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:</p> <p>$I_{\text{initial}} = \frac{1}{2} M R^2$</p> <p>Thus,</p> <p>$L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega$</p> <p>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:</p> <p>$$ I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2 $$</p> <p>The combined moment of inertia of the system after placing the second disc is:</p> <p>$$ I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 $$</p> <p>Thus, the final angular momentum of the system is:</p> <p>$$ L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega' $$</p> <p>By the conservation of angular momentum:</p> <p>$L_{\text{initial}} = L_{\text{final}}$</p> <p>This simplifies to:</p> <p>$$ \left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega' $$</p> <p>Solving for $\omega'$:</p> <p>$$ \omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega $$</p> <p>So, the new angular velocity of the system is:</p> <p>$\omega' = \frac{2}{3} \omega$</p> <p>Thus, the correct answer is:</p> <p>Option D: $\frac{2}{3} \omega$</p>