JEE Main 2024 — Rotational Motion Question with Solution

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 $${\omega }^{\prime }$$. 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:

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

By the conservation of angular momentum:

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

This simplifies to:

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

Solving for $${\omega }^{\prime }$$:

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

So, the new angular velocity of the system is:

$${\omega }^{\prime }=\frac{2}{3}\omega$$

Thus, the correct answer is:

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