Question

A round disc of moment of inertia $I_2$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $I_1$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is

• A. $\frac{I_2\omega}{I_1+I_2}$
• B. $\omega$
• C. $\frac{I_1\omega}{I_1+I_2}$
• D. $\frac{(I_1+I_2)\omega}{I_1}$

Answer 1

The Correct Option is C

The problem involves the application of the principle of conservation of angular momentum. Let's analyze the situation step-by-step to arrive at the solution.

1. Initially, only the first disc with moment of inertia $I_1$ is rotating with an angular velocity $\omega$.
2. The second disc with moment of inertia $I_2$ is initially at rest.
3. When the second disc is placed on top of the first disc, they start rotating together about the same axis.
4. The principle of conservation of angular momentum states that if no external torques are acting on a system, the initial angular momentum of the system will be equal to the final angular momentum.

Mathematically, this can be expressed as:

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

Where:

• Initial Angular Momentum, L_{\text{initial}} = I_1 \times \omega
• Final Angular Momentum, L_{\text{final}} = (I_1 + I_2) \times \omega_f

Equating initial and final angular momentum:

I_1 \omega = (I_1 + I_2) \omega_f

Solving for the final angular velocity \omega_f, we get:

\omega_f = \frac{I_1 \omega}{I_1 + I_2}

Thus, the correct answer is \frac{I_1 \omega}{I_1 + I_2}.

This solution efficiently uses the conservation of angular momentum to find the final angular velocity of the combined system of discs.