Question

A thin circular disc of mass \(M\) and radius \(R\) 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 \(\frac{M}{2}\) is placed gently on the first disc co-axially, then the new angular velocity of the system is:

• A. \(\frac{4}{5}\omega\)
• B. \(\frac{5}{4}\omega\)
• C. \(\frac{2}{3}\omega\)
• D. \(\frac{3}{2}\omega\)

Answer 1

The Correct Option is C

To solve this problem, we need to apply the principle of conservation of angular momentum. When no external torques are acting on a system, the angular momentum before and after an event must be the same.

Let's break down the problem step-by-step:

1. Initially, only the first disc with mass \(M\) and radius \(R\) is rotating with an angular velocity \(\omega\). The moment of inertia \(I_1\) for a disc rotating about its center is given by \(\frac{1}{2}MR^2\).
2. The initial angular momentum \(L_1\) is: \(L_1 = I_1 \cdot \omega = \frac{1}{2}MR^2 \cdot \omega\)
3. The second disc with mass \(\frac{M}{2}\) and radius \(R\) is placed gently on the first disc, co-axially, causing a new system to form.
4. The moment of inertia \(I_2\) for the second disc is: \(I_2 = \frac{1}{2} \cdot \frac{M}{2} \cdot R^2 = \frac{1}{4}MR^2\)
5. The total moment of inertia \(I_{\text{total}}\) for the new system is: \(I_{\text{total}} = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{4}MR^2 = \frac{3}{4}MR^2\)
6. By conservation of angular momentum, the total angular momentum before and after placing the second disc must be equal: \(L_1 = L_{\text{final}}\)
7. Thus, \(\frac{1}{2}MR^2 \cdot \omega = \frac{3}{4}MR^2 \cdot \omega_{\text{new}}\)
8. Solving for \(\omega_{\text{new}}\), we get: \(\omega_{\text{new}} = \frac{\frac{1}{2} \cdot MR^2 \cdot \omega}{\frac{3}{4} \cdot MR^2} = \frac{2}{3} \omega\)

Therefore, the new angular velocity of the system after the second disc is placed gently is \(\frac{2}{3} \omega\).

The correct answer is: \(\frac{2}{3}\omega\)

Answer 2

Using the law of conservation of angular momentum:
\[I_1\omega = I_2\omega_2.\]
The moment of inertia of the first disc:
\[I_1 = \frac{MR^2}{2}.\]
The combined moment of inertia of both discs:
\[I_2 = \frac{MR^2}{2} + \frac{1}{2} \left(\frac{MR^2}{2}\right) = \frac{3MR^2}{4}.\]
Applying conservation of angular momentum:
\[\frac{MR^2}{2} \times \omega = \frac{3MR^2}{4} \times \omega_2.\]
Simplifying:
\[\omega_2 = \frac{\frac{MR^2}{2}}{\frac{3MR^2}{4}} \times \omega = \frac{2}{3} \times \omega.\]
Thus, the new angular velocity of the system is:
\[\omega_2 = \frac{2}{3} \omega.\]