Question

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rad s^–1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rad s^–1)

• A. \(\frac{M}{M+m}\)
• B. \(\frac{M+2m}{2M}\)
• C. \(\frac{2M}{M+2m}\)
• D. \(\frac{2(M+2m)}{M}\)

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of conservation of angular momentum. When no external torque is acting on a system, the angular momentum of the system remains conserved.

Initially, we have a thin circular ring of mass \(M\) and radius \(R\) rotating with an angular velocity \(\omega = 2 \text{ rad s}^{-1}\). The moment of inertia of the ring about its central axis is given by:

I_i = MR^2

The initial angular momentum \(L_i\) is:

L_i = I_i \cdot \omega = MR^2 \cdot 2

Now, when two masses \(m\) are attached to the opposite ends of the diameter of the ring, the new moment of inertia \(I_f\) becomes:

I_f = MR^2 + 2m \left(\frac{R}{2}\right)^2

Here, both masses are situated at a distance \(\frac{R}{2}\) from the center of the ring (since they are added at the diameter end). Therefore, the new moment of inertia becomes:

I_f = MR^2 + 2m \cdot \frac{R^2}{4} = MR^2 + \frac{mR^2}{2}

By simplifying further:

I_f = (M + \frac{m}{2})R^2

According to the conservation of angular momentum:

L_i = L_f

Thus,

MR^2 \cdot 2 = (M + \frac{m}{2})R^2 \cdot \omega_f

Solving for the final angular velocity \(\omega_f\):

\omega_f = \frac{2M}{M + \frac{m}{2}}

Now substituting the condition \(m = 2\), we simplify:

\omega_f = \frac{2M}{M + 2m}

This shows that the correct answer is indeed \(\frac{2M}{M+2m}\) , matching the provided correct option.