Question

A thin circular ring of mass $ M $ and radius $ R $ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ \omega $. Four small spheres each of mass $ m $ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

• A. \( \left( \frac{M + 4m}{M} \right) \omega \)
• B. \( \frac{M}{4m} \omega \)
• C. \( \left( \frac{M}{M + 4m} \right) \omega \)
• D. \( \left( \frac{M}{M - 4m} \right) \omega \)

Hint:

In rotational motion, when additional masses are added to a rotating object, the total moment of inertia increases, causing the angular velocity to decrease. Use the conservation of angular momentum to calculate the final angular velocity.

Answer 1

The Correct Option is C

To solve the problem, we need to consider the conservation of angular momentum, which states that the initial angular momentum of the system must be equal to the final angular momentum of the system, provided no external torques act on it.

The initial angular momentum of the ring, \(L_i\), can be calculated as:

\(L_i = I_i \cdot \omega\)

where \(I_i\) is the initial moment of inertia of the ring, calculated as:

\(I_i = M \cdot R^2\)

So, the initial angular momentum is:

\(L_i = M \cdot R^2 \cdot \omega\)

When the four small spheres, each with mass \(m\), are added to the ring, the new moment of inertia \(I_f\) becomes:

\(I_f = I_i + 4 \cdot m \cdot R^2 = M \cdot R^2 + 4m \cdot R^2\)

\(I_f = (M + 4m) \cdot R^2\)

The final angular momentum \(L_f\), assuming conservation of angular momentum, is the same as the initial angular momentum:

\(L_f = L_i\)

Thus,

\(I_f \cdot \omega_f = M \cdot R^2 \cdot \omega\)

Substitute the expression for \(I_f\):

\((M + 4m) \cdot R^2 \cdot \omega_f = M \cdot R^2 \cdot \omega\)

Solve for \(\omega_f\):

\(\omega_f = \frac{M \cdot R^2 \cdot \omega}{(M + 4m) \cdot R^2}\)

The radius terms cancel out, giving:

\(\omega_f = \left( \frac{M}{M + 4m} \right) \omega\)

Thus, the new angular velocity of the ring with the spheres attached is:
\(\left( \frac{M}{M + 4m} \right) \omega\)

Answer 2

The initial moment of inertia of the ring is: \[ I_{\text{ring}} = M R^2 \] When the small spheres of mass \( m \) are placed at the opposite ends of two mutually perpendicular diameters, their moment of inertia with respect to the axis of rotation is: \[ I_{\text{spheres}} = 2 \times m R^2 \] The total moment of inertia of the system is: \[ I_{\text{total}} = I_{\text{ring}} + I_{\text{spheres}} = M R^2 + 2 m R^2 = (M + 2m) R^2 \] Since angular momentum is conserved, we have: \[ I_{\text{initial}} \omega_{\text{initial}} = I_{\text{final}} \omega_{\text{final}} \] Substitute the values: \[ M R^2 \omega = (M + 2m) R^2 \omega_{\text{final}} \] Simplifying, we get: \[ \omega_{\text{final}} = \left( \frac{M}{M + 4m} \right) \omega \]
Thus, the new angular velocity is \( \left( \frac{M}{M + 4m} \right) \omega \).