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