COMEDK · Physics · 8. Rotational Motion
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 \(\left(\dfrac{M+4 m}{M}\right) \omega\)
- B \(\left(\dfrac{M}{M-4 m}\right) \omega\)
- C \(\left(\dfrac{M}{M+4 m}\right) \omega\)
- D \(\dfrac{M}{4 m} \omega\)
Answer & Solution
Correct Answer
(C) \(\left(\dfrac{M}{M+4 m}\right) \omega\)
Step-by-step Solution
Detailed explanation
The initial moment of inertia of the ring about the axis passing through its centre and perpendicular to its plane is \(I_1 = MR^2\). The initial angular momentum of the system is \(L = I_1 \omega = MR^2 \omega\). When four small spheres of mass \(m\) are placed at the ends of…
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