Question

A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity \(\omega\). Six objects, each of mass m, are kept gently to the opposite ends of three diameters of the ring. The angular velocity of the ring will be

• A. \(\frac{(M+6m)\omega }{M}\)
• B. \(\frac{M\omega }{6m}\)
• C. \(\frac{M\omega }{M+6m}\)
• D. \(\frac{(M-6m)\omega }{M+6m}\)

Hint:

No external torque acts, so angular momentum of the ring plus masses stays constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The masses are placed gently, so no external torque acts and angular momentum is conserved.

Step 2: Moments of inertia
Initially \(I_{1}=Mr^{2}\).
The six masses sit on the rim at distance \(r\), so \(I_{2}=Mr^{2}+6mr^{2}\).

Step 3: Calculation
\(Mr^{2}\omega=(M+6m)r^{2}\omega'\)
\(\omega'=\dfrac{M\omega}{M+6m}\)

Final Answer:
The new angular velocity is \(\frac{M\omega}{M+6m}\). \[ \boxed{\text{(C) }\dfrac{M\omega}{M+6m}} \]