Question:
A thin circular ring of mass $M$ and radius $R$ is rotating about its central axis with a constant angular velocity $\omega$. Two objects, each of mass $m$, are gently attached to the opposite ends of a diameter of the ring. The new angular velocity of the ring is:
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[width=0.5\linewidth]{85.png}
A thin circular ring of mass $M$ and radius $R$ is rotating about its central axis with a constant angular velocity $\omega$. Two objects, each of mass $m$, are gently attached to the opposite ends of a diameter of the ring. The new angular velocity of the ring is:
[H]
[width=0.5\linewidth]{85.png}
Show Hint
Since angular momentum is conserved, increasing the moment of inertia must decrease the angular velocity. The ratio of new velocity to old is the inverse ratio of their moments of inertia: $\frac{M}{M+2m}$.
Updated On: May 31, 2026
- $\frac{M \omega}{M + 2m}$
- $\frac{(M + 2m)\omega}{M}$
- $\frac{M \omega}{M + m}$
- $\frac{(M - 2m)\omega}{M + 2m}$
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Since no external torque acts on the rotating ring system, the total angular momentum of the system is conserved: $I_1 \omega_1 = I_2 \omega_2$.
Step 2: Meaning
The initial moment of inertia of the ring is $I_1 = MR^2$. Adding two masses, each of mass $m$, at a distance $R$ from the center of rotation changes the final moment of inertia to $I_2$.
Step 3: Analysis
Find the final moment of inertia of the system: \[ I_2 = MR^2 + mR^2 + mR^2 = (M + 2m)R^2 \] Applying the conservation of angular momentum: \[ I_1 \omega_1 = I_2 \omega_2 \implies (MR^2)\omega = (M + 2m)R^2 \omega_2 \] \[ \omega_2 = \frac{M \omega}{M + 2m} \]
Step 4: Conclusion
The new angular velocity of the ring is $\frac{M \omega}{M + 2m}$. Final Answer: (A)
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