Question

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with an angular speed $\omega$. Two particles each of mass $m$ are now attached at diametrically opposite points. The angular speed of the ring will become

• A. $\dfrac{\omega M}{M+2m}$
• B. $\dfrac{\omega M}{M+m}$
• C. $\dfrac{\omega(M-2m)}{M}$
• D. $\dfrac{\omega(M-2m)}{M+2m}$

Hint:

When no external torque acts, angular momentum of the system remains conserved.

Answer 1

The Correct Option is A

Step 1: Moment of inertia before attaching masses.
For a thin circular ring:
\[ I_1 = Mr^2 \]

Step 2: Moment of inertia after attaching two masses.
Each particle of mass $m$ at radius $r$ contributes $mr^2$.
\[ I_2 = Mr^2 + 2mr^2 = (M+2m)r^2 \]

Step 3: Apply conservation of angular momentum.
\[ I_1 \omega = I_2 \omega' \]

Step 4: Solve for new angular speed.
\[ \omega' = \frac{Mr^2 \omega}{(M+2m)r^2} = \frac{\omega M}{M+2m} \]

Step 5: Conclusion.
The new angular speed is $\dfrac{\omega M}{M+2m}$.