Q178 / 180 · COMEDK 2023 (Morning Shift) · Physics
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$\left(\dfrac{M}{M-4 m}\right) \omega$
- Correct answerC$\left(\dfrac{M}{M+4 m}\right) \omega$
- D$\dfrac{M}{4 m} \omega$
Solution
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 two mutually perpendicular diameters, they are at a distance $R$ from the axis of rotation. The new moment of inertia of the system becomes $I_2 = MR^2 + 4(mR^2) = (M + 4m)R^2$.
Since no external torque acts on the system, the angular momentum is conserved, so $L_1 = L_2$.
$MR^2 \omega = (M + 4m)R^2 \omega'$, where $\omega'$ is the new angular velocity.
Solving for $\omega'$, we get $\omega' = \dfrac{MR^2 \omega}{(M + 4m)R^2} = \left( \dfrac{M}{M + 4m} \right) \omega$.
Answer: $\left(\dfrac{M}{M+4 m}\right) \omega$
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 two mutually perpendicular diameters, they are at a distance $R$ from the axis of rotation. The new moment of inertia of the system becomes $I_2 = MR^2 + 4(mR^2) = (M + 4m)R^2$.
Since no external torque acts on the system, the angular momentum is conserved, so $L_1 = L_2$.
$MR^2 \omega = (M + 4m)R^2 \omega'$, where $\omega'$ is the new angular velocity.
Solving for $\omega'$, we get $\omega' = \dfrac{MR^2 \omega}{(M + 4m)R^2} = \left( \dfrac{M}{M + 4m} \right) \omega$.
Answer: $\left(\dfrac{M}{M+4 m}\right) \omega$
Hence, the correct answer is option: 'C'.