As the masses are gently placed we can apply conservation of angular momentum.
Initial moment of Inertia I1=MR2
Final moment of inertia I2=MR2+(2m)R2
By conservation of angular momentum
I1ω1=I2ω2⇒MR2×2=(M+2m)R2×ω2
⇒ω2=M+2m2Mrads−1
A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rad s^-1 in a horizontal plane about an axis vertical to…
JEE Main 2022 — Physics Mechanics
2022mcqeasy
A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2rads−1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads−1).
Official previous-year question
Held on 26 Jun 2022 · Verified 6 Jul 2026.
Options
- A
(M+m)M
- B
2M(M+2m)
- C
(M+2m)2M
- D
M2(M+2m)
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