Rotational Motion

JEE Main level, single correct, medium

208 of 280

Question 208 of 280

A thin circular ring of mass MM and radius RR is rotating with a constant angular velocity 2rads−12rad{s}^{-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 mm be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads−1rad{s}^{-1}).

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Answer: (C) 2M(M+2m)\dfrac{2M}{\left(M+2m\right)}

As the masses are gently placed we can apply conservation of angular momentum.
Initial moment of Inertia I1=MR2{I}_{1}=M{R}^{2}
Final moment of inertia I2=MR2+(2m)R2{I}_{2}=M{R}^{2}+\left(2m\right){R}^{2}
By conservation of angular momentum
I1ω1=I2ω2{I}_{1}{\omega }_{1}={I}_{2}{\omega }_{2}⇒MR2×2=(M+2m)R2×ω2\Rightarrow M{R}^{2}\times 2=\left(M+2m\right){R}^{2}\times {\omega }_{2}
⇒ω2=2MM+2mrads−1\Rightarrow {\omega }_{2}=\dfrac{2M}{M+2m}rad{s}^{-1}

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