A ring of mass M and radius R is rotating with angular speed ω about… | MHT CET 2026 (19 May, Shift 2)

Q12MHT CET 2026 · 19 MAY · SHIFT 2Medium

Angular Momentum & its Conservation

A ring of mass $M$ and radius $R$ is rotating with angular speed $\omega$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$ . These masses can move radially outwards along two mass-less rods fixed on the ring as shown in figure. At some instant the angular speed of the system is $\frac{8\omega}{9}$ and one of the masses is at a distance of $\frac{3}{5}R$ from point $O$ . At this instant, the distance of the other mass from point $O$ is

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Answer: (d)

No external torque: $I_i\omega=I_f\cdot\frac{8\omega}{9}$ , with $I_i=MR^2$ (masses at O).
$I_f=MR^2+\frac M8\left(\frac{3R}{5}\right)^2+\frac M8x^2=\frac98MR^2$
$\frac{x^2}{8}=\frac{R^2}{8}-\frac{9R^2}{200}=\frac{16R^2}{200}\Rightarrow x^2=\frac{16R^2}{25}$
$x=\frac{4R}{5}$ , so the answer is (d).

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Angular Momentum & its Conservation question from MHT CET 2026 (19 May, Shift 2)

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