A disc of mass 1,kg and radius R is free to r | vedclass

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(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{	ext{disc}} = \frac{1}{2} mR^2$. The mo

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$5$

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(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{	ext{disc}} = \frac{1}{2} mR^2$. The moment of inertia of the particle at distance $R$ is $I_{	ext{particle}} = mR^2$. The total moment of inertia is $I = I_{	ext{disc}} + I_{	ext{particle}} = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.Using the law of conservation of mechanical energy,the loss in potential energy of the particle equals the gain in rotational kinetic energy of the system.The particle moves from the highest point to the lowest point,a vertical distance of $2R$.Loss in potential energy $= mg(2R) = 2mgR$.Gain in rotational kinetic energy $= \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{3}{2} mR^2 ight) \omega^2 = \frac{3}{4} mR^2 \omega^2$.Equating the two: $2mgR = \frac{3}{4} mR^2 \omega^2$.Solving for $\omega^2$: $\omega^2 = \frac{8g}{3R}$.Given $\omega = 4 \sqrt{\frac{x}{3R}}$,so $\omega^2 = 16 \left( \frac{x}{3R} ight) = \frac{16x}{3R}$.Equating the expressions for $\omega^2$: $\frac{16x}{3R} = \frac{8g}{3R} \implies 16x = 8g$.Substituting $g = 10 	ext{ m s}^{-2}$: $16x = 8(10) = 80$.$x = \frac{80}{16} = 5$.

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