A uniform solid cylinder of mass M and radius | vedclass

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Question

(A) By the law of conservation of energy,the loss in potential energy of the two masses equals the gain in kinetic energy of the system.Loss in potent

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$\frac{1}{R}\sqrt {8mgh/(M + 4m)} $

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(A) By the law of conservation of energy,the loss in potential energy of the two masses equals the gain in kinetic energy of the system.Loss in potential energy = $2 	imes (mgh) = 2mgh$.Gain in kinetic energy = Kinetic energy of the rotating cylinder + Kinetic energy of the two falling masses.$K.E. = \frac{1}{2} I \omega^2 + 2 	imes (\frac{1}{2} m v^2)$.For a solid cylinder,$I = \frac{1}{2} M R^2$. Also,$v = R \omega$.$K.E. = \frac{1}{2} (\frac{1}{2} M R^2) \omega^2 + m (R \omega)^2 = \frac{1}{4} M R^2 \omega^2 + m R^2 \omega^2 = R^2 \omega^2 (\frac{M}{4} + m) = R^2 \omega^2 (\frac{M + 4m}{4})$.Equating energy: $2mgh = R^2 \omega^2 (\frac{M + 4m}{4})$.$8mgh = R^2 \omega^2 (M + 4m)$.$\omega^2 = \frac{8mgh}{R^2 (M + 4m)}$.$\omega = \frac{1}{R} \sqrt{\frac{8mgh}{M + 4m}}$.

$A$ uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses $m$ are suspended with the help of two ropes wrapped around the cylinder. If the masses fall down through a distance $h$,the angular velocity of the cylinder will be

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