A flywheel having mass 3 kg and radius 5 m | vedclass

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(A) Let $M = 3 \ kg$ be the mass of the flywheel, $R = 5 \ m$ be its radius, and $m = 3 \ kg$ be the hanging mass. The moment of inertia of the flywhe

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let $M = 3 \ kg$ be the mass of the flywheel, $R = 5 \ m$ be its radius, and $m = 3 \ kg$ be the hanging mass. The moment of inertia of the flywheel is $I = \frac{1}{2}MR^{2}$.By the law of conservation of energy, the potential energy lost by the hanging mass equals the total kinetic energy gained by the system (flywheel + mass).$mgh = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$Since $v = \omega R$, we have $\omega = \frac{v}{R}$. Substituting $I$ and $\omega$:$mgh = \frac{1}{2}(\frac{1}{2}MR^{2})(\frac{v}{R})^{2} + \frac{1}{2}mv^{2} = \frac{1}{4}Mv^{2} + \frac{1}{2}mv^{2}$Given $m = 3 \ kg, M = 3 \ kg, h = 3 \ m, g = 10 \ m/s^{2}$:$3 	imes 10 	imes 3 = \frac{1}{4}(3)v^{2} + \frac{1}{2}(3)v^{2} = \frac{3}{4}v^{2} + \frac{2}{4}(3)v^{2} = \frac{9}{4}v^{2}$$90 = \frac{9}{4}v^{2} \implies v^{2} = 40 \ m^{2}/s^{2}$.The kinetic energy of the flywheel is $K.E._{flywheel} = \frac{1}{2}I\omega^{2} = \frac{1}{2}(\frac{1}{2}MR^{2})(\frac{v}{R})^{2} = \frac{1}{4}Mv^{2}$.$K.E._{flywheel} = \frac{1}{4} 	imes 3 	imes 40 = 30 \ J$.

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