A solid disc of radius 20 , text{cm} and mass | vedclass

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(D) Given: Radius $R = 20 \, 	ext{cm} = 0.2 \, 	ext{m}$,Mass $M = 10 \, 	ext{kg}$,Initial angular velocity $\omega_i = 600 \, 	ext{rpm} = \frac{60

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given: Radius $R = 20 \, 	ext{cm} = 0.2 \, 	ext{m}$,Mass $M = 10 \, 	ext{kg}$,Initial angular velocity $\omega_i = 600 \, 	ext{rpm} = \frac{600 	imes 2\pi}{60} \, 	ext{rad/s} = 20\pi \, 	ext{rad/s}$.Final angular velocity $\omega_f = 0 \, 	ext{rad/s}$,Time $\Delta t = 10 \, 	ext{s}$.The moment of inertia of the disc is $I = \frac{1}{2}MR^2 = \frac{1}{2} 	imes 10 	imes (0.2)^2 = 5 	imes 0.04 = 0.2 \, 	ext{kg m}^2$.The torque required is $	au = \frac{\Delta L}{\Delta t} = \frac{I(\omega_f - \omega_i)}{\Delta t}$.Magnitude of torque $|	au| = \frac{0.2 	imes (20\pi - 0)}{10} = \frac{4\pi}{10} = 0.4\pi = 4\pi 	imes 10^{-1} \, 	ext{Nm}$.Comparing with $x\pi 	imes 10^{-1} \, 	ext{Nm}$,we get $x = 4$.

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