A circular disc has a mass of 1 kg and radius | vedclass

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(D) The work done to stop the rotating disc is equal to the change in its rotational kinetic energy.Given: Mass $M = 1\ kg$,Radius $R = 40\ cm = 0.4\

Vedclass · Accepted Answer

$158$

Vedclass · Answer

(D) The work done to stop the rotating disc is equal to the change in its rotational kinetic energy.Given: Mass $M = 1\ kg$,Radius $R = 40\ cm = 0.4\ m$,Frequency $f = 10\ rev/s$.The moment of inertia of the disc about its central axis is $I = \frac{1}{2}MR^2 = \frac{1}{2} 	imes 1 	imes (0.4)^2 = 0.08\ kg\cdot m^2$.The angular velocity is $\omega = 2\pi f = 2 	imes \pi 	imes 10 = 20\pi\ rad/s$.The initial rotational kinetic energy is $K_i = \frac{1}{2}I\omega^2 = \frac{1}{2} 	imes 0.08 	imes (20\pi)^2$.$K_i = 0.04 	imes 400\pi^2 = 16\pi^2$.Using $\pi^2 \approx 9.87$,$K_i = 16 	imes 9.87 = 157.92\ J \approx 158\ J$.Since the final kinetic energy is $0$,the work done to stop the disc is $W = K_f - K_i = 0 - 158 = -158\ J$.The magnitude of the work done is $158\ J$.

$A$ circular disc has a mass of $1\ kg$ and radius $40\ cm$. It is rotating about an axis passing through its centre and perpendicular to its plane with a speed of $10\ rev/s$. The work done in joules in stopping it would be ...... $J$

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