A solid sphere of radius 4text{ cm} and mass | vedclass

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(D) Initial angular velocity $\omega_0 = 1200	ext{ rpm} = \frac{1200 	imes 2\pi}{60} = 40\pi	ext{ rad/s}$.Angular acceleration $\alpha = \frac{\Del

Vedclass · Accepted Answer

$0.0128\pi	ext{ Nm}$,$100$

Vedclass · Answer

(D) Initial angular velocity $\omega_0 = 1200	ext{ rpm} = \frac{1200 	imes 2\pi}{60} = 40\pi	ext{ rad/s}$.Angular acceleration $\alpha = \frac{\Delta\omega}{\Delta t} = \frac{0 - 40\pi}{10} = -4\pi	ext{ rad/s}^2$.Moment of inertia of a solid sphere $I = \frac{2}{5}mR^2 = \frac{2}{5} 	imes 5 	imes (0.04	ext{ m})^2 = 2 	imes 0.0016 = 0.0032	ext{ kg m}^2$.Torque applied $	au = I|\alpha| = 0.0032 	imes 4\pi = 0.0128\pi	ext{ Nm}$.Angular displacement $	heta = \omega_0 t + \frac{1}{2}\alpha t^2 = 40\pi(10) + \frac{1}{2}(-4\pi)(10)^2 = 400\pi - 200\pi = 200\pi	ext{ rad}$.Number of rotations $N = \frac{	heta}{2\pi} = \frac{200\pi}{2\pi} = 100$.

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