A solid disc has a diameter of 0.5 m and a m | vedclass

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(D) The moment of inertia of a solid disc is $I = \frac{1}{2}MR^2$.Given $M = 16 \ kg$ and $R = \frac{0.5}{2} = 0.25 \ m = \frac{1}{4} \ m$.$I = \frac

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$\frac{\pi}{4} \ N \cdot m$

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(D) The moment of inertia of a solid disc is $I = \frac{1}{2}MR^2$.Given $M = 16 \ kg$ and $R = \frac{0.5}{2} = 0.25 \ m = \frac{1}{4} \ m$.$I = \frac{1}{2} 	imes 16 	imes (\frac{1}{4})^2 = 8 	imes \frac{1}{16} = 0.5 \ kg \cdot m^2$.The final angular velocity is $\omega = 120 \ rpm = \frac{120 	imes 2\pi}{60} \ rad/s = 4\pi \ rad/s$.The initial angular velocity is $\omega_0 = 0$.The angular acceleration is $\alpha = \frac{\omega - \omega_0}{t} = \frac{4\pi - 0}{8} = \frac{\pi}{2} \ rad/s^2$.The required torque is $	au = I\alpha$.$	au = 0.5 	imes \frac{\pi}{2} = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4} \ N \cdot m$.

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

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