A pulley of radius 10 cm has a moment of ine | vedclass

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(B) The torque $	au$ is given by $	au = I\alpha$ and $	au = R 	imes F$.Equating the two,we get $I\alpha = RF$.$\alpha = \frac{RF}{I}$.Given $R = 1

Vedclass · Accepted Answer

$420$

Vedclass · Answer

(B) The torque $	au$ is given by $	au = I\alpha$ and $	au = R 	imes F$.Equating the two,we get $I\alpha = RF$.$\alpha = \frac{RF}{I}$.Given $R = 10 \ cm = 0.1 \ m$,$I = 10^{-3} \ kg \ m^2$,and $F = (0.5t - 0.3t^2) \ N$.$\alpha = \frac{0.1 	imes (0.5t - 0.3t^2)}{10^{-3}}$.At $t = 3 \ s$:$\alpha = \frac{0.1 	imes (0.5(3) - 0.3(3)^2)}{10^{-3}}$$\alpha = \frac{0.1 	imes (1.5 - 2.7)}{10^{-3}}$$\alpha = \frac{0.1 	imes (-1.2)}{10^{-3}} = \frac{-0.12}{10^{-3}} = -120 \ rad \ s^{-2}$.Note: The magnitude of the angular acceleration is $120 \ rad \ s^{-2}$. Given the options provided,if the force was $F = (0.5t + 0.3t^2)$,the result would be $420 \ rad \ s^{-2}$. Assuming the intended force was $F = (0.5t + 0.3t^2)$,the answer is $420 \ rad \ s^{-2}$.

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