A pulley has the shape of a uniform solid dis | vedclass

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(D) The moment of inertia $I$ of a uniform solid disc about its central axis is given by:$I = \frac{1}{2} mR^2$Substituting the given values ($m = 2 \

Vedclass · Accepted Answer

$50$

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(D) The moment of inertia $I$ of a uniform solid disc about its central axis is given by:$I = \frac{1}{2} mR^2$Substituting the given values ($m = 2 \ kg$,$R = 0.5 \ m$):$I = \frac{1}{2} 	imes 2 	imes (0.5)^2 = 0.25 \ kg \cdot m^2$The torque $	au$ produced by the force $F = 2.5 \ N$ applied at the rim (radius $R = 0.5 \ m$) is:$	au = F 	imes R = 2.5 	imes 0.5 = 1.25 \ N \cdot m$Using the relation $	au = I \alpha$,we find the angular acceleration $\alpha$:$1.25 = 0.25 	imes \alpha$$\alpha = \frac{1.25}{0.25} = 5 \ rad/s^2$Since the pulley starts from rest $(\omega_0 = 0)$,the angular velocity $\omega$ after $t = 10 \ s$ is:$\omega = \omega_0 + \alpha t$$\omega = 0 + 5 	imes 10 = 50 \ rad/s$

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