Consider a 20, kg uniform circular disk of ra | vedclass

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(C) Given: Mass $m = 20\, kg$,Radius $R = 0.2\, m$,Force $F = 20\, N$,Initial angular speed $\omega_0 = 0$,Final angular speed $\omega = 50\, rad/s$.T

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$20$

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(C) Given: Mass $m = 20\, kg$,Radius $R = 0.2\, m$,Force $F = 20\, N$,Initial angular speed $\omega_0 = 0$,Final angular speed $\omega = 50\, rad/s$.The torque $	au$ applied by the force $F$ is $	au = F \cdot R$.The moment of inertia of the disk about its center is $I = \frac{1}{2} mR^2$.Using $	au = I\alpha$,the angular acceleration $\alpha$ is:$\alpha = \frac{F \cdot R}{\frac{1}{2} mR^2} = \frac{2F}{mR} = \frac{2 	imes 20}{20 	imes 0.2} = \frac{2}{0.2} = 10\, rad/s^2$.Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\Delta	heta$:$(50)^2 = 0^2 + 2(10)\Delta	heta$$2500 = 20\Delta	heta$$\Delta	heta = 125\, rad$.Since one revolution corresponds to $2\pi \approx 6.28\, rad$,the number of revolutions $n$ is:$n = \frac{\Delta	heta}{2\pi} = \frac{125}{6.28} \approx 19.90$.Rounding to the nearest integer,$n = 20$.

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