A uniform circular disc of mass 1.5 kg and r | vedclass

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(A) The torque $	au$ produced by each force $F$ about the center of the disc is $	au = F \cdot r_{\perp}$,where $r_{\perp}$ is the perpendicular dis

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(A) The torque $	au$ produced by each force $F$ about the center of the disc is $	au = F \cdot r_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the center to the line of action of the force.For an equilateral triangle inscribed in a circle of radius $R$,the distance from the center to each side is $R \cos(60^{\circ}) = R/2$.Thus,the torque due to one force is $	au = F \cdot (R/2)$.Since there are three such forces acting in the same rotational sense,the total torque is $	au_{	ext{net}} = 3 \cdot F \cdot (R/2) = 1.5 \cdot F \cdot R$.Given $F = 0.5 \ N$ and $R = 0.5 \ m$,$	au_{	ext{net}} = 1.5 \cdot 0.5 \cdot 0.5 = 0.375 \ N \cdot m$.The moment of inertia of the disc about its central axis is $I = \frac{1}{2} M R^2 = \frac{1}{2} \cdot 1.5 \cdot (0.5)^2 = 0.1875 \ kg \cdot m^2$.Using the angular impulse-momentum theorem,$	au_{	ext{net}} \cdot t = I \cdot \omega$,where $t = 1 \ s$.$\omega = \frac{	au_{	ext{net}} \cdot t}{I} = \frac{0.375 \cdot 1}{0.1875} = 2 \ 	ext{rad } s^{-1}$.

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