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(C) At $t=0$,$\omega=0$. At $t=\sqrt{\pi} 	ext{ s}$,$\omega = \alpha t = \frac{2}{3} \sqrt{\pi} 	ext{ rad/s}$.The linear velocity of the center of m

Vedclass · Accepted Answer

$0.52$

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(C) At $t=0$,$\omega=0$. At $t=\sqrt{\pi} 	ext{ s}$,$\omega = \alpha t = \frac{2}{3} \sqrt{\pi} 	ext{ rad/s}$.The linear velocity of the center of mass is $v_{cm} = \omega R = \frac{2}{3} \sqrt{\pi} 	imes 1 = \frac{2}{3} \sqrt{\pi} 	ext{ m/s}$.The angle rotated by the disk is $	heta = \frac{1}{2} \alpha t^2 = \frac{1}{2} 	imes \frac{2}{3} 	imes (\sqrt{\pi})^2 = \frac{\pi}{3} = 60^{\circ}$.At this instant,the stone is at a height $y = R - R \cos 	heta = 1 - \cos 60^{\circ} = 1 - 0.5 = 0.5 	ext{ m}$ from the ground.The velocity of the stone relative to the ground is the vector sum of the velocity of the center of mass and the tangential velocity relative to the center of mass. The vertical component of this velocity is $v_y = v_{cm} \sin 	heta = (\frac{2}{3} \sqrt{\pi}) \sin 60^{\circ} = \frac{2}{3} \sqrt{\pi} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3\pi}}{3} 	ext{ m/s}$.The additional height gained by the stone after detaching is $h = \frac{v_y^2}{2g} = \frac{(\frac{\sqrt{3\pi}}{3})^2}{2 	imes 10} = \frac{3\pi / 9}{20} = \frac{\pi}{60} 	ext{ m}$.The total maximum height from the plane is $H = y + h = 0.5 + \frac{\pi}{60} = \frac{1}{2} + \frac{\pi/6}{10}$.Comparing with $\frac{1}{2} + \frac{x}{10}$,we get $x = \frac{\pi}{6} \approx \frac{3.14}{6} \approx 0.523$. Thus,$x \approx 0.52$.

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