A uniform disk of mass m and radius R rolls w | vedclass

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(B) Using the conservation of energy,the potential energy at the top equals the total kinetic energy at the bottom:$mgl\sin	heta = \frac{1}{2}mv^2 +

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$\sqrt{3m^2R^2gl\sin	heta}$

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(B) Using the conservation of energy,the potential energy at the top equals the total kinetic energy at the bottom:$mgl\sin	heta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = \frac{1}{2}mR^2$ and $\omega = \frac{v}{R}$ for rolling without slipping:$mgl\sin	heta = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$Solving for $v$,we get $v = \sqrt{\frac{4gl\sin	heta}{3}}$.The angular momentum $L$ about the point of contact is given by $L = I_{cm}\omega + mvR$:$L = (\frac{1}{2}mR^2)(\frac{v}{R}) + mvR = \frac{1}{2}mvR + mvR = \frac{3}{2}mvR$Substituting the value of $v$:$L = \frac{3}{2}mR\sqrt{\frac{4gl\sin	heta}{3}} = \sqrt{\frac{9}{4}m^2R^2 \cdot \frac{4gl\sin	heta}{3}} = \sqrt{3m^2R^2gl\sin	heta}$.

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