A solid cylinder of mass m and radius R rolls | vedclass

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(B) Since the cylinder rolls without slipping, the work done by friction is zero. Therefore, we can apply the principle of conservation of mechanical

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$20$

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(B) Since the cylinder rolls without slipping, the work done by friction is zero. Therefore, we can apply the principle of conservation of mechanical energy.At the top, the total energy is purely potential: $E_i = mgh$.At the bottom, the total energy is the sum of translational and rotational kinetic energy: $E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid cylinder, the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling, $\omega = \frac{v}{R}$.Equating $E_i = E_f$:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$v^2 = \frac{4gh}{3}$$v = \sqrt{\frac{4 	imes 10 	imes 30}{3}} = \sqrt{400} = 20 \ m \ s^{-1}$.

$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $30 \ m$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $[$use $g=10 \ m \ s^{-2}]$ (in $m \ s^{-1}$)

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