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

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(B) For a solid cylinder rolling without slipping,the total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy: $K.E. =

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$\sqrt{\frac{4gh}{3}}$

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(B) For a solid cylinder rolling without slipping,the total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy: $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ Since the moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2$ and the condition for rolling without slipping is $v = R\omega$ (or $\omega = v/R$): $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$ $K.E. = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ By the law of conservation of energy,the loss in potential energy $(P.E.)$ equals the gain in kinetic energy: $mgh = \frac{3}{4}mv^2$ $v^2 = \frac{4gh}{3}$ $v = \sqrt{\frac{4gh}{3}}$

$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is

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