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

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(C) The potential energy of the solid cylinder at height $h$ is $U = Mgh$.When the cylinder reaches the bottom,its total kinetic energy $(K.E.)$ is th

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

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(C) The potential energy of the solid cylinder at height $h$ is $U = Mgh$.When the cylinder reaches the bottom,its total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy:$K.E. = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$Since the cylinder rolls without slipping,$\omega = \frac{v}{R}$. For a solid cylinder,the moment of inertia $I = \frac{1}{2} M R^2$.Substituting these into the kinetic energy equation:$K.E. = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{1}{2} M R^2) (\frac{v}{R})^2$$K.E. = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$By the law of conservation of energy,the potential energy at the top equals the kinetic energy at the bottom:$Mgh = \frac{3}{4} M v^2$Solving for $v$:$v^2 = \frac{4}{3} gh$$v = \sqrt{\frac{4}{3} gh}$

$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?

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