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

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(C) For a body rolling without slipping down an inclined plane,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{

Vedclass · Accepted Answer

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

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(C) For a body rolling without slipping down an inclined plane,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the body is a solid cylinder,its moment of inertia about the central axis is $I = \frac{1}{2}MR^2$ and the condition for rolling without slipping is $v = R\omega$.Substituting these into the energy equation:$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$$Mgh = \frac{3}{4}Mv^2$$v^2 = \frac{4}{3}gh$$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 velocity of its center of mass when the cylinder reaches the bottom of the incline?

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