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

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Question

(B) For a body rolling without slipping down an inclined plane,the conservation of mechanical energy states that the potential energy at the top equal

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a body rolling without slipping down an inclined plane,the conservation of mechanical energy states that the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom.$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$Since the body is a solid cylinder,its moment of inertia about the central axis is $I = \frac{1}{2}MR^2$.For rolling without slipping,$\omega = \frac{v}{R}$.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 speed of its centre of mass when the cylinder reaches its bottom?

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