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Question

(B) By the law of conservation of mechanical energy,the potential energy at the top is equal to the sum of translational and rotational kinetic energy

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) By the law of conservation of mechanical energy,the potential energy at the top is equal to 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 cylinder rolls without slipping,$v = R\omega$,so $\omega = \frac{v}{R}$.The moment of inertia of a solid cylinder about its central axis is $I = \frac{1}{2}mR^2$.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$$v^2 = \frac{4gh}{3}$$v = \sqrt{\frac{4gh}{3}}$

$A$ solid cylinder rolls without slipping down an inclined plane of height $h$. The velocity of the cylinder when it reaches the bottom is

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