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(B) The total kinetic energy of a rolling disc is the sum of its translational and rotational kinetic energy.Total Kinetic Energy $K = K_{	ext{trans}

Vedclass · Accepted Answer

$\frac{3v^2}{4g}$

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(B) The total kinetic energy of a rolling disc is the sum of its translational and rotational kinetic energy.Total Kinetic Energy $K = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.For a disc,the moment of inertia $I = \frac{1}{2}MR^2$ and the rolling condition is $\omega = \frac{v}{R}$.Substituting these values,$K = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$.By the law of conservation of energy,this kinetic energy is converted into gravitational potential energy at the maximum height $h$.$Mgh = \frac{3}{4}Mv^2$.Solving for $h$,we get $h = \frac{3v^2}{4g}$.

$A$ disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v$,the height to which the disc will rise will be

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