A disc of mass M and radius R rolls on a hori | vedclass

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(B) The total kinetic energy of the rolling disc on the horizontal surface is the sum of its translational and rotational kinetic energies.$K = K_{tra

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The total kinetic energy of the rolling disc on the horizontal surface is the sum of its translational and rotational kinetic energies.$K = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$For a disc,the moment of inertia $I = \frac{1}{2}MR^2$ and for pure rolling,$\omega = \frac{v}{R}$.Substituting these values,we get:$K = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$When the disc rolls up the inclined plane,its total kinetic energy is converted into potential energy at the maximum height $h$.$Mgh = \frac{3}{4}Mv^2$$h = \frac{3v^2}{4g}$Thus,the height to which the disc will rise is $\frac{3v^2}{4g}$.

$A$ 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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