A disc of radius R and mass M is rolling hori | vedclass

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(C) The disc is rolling on a horizontal surface without slipping,so it possesses both translational kinetic energy $(K_t = \frac{1}{2} Mv^2)$ and rota

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$\frac{1}{2} \frac{v^2}{g}$

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(C) The disc is rolling on a horizontal surface without slipping,so it possesses both translational kinetic energy $(K_t = \frac{1}{2} Mv^2)$ and rotational kinetic energy $(K_r = \frac{1}{2} I\omega^2)$.When the disc moves up a smooth inclined surface,there is no friction. Therefore,the torque due to friction is zero,and the rotational kinetic energy remains constant throughout the motion.Only the translational kinetic energy is converted into gravitational potential energy as the disc climbs the incline.Applying the law of conservation of energy for the translational part:$\frac{1}{2} Mv^2 = Mgh$Solving for $h$:$h = \frac{v^2}{2g}$

$A$ disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

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