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Question

(D) To climb the inclined surface to a height $h$ while rolling without slipping,the total kinetic energy of the rolling sphere at the bottom must be

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$\sqrt {\frac{10}{7}gh} $

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(D) To climb the inclined surface to a height $h$ while rolling without slipping,the total kinetic energy of the rolling sphere at the bottom must be equal to the potential energy at the top.The total kinetic energy of a rolling body is given by $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the sphere rolls without slipping,$\omega = \frac{v}{R}$. For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$.Substituting these values,$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.Equating this to the potential energy at height $h$,we get $\frac{7}{10}mv^2 = mgh$.Solving for $v$,we get $v^2 = \frac{10}{7}gh$,which implies $v = \sqrt{\frac{10}{7}gh}$.

$A$ solid sphere is rolling on a surface as shown in the figure,with a translational velocity $v \, ms^{-1}$. If it is to climb the inclined surface continuing to roll without slipping,then the minimum velocity for this to happen is:

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