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Question

(A) According to the law of conservation of energy,the total initial kinetic energy (translational + rotational) must be equal to the potential energy

Vedclass · Accepted Answer

$v \ge \sqrt{10gh/7}$

Vedclass · Answer

(A) According to the law of conservation of energy,the total initial kinetic energy (translational + rotational) must be equal to the potential energy at the maximum height $h$.Total Kinetic Energy $K = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and for pure rolling,$\omega = v/R$.Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.Equating this to the potential energy at height $h$ $(PE = mgh)$:$\frac{7}{10}mv^2 = mgh$.Solving for $v$,we get $v^2 = \frac{10gh}{7}$,so $v = \sqrt{\frac{10gh}{7}}$.Since the sphere must reach at least height $h$,the minimum velocity is $v \ge \sqrt{\frac{10gh}{7}}$.

$A$ solid sphere is rolling on a frictionless surface. It moves with a translational velocity $v \ m/s$ and climbs an inclined plane as shown in the figure. What should be the minimum value of $v$?

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