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(C) For a solid sphere rolling without slipping down an inclined plane, the total kinetic energy $K$ at the bottom is the sum of translational and rot

Vedclass · Accepted Answer

$\frac{5 h}{7}$

Vedclass · Answer

(C) For a solid sphere rolling without slipping down an inclined plane, the total kinetic energy $K$ at the bottom is the sum of translational and rotational kinetic energy.$K = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since $I = \frac{2}{5}mR^2$ and $v = R\omega$, we have $K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.By conservation of energy, the potential energy at height $h$ equals the total kinetic energy at the bottom:$mgh = \frac{7}{10}mv^2 \implies v^2 = \frac{10}{7}gh$.When the ball is thrown vertically upwards with velocity $v$, it undergoes motion under gravity. At the maximum height $h'$, the final velocity is $0$. Using the kinematic equation $v_f^2 = v_i^2 - 2gh'$:$0 = v^2 - 2gh' \implies h' = \frac{v^2}{2g}$.Substituting $v^2 = \frac{10}{7}gh$:$h' = \frac{10/7 gh}{2g} = \frac{5}{7}h$.

Similar Questions

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