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Question

(A) For a solid sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy.$K = \frac{1}{2

Vedclass · Accepted Answer

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

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(A) For a solid sphere rolling without slipping,the total kinetic energy $K$ is the sum of translational and rotational kinetic energy.$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$ for rolling without slipping:$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$By the law of conservation of energy,the initial kinetic energy must be equal to the potential energy at height $h$ (assuming it just reaches the top with zero velocity):$\frac{7}{10}mv^2 = mgh$$v^2 = \frac{10}{7}gh$$v = \sqrt{\frac{10}{7}gh}$

$A$ solid sphere rolls on a surface with a translational velocity $v \ m/s$. It climbs up a curved surface without slipping. The minimum value of $v$ required to reach the height $h$ is:

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