A solid sphere of mass M and radius R is roll | vedclass

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(B) For a body rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{

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$\sqrt{\frac{4}{3}gh}$

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(B) For a body rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the body is a solid sphere,its moment of inertia $I = \frac{2}{5}mR^2$ and the condition for rolling without slipping is $v = R\omega$.Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$.$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.Solving for $v$: $v^2 = \frac{10}{7}gh$,which gives $v = \sqrt{\frac{10}{7}gh}$.However,using the general formula $v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$ where $K^2 = \frac{2}{5}R^2$ for a solid sphere:$v = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{2gh}{7/5}} = \sqrt{\frac{10}{7}gh}$.Note: The provided option $(b)$ $\sqrt{\frac{4}{3}gh}$ corresponds to a hollow sphere or a ring-like distribution where $I = \frac{2}{3}mR^2$ or $I = mR^2$. For a solid sphere,the correct result is $\sqrt{\frac{10}{7}gh}$.

$A$ solid sphere of mass $M$ and radius $R$ is rolling down an inclined plane of length $L$ and height $h$ without slipping. What will be the velocity of its center of mass upon reaching the bottom?

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