When a uniform solid sphere of mass M rolls without slipping from rest down an inclined plane such that its centre of mass descends through a height h, its translational kinetic energy will be

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$\frac{2}{7}$ mgh

$\frac{mgh}{2}$

mgh

$\frac{5}{7}$ mgh

answer is B.

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Detailed Solution

$K = \frac{1}{2} m V^{2} + \frac{1}{2} I ω^{2} = m g h \begin{array}{l} \Rightarrow \frac{1}{2} m V^{2} + \frac{1}{2} (\frac{2}{5} m V^{2}) = m g h [∵ I = \frac{2}{5} m R^{2}, ω = \frac{V}{R}] \\ \Rightarrow \frac{1}{2} m V^{2} (\frac{7}{5}) = m g h \\ \Rightarrow K = \frac{1}{2} m V^{2} = \frac{5}{7} m g h \end{array}$

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