A uniform sphere of mass $m$ and radius $r$ rolls without sliding over a horizontal plane, rotating about a horizontal axle $O A$ as shown in Figure. In the process, the centre of the sphere moves with velocity $v$ along a circle of radius $R$ . Find the kinetic energy of the sphere.

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$\frac{7 m}{10} v^{2} [1 + \frac{2 r^{2}}{R^{2}}]$

$\frac{7 m}{10} v^{2} [1 + \frac{2 r^{2}}{7 R^{2}}]$

$\frac{m}{10} v^{2} [1 + \frac{2 r^{2}}{7 R^{2}}]$

$\frac{7 m}{5} v^{2} [1 + \frac{2 r^{2}}{7 R^{2}}]$

answer is B.

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Detailed Solution

The kinetic energy of the sphere due to its rotation about own axis and its motion along a circular path with velocity $v$ .

$∴ K . E = \frac{1}{2} I_{1} {ω_{1}}^{2} + \frac{1}{2} I_{2} {ω_{2}}^{2}$

where $I_{1} =_{} \frac{2}{5} m r^{2}$ and $I_{2} = (\frac{2}{5} m r^{2} + m R^{2})$

$ω_{1} = \frac{v}{r}$ and $ω_{2} = \frac{v}{R}$

Substituting these values in the above equation, we get

$∴ K . E = \frac{7 m}{10} v^{2} [1 + \frac{2 r^{2}}{7 R^{2}}]$

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