A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is [2025]

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Solution

Q.
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is [2025]

1 $\frac{2}{5}$

2 $\frac{5}{2}$

3 $\frac{3}{4}$

4 $\frac{4}{3}$

Ans.
(2)

${KE}_{(T)} = \frac{1}{2} m v^{2}$

${KE}_{(R)} = \frac{1}{2} \cdot \frac{2}{5} m R^{2} \cdot \frac{v^{2}}{R^{2}} = \frac{1}{2} m v^{2} (\frac{2}{5})$

So, $\frac{{KE}_{(T)}}{{KE}_{(R)}} = \frac{5}{2}$

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