A solid sphere is rolling without slipping on a horizontal plane....

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 :

Options

A $\frac{3}{4}$
B $\frac{4}{3}$
C $\frac{5}{2}$
D $\frac{2}{5}$

Correct answer

C. $\frac{5}{2}$

Step-by-step solution

\begin{aligned} & \mathrm{KE}_{(T)}=\frac{1}{2} m v^2 \\ & \mathrm{KE}_{(R)}=\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2}=\frac{1}{2} m v^2\left(\frac{2}{5}\right) \end{aligned}

So,

\frac{\mathrm{KE}_{(T)}}{\mathrm{KE}_{(R)}}=\frac{5}{2}

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