Question

A body is rolling down an inclined plane. If kinetic energy of rotation is $40\%$ of kinetic energy in translatory state, then the body is a

• A. ring
• B. cylinder
• C. hollow ball
• D. solid ball

Answer 1

Answer: (D)

Rotational kinetic energy is
$K_{R} = \frac{1}{2}I\omega^{2} = \frac{1}{2}Mk^{2}\left(\frac{v}{R}\right)^{2} \quad\left(\because I= Mk^{2} {\text{and}} \, v= R\omega\right) $
$= \frac{1}{2}Mv^{2} \left(\frac{k^{2}}{R^{2}}\right) $
Translational kinetic energy is
$ K_{T} = \frac{1}{2}Mv^{2} $
As per question, $K_{R} = 40\% K_{T} $
$\therefore \frac{1}{2}Mv^{2} \left(\frac{k^{2}}{R^{2}}\right) $
$= 40\% \frac{1}{2}Mv^{2}$ or $\frac{k^{2}}{R^{2}} = \frac{40}{100} = \frac{2}{5}$
For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5} $
Hence, the body is solid ball.