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
$K_R=\frac{1}{2}I{\omega }^2=\frac{1}{2}M{k}^2{\left(\frac{v}{R}\right)}^2\left(\because I=M{k}^2\text{and}v=R\omega \right)$
$=\frac{1}{2}M{v}^2\left(\frac{k^2}{R^2}\right)$
Translational kinetic energy is
$K_T=\frac{1}{2}M{v}^2$
As per question, $K_R=40\%K_T$
$\therefore \frac{1}{2}M{v}^2\left(\frac{k^2}{R^2}\right)$
$=40\%\frac{1}{2}M{v}^2$ 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.