Question

A body rolls down an inclined plane. If its kinetic energy of rotation is $40\%$ of it's kinetic energy of translation motion, then the body is

• A. Hollow cylinder
• B. Ring
• C. Disc
• D. Solid sphere

Answer 1

The correct option is D Solid sphere

$\because K{E}_{Rot}=40\%$ of $K{E}_{Trans}$

$\Rightarrow \frac{K{E}_{Rot}}{K{E}_{Trans}}=\frac{0.4\times K{E}_{Trans}}{K{E}_{Trans}}=0.4...\left(i\right)$
Now from the formula of rotational kinetic energy:
$\frac{K{E}_{Rot}}{K{E}_{Trans}}=\frac{\frac{1}{2}m{v}^2\left(\frac{K^2}{R^2}\right)}{\frac{1}{2}m{v}^2}=\frac{K^2}{R^2}...\left(ii\right)$
where, $K$ is radius of gyration of body.
From Eq $\left(i\right)$ and Eq $\left(ii\right)$,
$K^2=0.4R^2=\frac{2}{5}{R}^2$
Moment of inertia of the body about axis or rotation through it's centre is:
$I=m{K}^2$
$\Rightarrow I=\frac{2}{5}m{R}^2$
$\therefore$ The body is a solid sphere.