Question

A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a

• A. solid sphere
• B. hollow sphere
• C. solid cylinder
• D. hollow cylinder

Answer 1

Answer: (D)

When a body rolls down an inclined plane, it is accompanied by rotational and translational kinetic energies.
Rotational kinetic energy, $K_R=\frac{1}{2}I{\omega }^2$
where, $I$ is the moment of inertia and $\omega$ is the angular velocity.
Translational kinetic energy for pure rolling,
$=K_T=\frac{1}{2}m{v}_{CM}^2=\frac{1}{2}m(r\omega {)}^2$
$\left(\because v_{CM}=r\omega \right)$
where, $m$ is mass of the body, $v_{CM}$ is the velocity and $\omega$ is the angular velocity.
Given, translational kinetic energy = rotational kinetic
$\therefore \frac{1}{2}m\left(r^2{\omega }^2\right)=\frac{1}{2}I{\omega }^2$
$\Rightarrow I=m{r}^2$
We know that, $m{r}^2$ is the moment of inertia of hollow cylinder about its axis, where $m$ is the mass of hollow cylindrical body and $r$ is the radius of the cylinder.