Question

A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum
height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is

• A. ring
• B. solid sphere
• C. hollow sphere
• D. disc

Answer 1

The correct option is D disc

By the concept of energy conservation
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mg\left(\frac{3v^2}{4g}\right)$
For rolling motion $v=R\omega \therefore \frac{1}{2}m{v}^2+\frac{1}{2}I\frac{v^2}{R^2}=\frac{3}{4}m{v}^2\therefore \frac{1}{2}I\frac{v^2}{R^2}=\frac{3}{4}m{v}^2-\frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2\frac{1}{2}I\frac{v^2}{R^2}=\frac{1}{4}m{v}^2\Rightarrow I=\frac{1}{2}m{R}^2$
This is the formula of the moment of inertia of the disc.