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 $3v^2/4g$ with respect to the initial position. The object is

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

Answer 1

Answer: (D)

From conservation of energy, we get
$\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{CM}{\omega }^2=mgh$
$\frac{1}{2}m{v}^2\left(1+\frac{k^2}{r^2}\right)=mg\left(\frac{3v^2}{4g}\right)$ where $k$ is the radius of gyration.

Now, $1+\frac{k^2}{r^2}=\frac{3}{2}$
$\Rightarrow \frac{k^2}{r^2}=\frac{1}{2}$
Therefore, the object is a disc.