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 $3 v^{2} / 4 g$ 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
$\frac{1}{2} m v^{2}+\frac{1}{2} I_{C M} \omega^{2}=m g h$
$\frac{1}{2} m v^{2}\left(1+\frac{k^{2}}{r^{2}}\right)=m g\left(\frac{3 v^{2}}{4 g}\right)$ 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.