Question

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

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

Answer 1

The correct option is D disc

Applying mechanical energy conservation, since work done by friction $W_f=0$ in pure rolling,
${\text{Loss in KE}}_{\text{Trans}}+{\text{Loss in KE}}_{\text{Rot}}=\text{Gain in PE}$
or $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$
[$\because v=R\omega$ for pure rolling]
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right)=mgh$
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right)=mg\left(\frac{3v^2}{4g}\right)$
$\Rightarrow \frac{1}{2}I\left(\frac{v^2}{R^2}\right)=\frac{3m{v}^2}{4}-\frac{m{v}^2}{2}$
$\Rightarrow I=\frac{m{v}^2}{4}\times \frac{2R^2}{v^2}$
$\therefore I=\frac{m{R}^2}{2}$
Hence the object is a disc.