Question

A small object of uniform density rolls up a curved surface without slipping, with an initial velocity $v\text{m/s}$. It reaches up to a maximum height of $\frac{v^2}{g}\text{m}$ 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 A Ring

Let the mass of object be $m\text{kg}$ and radius be $r\text{m}$
Initially,
$v_i=v,{\omega }_i=\omega$
For pure rolling$\Rightarrow v=r\omega ...\left(i\right)$
When object reaches maximum height $(h=\frac{v^2}{g})$, it stops completely, so $v_f=0,{\omega }_f=0$
$\therefore$Applying mechanical energy conservation from initial to final position:
$\text{Gain in}P{E}_{\text{gravitational}}=\text{Loss in}K{E}_{\text{trans}}+\text{Loss in}K{E}_{\text{rot}}$
$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{\text{CM}}{\omega }^2...\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$,
$\Rightarrow mg\left(\frac{v^2}{g}\right)=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_{\text{CM}}\times \left(\frac{v^2}{r^2}\right)$

$m=\frac{m}{2}+\frac{I_{\text{CM}}}{2r^2}$
$\frac{m}{2}=\frac{I_{\text{CM}}}{2r^2}$

$\therefore I_{\text{CM}}=m{r}^2$
The moment of inertia of object passing through its centre or centre of mass matches with that of a ring.
$\Rightarrow$ Object in question is a ring.