Question

In the figure shown a ring A is rolling without sliding with a velocity v on the horizontal surface of the body B (of the same mass as A). All surfaces are smooth. B has no initial velocity. What will be the maximum height (from initial position) reached by A on B

• A. $\frac{3v^2}{4g}$
• B. $\frac{v^2}{4g}$
• C. $\frac{v^2}{2g}$
• D. $\frac{v^2}{3g}$

Answer 1

The correct option is A $\frac{3v^2}{4g}$

Let the ring will rise to a height $h$

From energy conservation we have,

K.E of linear motion+K.E of rotational motion = Potential energy of body at height $h$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2=mgh$

Here $m=$Mass of ring.

$I=$ moment of inertia of ring$=\frac{1}{2}m{r}^2$

$\omega =$ Angular velocity of ring$=\frac{V}{R}$

$V=$ Velocity of ring.

$R=$Radius of ring.

So we can write,

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)\frac{V^2}{R^2}=mgh$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{4}m{V}^2=mgh$

$\Rightarrow \frac{3V^2}{4}=gh$

$\Rightarrow h=\frac{3V^2}{4g}$