Question

A disc of radius $0.1m$ rolls without sliding on a horizontal surface with a velocity of $6m/s^{-1}$.Then, it ascends a smooth continuous track as shown in the figure. Given, $g=10m{s}^{-1}$, the height up to which it will ascend is $(g=10m/s^2)$

• A. $2.4m$
• B. $0.9m$
• C. $2.7m$
• D. $1.8m$

Answer 1

The correct option is C $2.7m$

When the disc rolls without sliding, then the kinetic energy of the disc is,

$K=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right)\left(\frac{v^2}{r^3}\right)$

$⟹K=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2$

$⟹\frac{3}{4}m{v}^2$

So, the height reached by the disc is,

$mgh=\frac{3}{4}m{v}^2$

$⟹h=\frac{3v^2}{4g}$

Given,

Velocity, $v=6m/s$

Then,

Height, $h=\frac{3\times 6^2}{4\times 10}=2.7m$