Question

A thin metal disc of radius of $0.25m$ and mass $2kg$ starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is $4J$ at the foot of inclined plane, then the linear velocity at the same point, is in $m/s$

• A. $2$
• B. $2\sqrt{2}$
• C. $2\sqrt{3}$
• D. $3\sqrt{2}$

Answer 1

Answer: (B)

Rotational kinetic energy $=\frac{1}{2}I{\omega }^2$
$\therefore$ Rotational $KE=\frac{1}{2}\left[\frac{1}{2}m{r}^2\right]\frac{v^2}{r^2}$
$\left(\text{where}I=\frac{1}{2}m{r}^2\right)$
$4=\frac{1}{2}\left[\frac{1}{2}(2)r^2\right]\frac{v^2}{r^2}$
$\Rightarrow v^2=8$
$v=2\sqrt{2}m/s$