Question

A thin metal disc of radius of $0.25\, m$ and mass $2\, kg$ starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is $4\, J$ 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 $K E=\frac{1}{2}\left[\frac{1}{2} m r^{2}\right] \frac{v^{2}}{r^{2}}$
$\left(\text{where} I=\frac{1}{2} m r^{2}\right)$
$4=\frac{1}{2}\left[\frac{1}{2}(2) r^{2}\right] \frac{v^{2}}{r^{2}}$
$\Rightarrow v^{2}=8$
$v=2 \sqrt{2} m / s$