Q. 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 ${ms}^{- 1}$.
A) 2
B) $2\sqrt{2}$
C) $2\sqrt{3}$
D) $3\sqrt{2}$
Explanation
(b) Rotational kinetic energy $= \frac{1}{2}I\omega^{2}$
$\therefore$ Rotational KE $= \frac{1}{2}\left\lbrack \frac{1}{2}mr^{2} \right\rbrack\frac{v^{2}}{r^{2}}$ (where $I = \frac{1}{2}mr^{2}$)
$4 = \frac{1}{2}\left\lbrack \frac{1}{2}(2)r^{2} \right\rbrack\frac{v^{2}}{r^{2}}$
$\Longrightarrow$ $v^{2} = 8$
$v = 2\sqrt{2}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.