Q. A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is
A) $1.2\ ms^{- 1}$
B) $2\sqrt{2}\ ms^{- 1}$
C) $20\ ms^{- 1}$
D) $2\ ms^{- 1}$
Explanation
(b) Here, $r = 0.5\ m,\ m = 2\ kg$
Rotational KE$= \frac{1}{2}I\omega^{2} = \frac{1}{2} \times \left( \frac{1}{2}mr^{2} \right)\omega^{2}$
$4 = \frac{1}{4}mv^{2} = \frac{1}{4} \times 2v^{2}$
$v = \sqrt{8} = 2\sqrt{2}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.