Q. The total kinetic energy of a body of mass $10\ kg$ and radius $0.5\ m$ moving with a velocity of $2\ m/s$ without slipping is $32.8\ joule.$ The radius of gyration of the body is
A) $0.25\ m$
B) $0.2\ m$
C) $0.5\ m$
D) $0.4\ m$
Explanation
(d) Total kinetic energy $= \frac{1}{2}mv^{2}\left( 1 + \frac{K^{2}}{R^{2}} \right) = 32.8\ J$
$\Rightarrow \frac{1}{2} \times 10 \times (2)^{2}\left( 1 + \frac{K^{2}}{(0.5)^{2}} \right) = 32.8 \Rightarrow K = 0.4\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.