Q. A wheel of mass 8 kg and radius 40 cm is rolling on a horizontal road with angular velocity of $15\ rad\ s^{- 1}$. The moment of inertia of the wheel about its axis is $0.64\ kg\ m^{- 2}$. Total KE of wheel is
A) 288 J
B) 216 J
C) 72 J
D) 144 J
Explanation
(b) Here, $m = 8\ kg,\ r = 40\ cm = \frac{2}{5}m$,
$\omega = 12\ rad\ s^{- 1},\ I = 0.64\ kg\ m^{2}$
$Total\ KE = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$
$= \frac{1}{2}I\omega^{2} + \frac{1}{2}mr^{2}\omega^{2}$
$= \frac{1}{2} \times 0.64 \times 15^{2} + \frac{1}{2} \times 8 \times \left( \frac{2}{5} \right)^{2} \times 15^{2} = 216\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.