Q. The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $h$ from rest without sliding is
A) $\sqrt{\frac{10}{7}g\ h}$
B) $\sqrt{\frac{4}{3}g\ h}$
C) $\sqrt{g\ h}$
D) $\sqrt{\frac{6\ g\ h}{5}}$
Explanation
(a) $mgh = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$
$= \frac{1}{2}\left( \frac{2}{5}mr^{2} \right)\omega^{2} + \frac{1}{2}mv^{2} = \frac{7}{10}mv^{2}$
$\therefore\ v = \sqrt{\frac{10}{7}\ gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.