Q. The bodies situated on the surface of earth at its equator, becomes weightless, when the earth has KE about it axis
A) $mgR$
B) $2\ mgR/5$
C) $\ MgR/5$
D) $5MgR/2$
Explanation
(c) When there is a weightlessness in the body at the equator, then $g^{'} = r - R\omega^{2} = 0$
or $\omega = \sqrt{g/R}$ and linear velocity
$= \omega R = \left( \sqrt{g/R} \right)R = \sqrt{gR}$
$\therefore$ KE of rotation of earth $= \frac{1}{2}I\omega^{2}$
$= \frac{1}{2} \times \frac{2}{5}MR^{2} \times \omega^{2}$
$= \frac{2}{5}M(\omega R)^{2} = \frac{1}{5}MgR$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.