Question

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. 5 MgR/2

Answer 1

The correct option is C

MgR/5

when there is a weightlessness in the body at the equator, then $g^{\prime }=r-R{\omega }^2=0$
$\omega =\sqrt{\frac{g}{R}}$ and
or
linear velocity $=\omega R=\sqrt{\left(\frac{g}{R}\right)}R=\sqrt{gR}$
$\therefore$ KE of rotation of earth $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{2}{5}M{R}^2\times {\omega }^2$
$\frac{2}{5}M(\omega R{)}^2=\frac{1}{5}MgR$