Question

If a man at the equator would weigh $\left(3/5\right)th$ of his weight, the angular speed of the earth is:

• A. $\sqrt{\frac{2}{5}\frac{g}{R}}$
• B. $\sqrt{\frac{g}{R}}$
• C. $\sqrt{\frac{R}{g}}$
• D. $\sqrt{\frac{2}{5}\frac{R}{g}}$

Answer 1

The correct option is A $\sqrt{\frac{2}{5}\frac{g}{R}}$

$m{g}^{\prime }=mg-mR{\omega }^{2}co{s}^2\varphi$

Now, $\frac{3}{5}mg=mg-mR{\omega }^2$

or $mR{\omega }^2=mg-\frac{3}{5}mg$

or $mR{\omega }^2=\frac{2}{5}mg$

or $\omega =\sqrt{\frac{2g}{5R}}$