Question

What should be the angular speed with which the earth have to rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$ as much as present ?

• A. $\sqrt{\frac{2g}{5R}}$
• B. $\sqrt{\frac{5g}{2R}}$
• C. $\sqrt{\frac{g}{R}}$
• D. $\sqrt{\frac{7g}{2R}}$

Answer 1

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

The acceleration due to gravity $g^{\prime }$ considering the effect of rotation at the equator is given by

$g^{\prime }=g(1-{\omega }^{2}R)$

where,
$\omega =$ angular velocity of the Earth
$R=$ is the radius of the Earth

Hence, the weight is given by

$\Rightarrow m{g}^{\prime }=mg-m{\omega }^{2}R$

From question, $m{g}^{\prime }=\frac{3}{5}mg$

$\therefore \frac{3}{5}mg=mg-m{\omega }^{2}R$

$\Rightarrow \frac{2}{5}mg=m{\omega }^{2}R$

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

Hence, option $\left(a\right)$ is correct