Question

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

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

Answer 1

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

Here we have to determine the speed with the earth would have to rotate on its axis so that a person on the equator would weight 3/5 as much as the present.

Solution:

Actual weight on the equator = W = mg

where 'm' is the mass and 'g' is the gravity.

According to the given condition,

Weight on the equator = W' = mg'

Weight on the equator = W' = 3/5 mg ........... (1)

We know that, λ = 0 at the equator.

Now,

mg' = mg - m R ω² cos λ

$\frac{3}{5}mg=mg-mR\omega ²cos0$ (∵ λ = 0 )

$\Rightarrow \frac{3}{5}mg=mg-mR\omega ²\left(1\right)$ (∵ cos 0 =1)

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

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

$\Rightarrow mR\omega ²=(1-\frac{3}{5})mg$

$\Rightarrow mR\omega ²=2/5mg$

$\Rightarrow R\omega ²=2/5g$

$\Rightarrow \omega ²=2g/5R$

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

Hence correct option A is correct