First slide

Acceleration due to gravity and its variation

Question

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

Moderate

A

$\sqrt{\frac{2}{5} \frac{g}{R}}$
B

$\sqrt{\frac{g}{R}}$
C

$\sqrt{\frac{R}{g}}$
D

$\sqrt{\frac{2}{5} \frac{R}{g}}$

Solution

$\begin{array}{l} 3 / 5 mg = mg - {mRω}^{2} \\ ω^{2} R = g - \frac{3}{5} g \Rightarrow ω = \sqrt{\frac{2}{5} \frac{g}{R}} \end{array}$

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