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 $\frac{3}{5}$ th as much as present?
✓
Answer
(a)
$w=w-m \omega^2 R$
$\Rightarrow m g_e=m g-m \omega^2 R$
$m g_e=\frac{3}{5} m g$
$\Rightarrow m \omega^2 R=\frac{2}{5} m g$
$\Rightarrow \omega=\sqrt{\frac{2 g}{5 R}}$
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