Question

What should be the angular velocity of rotation of earth about its own axis, so that the weight of a person on equator reduces to $\frac{3}{5}$ of its present value $(R=6400km)$

• A. $7.8\times {10}^{-4}rad/s$
• B. $7.8rad/s$
• C. $0.8\times {10}^{-4}rad/s$
• D. $1rad/s$

Answer 1

The correct option is A $7.8\times {10}^{-4}rad/s$

As the speed of rotation varies, the value of gravity also changes according to the relation
$g_e=g-{\omega }^{2}R$
Weight of the person decreases implies the value of $g$ decreases.
$\Rightarrow \frac{3}{5}g=g-{\omega }^{2}R$
$\Rightarrow {\omega }^{2}R=g-\frac{3}{5}g=\frac{2}{5}g$
$\Rightarrow \omega =\sqrt{\frac{2}{5}\frac{g}{R}}=\sqrt{\frac{2}{5}\frac{9.81}{6400\times {10}^3}}=0.078\times {10}^{-2}=7.8\times {10}^{-4}$ rad/s