Q. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60{^\circ}$ latitude becomes zero is (Radius of earth $= 6400\ km.$ At the poles $g = 10\ ms^{- 2}$)
A) $2.5 \times 10^{- 3}rad/s$
B) $5.0 \times 10^{- 1}rad/s$
C) $1 \times 10^{1}rad/s$
D) $7.8 \times 10^{- 2}rad/s$
Explanation
(a) $g^{'} = g - \omega^{2}R\cos^{2}{\lambda \Rightarrow 0 = g - \omega^{2}R\cos^{2}{60{^\circ}}}$
$0 = g - \frac{\omega^{2}R}{4} \Rightarrow \omega = 2\sqrt{\frac{g}{R}} = \frac{1}{400}\frac{rad}{\sec} = 2.5 \times 10^{- 3}\frac{rad}{\sec}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.