The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms^–2) : 

2.5 x 10^-3 rad/sec

Effective acceleration due to gravity due to rotation of earth

\(\Rightarrow 0 = g - \omega^2 R cos^2 60^o\)

\(\Rightarrow \frac{\omega^2 R}{4} = g\)

\(\Rightarrow \omega = \sqrt{\frac{4g}{R}}\)

\(\Rightarrow \omega = 2\sqrt{\frac{g}{R}}\)

\(\Rightarrow \omega = \frac{2}{800} rad \text{ } sec^{-1}\)

\(g' = 0 \text{ and } \lambda = 60^o\)

\(\Rightarrow \omega = \frac{1}{400} \)

\(\Rightarrow \omega = 2.5 \text{ x } 10^{-3} rad \text{ } sec^{-1}\)