Question

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on ${60}^o$ latitude becomes zero is (Radius of earth = 640 km. At the poles g=10$m{s}^{-2}$)

• A.
• B.
• C.
• D. $7.8\times {10}^{-2}rad/sec$

Answer 1

The correct option is A

Effective acceleration due to gravity due to rotation of earth $g^{\prime }=g-{\omega }^{2}Rco{s}^2\lambda$
$0=g-{\omega }^{2}Rco{s}^2{60}^o;\frac{{\omega }^{2}R}{4}=g\Rightarrow \omega =\sqrt{\frac{4g}{R}}=2\sqrt{\frac{g}{R}}=\frac{2}{800}\frac{rad}{sec}[As{g}^{\prime }=0and\lambda ={60}^o]$
$\omega =\frac{1}{400}=2.5\times {10}^{-3}\frac{rad}{sec}$.