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Q.
The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to
AMUAMU 2012
Solution:
Acceleration due to gravity $g=g-\omega^{2} R \cos ^{2} \lambda$
$0=g-\omega^{2} R \cos ^{2} 60^{\circ} 0=g-\frac{\omega^{2} R}{4}$
$\omega=2 \sqrt{\frac{g}{R}}=2 \sqrt{\frac{10}{6400 \times 100}}$
Angular velocity $\omega=\frac{1}{400}$
$=2.5 \times 10^{-3} rad / s$ (Take $g=10\, m / s ^{2}$
for the acceleration due to gravity if the earth were at rest and radius of earth
equal to $6400\, km$.)