First slide
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Question

The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to [AMU 2012J (Take, g = 10 ms-2 for the acceleration due to gravity, if the earth were at rest, radius of earth equal to 6400 km and  λ = 60)

Difficult
Solution

Acceleration due to gravity, 

g=g-ω2Rcos2λ

  0=g-ω2Rcos260°

0=g-ω2R4

  ω=2gR=2106400×1000

Angular velocity,  ω=1400=2.50×10-3rads-1

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