Find the value of angular velocity of axial rotation of the earth, such that weight of a person at equator becomes 3/4 of its weight at pole, Radius of the earth at equator is 6400 km.

Explanation

Let acceleration due to gravity at pole is g and at equator it is g'. If mass of the person is `m` and angular velocity of rotation of the Earth is `omega`, then
`g'=g-omega^(2)RrArr mg'=mg-momega^(2)R`
`rArr (3)/(4)mg=mg-momega^(2)R`
`rArr omega^(2)R=(1)/(4)g`
`rArr omega=sqrt((g)/(4R))=sqrt((9.8)/(4xx6400xx10^(3)))`
`rArr omega=(1)/(2)xx(1)/(80)xx(1)/(10^(2))xxsqrt(98)=62xx10^(-4)"rads"^(-1)`.