Determine the speed with which the earth would have to rotate on its axis , so that a person on the equator would weigh `(3)/(5)` th as much as the person. Take `R = 6400 km`.

Explanation

The apparent weight of a person on equator due to rotation of earth is
`W' = W - mR omega^(2) cos^(2) 0^(@) = W - mR omega^(2)`
So `(3)/(5) mg = mg -mR omega^(2)`
or `mR omega^(2) = mg - (3)/(5) mg = (2)/(5) mg`
or `omega = sqrt((2g)/(5R)) = sqrt((2 xx 9.8)/(5 xx (6400 xx 10^(3)))`
`= 7.826 xx 10^(-4) rad s^(-1)`