Determine the speed with which the Earth would have to rotate on is axis. So that a person on the equator would weigh `(3^(th))/(5)` as much as at present. Take the equatorial radius as 6400 km. `g=9.8m//s^(2)`

Explanation

Here w = ?, `mg'=(3)/(5)mg`,
`R=6400km=6.4xx10^(6)m`
Acceleration due to gravity at the equator is
`g'=g-Romega^(2)`
(or) `mg'=mg-mRomega^(2)`
(or) `(3)/(5)mg=mg-mRomega^(2)`
(or) `mRomega^(2)=mg-(3)/(5)mg=(2)/(5)mg`
(or) Speed `omega=sqrt((2g)/(5R))=sqrt((2xx9.8)/(5xx6.4xx10^(6)))`
`=7.8xx10^(-4)rads^(-1)`