If the earth has no rotational motion, the weight of a person on the equation is W. Detrmine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight `(3)/(4)` W. Radius of the earth is 6400 km and g = 10 `m//s^(2)`.

Options

A. `1.1 xx 10^(-3) "rad/s"`
B. `0.83 xx 10^(-3) "rad/s"`
C. `0.63 xx 10^(-3) "rad/s"`
D. `0.28 xx 10^(-3) "rad/s"`

Answer

C

Explanation

At equator
`g. = g - R omega^(2)`
`mg. = mg - R omega^(2) m`
`W. = W - R omega^(2) m`
When earth is at rest, W = mg
When earth has rotational motion,
`(3W)/(4) = W - R omega^(2)m`
`-(W)/(4) = -R omega^(2) m`
`(mg)/(4) = -R omega^(2)m`
`omega = sqrt((g)/(4R)) = sqrt((10)/(4 xx 6,400 xx 10^(3)))`
`omega = 0.63 xx 10^(-3)` rad/s