If a man at the equator would weight (3/5)th of his weight, the angular speed of the earth is:

Options

A. `(2v)/(3d)`
B. `(2v)/d`
C. `v/d`
D. `(3v)/(2d)`

Answer

A

Explanation

By conservation of angular momentum about pivot
`L=Iomega`
`(mvd)/2=[(Md^(2))/12+m(d/2)^(2)]omega`
`=((md^(2))/2+(md^(2))/4)omega`
`=3/4md^(2)omegaimplies 2/3v/d=omega`