A person stands on a spring balance at the equator. a) By what fraction is the balance reading less than his true weight? b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the lenght of the day in this case?

Answer

A, B, C

Explanation

a. Net force on the spring balance,
`R=mg=momega^2r`
So, fraction less than the true weight (3mg) is
`=(mg-(mg-momega^2r))/(mg)`
`(omega^2)/g((2pi)/(24xx3600))^2
`=(6500xx10^3)/10`
when the balance reading is half the true weight
`(mg-momega^2r)/(mg)=1/2`

omega^2r=g/2`
`rarr omega=sqrt(g/(2r)`
`=sqrt(10/(2xx6400xx10^3))` rad/sec
`Duration of the day is `
`T=(2pi)/omega`
`=2pixxsqrt((2xx6400xx10^3)/9.8` sec
`2pisqrt((64xx10^5)/49)sec`
`=(2pixx8000)/(7xx3600)hr=2hr`