How much tangential force would be needed to stop the earth in one year, if it were rotating with angular velocity of `7.3 xx 10^(-5) "rad s"^(-1)` ? Given the moment of inertia of the earth `=9.3 xx 10^(37) "kg m"^(2)` and radius of the earth `=6.4 xx 10^(6)` m.

Answer

`3.36`

Explanation

Here `I=9.3xx10^(37)kgm^(2)`, `R=6.4xx10^(6)m`,
`omega_(0)=7.5xx10^(-5)rads^(-1)`
`t=1` year `=365xx24xx3600s`
As `omega=omega_(0)+alphat`
`:.alpha=(omega-omega_(0))/(t)=(0-7.3xx10^(-5))/(365xx24xx3600)`
`=-(7.3xx10^(-6))/(365xx24xx3600)rads^(-2)`
Torque, `tau=Ialpha=9.3xx10^(37)xx(7.3xx10^(-5))/(365xx24xx3600)Nm` [Omitting `-ve` sign]
Let F be the tangential force needed to stop the earth. Then
`tau=FR`
or `F=(tau)/(R)=(9.3xx10^(37)xx7.3xx10^(-5))/(365xx24xx3600xx6.4xx10^(6))`
`=3.363xx10^(19)N`