The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kg`m^(2)` . Find torque required to bring it to rest in 20 s.

Explanation

`n_(1)`=300 rpm=`(300)/(60)=5rps,1=0.3 kgm^(2)`
`omega_(1)=2pin_(1)=2pixx5=10pi,omega_(2)=2pin_(2)=0`
`therefore tau=(0.3xx(0-10pi))/(20)=(-0.3xx10pi)/(20)=-0.15 pi=-0.471 nm`