A solid flywheel of 20 kg mass and 120 mm radius revolves at `600 "revmin"^(-1)`. With what force must a brake lining be pressed against it for the flywheel to stop in 3s, if the coefficient of friction is 0.1 ?

Explanation

Given, revolutions per minute, `n_(0)=600 "rev min"^(-1)`
`=(600)/(60)"rev s"^(-1)`
Revolutions per second, `n_(0)=10 "rev s"^(-1)`
So, initial angular velocity, `omega_(0)=2pin_(0)=(2pi)(10)=20pi "rad s"^(-1)`
Let `alpha` be the constant angular retardation, then applying
`omega = omega_(0)=alpha t`
or `0=(20 pi)-3(alpha)`
or `alpha=(20)/(3)pi "rad s"^(-2)`
Further, `alpha=(tau)/(I)`
Here, torque `tau=muNR" "("R=radius")`
or `tau=muFR" "("F=applied force")`
`=N=F and I=(1)/(2)mR^(2)`
From the above equations,
`(20)/(3)pi=(muFR)/((1)/(2)mR^(2))=(2muF)/(mR)`
or Force, `F=(10pimR)/(3mu)`
Substituting the values, we have
`F=(10xx22xx20xx0.12)/(3xx7xx0.1)`
or `F=2.51.43 N`.