Physics
posted by Tom on .
A 2.50 kg wheel has a diameter of 20.0 cm and is rotating at 3600 rpm. A force acts on it causing it to come to a rest in 10.0 seconds. What is the magnitude of the force?

m=2.5 kg. R=D/2=0.1 m,
n₀=3600 rpm=3600/60=60 rev/s
t=10 s. F=?
2πn=2πn₀εt.
Final n=0
2πn₀ = εt.
ε =2πn₀/t.
Newton’s 2 law for rotation:
Iε=M.
If the wheel is disc I=mR²/2,
M=FR.
Then
(mR²/2) •(2πn₀/t)=F•R
F=mRπn₀/t=2.5•0.1•3.14•60/10 = 4.71 N