Posted by **Tom** on Monday, November 26, 2012 at 10:09pm.

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?

- Physics -
**Elena**, Tuesday, November 27, 2012 at 9:26am
m=2.5 kg. R=D/2=0.1 m,

n₀=3600 rpm=3600/60=60 rev/s

t=10 s. F=?

2πn=2πn₀-εt.

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

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