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

## Answer This Question

## Related Questions

- Physics - The combination of an applied force and a frictional force produces a ...
- Physics - Rotating initially at 1800 rpm, a wheel with a diameter of 74.0 cm is ...
- Physics - The combination of an applied force produces a constant total torque ...
- physics - The flywheel of an engine has a moment of inertia of 24 kg*m^2. a) if ...
- physics - The flywheel of an engine has a moment of inertia of 24 kg*m^2. a) if ...
- physics - A grinding wheel of radius 0.370 m rotating on a frictionless axle is ...
- Physics - A wheel 31.7 cm in diameter accelerates uniformly from 238 rpm to 347 ...
- Physics - A tangential force acts on the rim of a 2.0 kilogram disk-shaped wheel...
- Physics - A tangential force acts on the rim of a 2.0 kilogram disk-shaped wheel...
- PHYSICS HELP PLEASE!! - A tangential force acts on the rim of a 2.0 kilogram ...

More Related Questions