A bicycle wheel has a moment of inertia of 0.25 kg m² about ots axle. The wheel rotates at 120 rpm . Calculate the angular momentum of the wheel and the rotational kinetic energy of the wheel?

1 Answer
Jul 14, 2015

I found:
L=3.14kgm^2/sL=3.14kgm2s
K_(rot)=19.72JKrot=19.72J

Explanation:

The angular momentum is:
L=IomegaL=Iω
while the rotational kinetic energy is:
K_(rot)=1/2Iomega^2Krot=12Iω2
Where:
I=I= moment of inertia;
omega=ω=angular velocity.

In your case:
120120rpm corresponds to 120120 rotations of 2pi2π in 1min=60sec or 120xx(2pi)/60=12.56 (rad)/s
with this in mind:
L=0.25*12.56=3.14kgm^2/s
K_(rot)=1/2*0.25(12.56)^2=19.72J