Question

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?

Answer 1

I found:
`L=3.14kgm^2/s`
`K_(rot)=19.72J`

Explanation:

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

In your case:
`120`rpm corresponds to `120` rotations of `2pi` 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`