A bicycle wheel has almost all its mass M located in the outer rim at radius R. What is the moment of inertia of the bicycle wheel about its center of mass?

(Hint: It's helpful to think of dividing the wheel into the atoms it is made of and think about how much each atom contributes to the moment of inertia.)

Let m represent the mass of one atom in the rim. The moment of inertia is

$I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $m_{3}r^{3}_{\perp3}$ + $m_{4}r^{4}_{\perp4} + \cdot \cdot \cdot$

$I = m_{1}R^{2} + m_{2}R^{2} + m_{3}R^{2} + m_{4}R^{2} + \cdot \cdot \cdot$

$I = [m_{1} + m_{1} + m_{1} + m_{1} + \cdot \cdot \cdot]R^2$

$I = MR^2$

We've assumed that the mass of the spokes is negligible compared to the mass of the rim, so that the total mass os just the mass of the atoms in the rim.