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$

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

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

For two masses, $I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$. The distance between masses is d, so the distance of each object from the center of mass is $r_{\perp1} = r_{\perp2} = d/2$.

Therefore:

$I = M(d/2)^{2} + M(d/2)^{2} = 2M(d/2)^{2}$

$I = 2 \cdot (2.3$ x $10^{-26}kg)(0.75$ x $10^{-10}m)^2$

$I = 2.6$ x $10^{-46} kg \cdot m^2$