In the figure which is in the representations section you observe that a wheel is mounted on a stationary axel, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel. During the time that the wheel's rotation changes from 4 revolutions per second to 7 revolutions per second, how much work do you do?

System: Wheel and string

Surroundings: Your hand, axle, Earth

From the Energy Principle:

$E_{f} = E{i} + W$

$\frac{1}{2}I\omega^{2}_{f} = \frac{1}{2}I\omega^{2}_{i} + W$

$W = \frac{1}{2}I(\omega^{2}_{f} - \omega^{2}_{i})$

What is the moment of inertia $I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $m_{3}r^{3}_{\perp3}$ + $m_{4}r^{4}_{\perp4} + \cdot \cdot \cdot$ ? Group this sum into a part that includes just the atoms of the inner ring and another part that includes just the atoms of the outer ring:

$I = (m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $\cdot \cdot \cdot)_{inner}$ + $(m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2})_{outer}$ + $\cdot \cdot \cdot$

$I = I_{inner} + I_{outer}$

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.