Moment of Inertia

Engineering Physics | AQA A-Level Physics

Key Definition

Moment of inertia ($I$): The resistance to a change of rotational motion, depending on the distribution of mass around a chosen axis of rotation. Measured in kg m$^2$.

What determines moment of inertia

In linear motion, the resistance to a change of motion is called inertia: the larger the mass, the greater its inertia.
In rotational motion, we use moment of inertiaThe rotational equivalent of mass. It depends not just on how much mass there is, but on how that mass is distributed relative to the axis of rotation., which is the rotational equivalent of mass.
The moment of inertia depends on two factors: the total mass ($m$) and how that mass is distributed about the axis of rotation ($r$).
For example, a springboard diver who tucks their legs in closer to their chest decreases their moment of inertia, because more of their mass is distributed over a smaller distance. This makes it easier for them to rotate.

Calculating moment of inertia

For a point mass, the moment of inertia is: $$I = mr^2$$
Where: $I$ = moment of inertia (kg m$^2$), $m$ = mass of the object (kg), $r$ = distance from the axis of rotation (m).
For an extended object, the moment of inertia is the summation of $mr^2$ for all particles: $$I = \sum mr^2$$

Axis of rotation matters

Crucially, the moment of inertia of an object can change depending on the chosen axis of rotation.
A thin rod, for example, has different moments of inertia depending on whether it rotates about its vertical axis ($I = \tfrac{1}{2}mr^2$), its centre of mass ($I = \tfrac{1}{12}mL^2$), or one end ($I = \tfrac{1}{3}mL^2$).
There is an infinite range of possible axes, and therefore an infinite set of possible moments of inertia.

Worked example: dumbbell

Two solid spheres (mass 750 g, radius 4 cm) are attached to the ends of a thin rod (mass 20 g). Each sphere's centre of mass is 22 cm from the axis. The rod length is $L = 2 \times (22 - 4) = 36$ cm = 0.36 m.
The total moment of inertia is: $$I = 2 \times \left(\tfrac{2}{5}m_{\text{sphere}}r^2\right) + \tfrac{1}{12}m_{\text{rod}}L^2$$ $$I = 2 \times \left(\tfrac{2}{5} \times 0.75 \times 0.22^2\right) + \left(\tfrac{1}{12} \times 0.02 \times 0.36^2\right) = 0.029 \text{ kg m}^2$$
The rod contributes only about 0.7% of the total moment of inertia.

Common Mistake

You will never be expected to memorise the moment of inertia formulae for different shapes. They will always be given in the exam question. The key part is knowing how to use them and how to add them together for composite objects.