How to Calculate Moment of Inertia for Common Shapes

The moment of inertia measures how hard something is to set spinning. For the same mass it grows as the mass sits further from the axis. Choose a shape and give the mass and size to obtain the moment of inertia and the rotational kinetic energy.

The moment of inertia measures how hard something is to set spinning. It plays the part that mass plays in straight-line motion, and for the same mass it grows as that mass sits further from the axis.

I = k m r^2

Here $m$ is the mass, $r$ the radius, or the length for a rod, $k$ a coefficient fixed by the shape and $I$ the moment of inertia. Every shape takes this same form, and only the coefficient changes.

The coefficient for each shape

A thin ring about its centre gives 1
A disc or cylinder about its centre gives one half
A thin spherical shell gives two thirds
A solid sphere gives two fifths
A thin rod about its centre gives one twelfth
A thin rod about one end gives one third

The ring has the largest coefficient because every part of its mass sits at the full distance $r$ from the axis. A solid sphere falls to two fifths, since much of its mass lies close in. The same rod is four times harder to spin about its end than about its centre.

Rotational energy

E = \dfrac{1}{2} I \omega^2, \qquad L = I \omega

Here $\omega$ is the angular velocity, $E$ the rotational kinetic energy and $L$ the angular momentum. These are $\frac{1}{2} m v^2$ and $m v$ with the mass replaced by the moment of inertia and the speed by the angular velocity.

Worked example

A disc of 2 kg and radius 0.3 m turning at 10 rad/s has a moment of inertia of 0.09 kg·m².

The coefficient is one half, so 0.5 × 2 × 0.3² = 0.09. The rotational kinetic energy is 0.5 × 0.09 × 10² = 4.5 J and the angular momentum is 0.09 × 10 = 0.9.

Points to watch

Every shape here is taken as having its mass spread evenly through it.

Moving the axis changes the coefficient. For an axis not in the list, use the parallel axis theorem: shifting the axis a distance $d$ from the centre of mass adds $m d^2$ to the moment of inertia. It also accounts for the step from one twelfth to one third for the rod.