Rotational Energa & Moment of Inertia

The total energy contained in the rotation of a rigid body around its center of mass can be calculated by dividing it into $N$ small cubes with mass $\Delta m_i$, and computing the kinetic energy of each cube. This method is similar to finding the center of mass. Then all the energies are summed. Each of these cubes is at a distance $\vec{r}_i$ from the center of mass. This results in:

$$
E_\mathrm{rot} = \sum_{i=1}^{N}\frac{1}{2} \Delta m_i v_i^2 = \sum_{i=1}^{N}\frac{1}{2} \Delta m_i \omega^2 r_i^2
$$

In the last step, the relation $v = \omega r$ was used, since the angular velocity is the same everywhere. The most accurate value is obtained when the individual cubes become infinitesimally small, i.e., when taking the limit $\Delta m \rightarrow 0$. In this case, the sum becomes an integral:

$$
E_\mathrm{rot} = \int \frac{1}{2} \omega^2 r^2\, \mathrm{d}m
$$

The quantity

$$
I = \int r^2\,\mathrm{d}m
$$

is called the moment of inertia.

Inserting the definition of the moment of inertia gives a simple formula for the rotational energy:

$$
E_\mathrm{rot} = \frac{1}{2}I\omega^2
$$

Note that a rolling object possesses both rotational and kinetic energy. In this case, the total energy is always the sum of both forms:

$$
E_\mathrm{roll} = E_\mathrm{rot} + E_\mathrm{kin}
$$

For a point mass $m$ at a distance $r$ from the axis of rotation, the definition of the moment of inertia simplifies to:

$$
I = mr^2
$$

In the case of continuous mass distributions, the calculation is generally much more complex. However, for many geometric bodies, the integral of the moment of inertia can still be solved analytically. The following table provides an overview of calculated moments of inertia for cylinders, spheres, and a long rod:

If the integral can no longer be solved analytically, numerical calculations or simulations are usually helpful. However, the moment of inertia of very complex shapes generally has to be determined experimentally.