# [Physics] Physical meaning of the moment of inertia about an axis

angular velocityclassical-mechanicsinertiamoment of inertiarigid-body-dynamics

In the context of rigid bodies, the inertia tensor is defined as the linear map that takes angular velocity to angular momentum, that is, the linear map $I : \mathbb{R}^3\to \mathbb{R}^3$ such that

$$\mathbf{L}=I\boldsymbol{\omega}.$$

Now, given one unit vector $\hat{\mathbf{n}}$ characterizing the direction of a line, one can define

$$I_{\mathbf{n}}=\hat{\mathbf{n}}\cdot I(\hat{\mathbf{n}}),$$

which is the moment of inertia about that axis.

In that setting, if $\boldsymbol{\omega}= \omega \ \hat{\mathbf{n}}$ one gets, for instance, the nice looking formula for kinetic energy:

$$T = \dfrac{1}{2}I\omega^2,$$

where $I$ is the moment of inertia about the axis of rotation.

Now, although I grasp mathematically what is going on, I have no idea whatsoever about the physical meaning of the moment of inertia about an axis.

What is the physical meaning of the moment of inertia about an axis? What it really is, and how this physical significance relates to the actual mathematical definition I gave?

The 3×3 mass moment of inertia represents a tensor that expresses a single radius of gyration for each plane passing through the center of of mass.

What is a radius of gyration?

The radius of gyration (RGYR) expresses the distribution of mass around the rotation axis (perpendicular to said plane) as an equivalent ring or cylinder with the entire mass on a single radius from the axis.

But it does more. It defines also where the percussion axis is for a given rotation away from the center of mass.

What is a percussion axis?

The percussion axis, is commonly referred to as the sweet spot in sports is the axis in space which when impacted induces a particular rotation.

How?

In 2D it is kind of magical. Suppose you have a rigid body with radius of gyration $r_G$ and you want to rotate it about a pivot located a distance $c$ from the center of mass. Here is the sketch on the plane perpendicular to the rotation

I have drawn the radius of gyration from the center of mass.

1. Draw construction lines from the rotation point tangent to the radius of gyration and connect the tangent points
2. Mirror this line about the center of mass

You have now the percussion axis defined

$$p = \frac{I_{\rm plane}}{m c} = \frac{r_G^2}{c}$$

Notice that the percussion axis is purely a geometrical construct once the radius of gyration is known. The above is related to a pole-polar mapping in geometry.

Lemma

The radius of gyration on a plane can map each point on the plane (rotation center) to a unique line on the plane (percussion axis) and vise versa. If the rotation point is at infinity (a pure translation) then the percussion axis passes through the center of mass (a force through CM translates a body). Additionally, if the rotation point is at the center of mass then the percussion axis is at infinity which represents a pure torque on the body. Hence a pure torque will always rotate a body about its center of mass.

In 3D the 3×3 mass moment tensor represents three radiii of gyration and a mass.

$$\boldsymbol{\rm I} = m \begin{vmatrix} r_y^2+r_z^2 & -r_x r_y & -r_x r_z \\ -r_x r_y & r_x^2+r_z^2 &-r_y r_z \\ -r_x r_z & -r_y r_z & r_x^2+r_y^2 \end{vmatrix}$$

The above is reduced into three principal radii of gyration about some rotated coordinate system which eliminates the cross terms (non diagonal terms).

Proof of the percussion geometry

Consider the triangle ABC from the rotation to the tangent point and the center of mass.

The angle $\theta$ is found by $\sin \theta = \frac{r_G}{c}$

Now consider the small triangle BDC which has a side $p = r_G \sin \theta$ because it is similar to ABC.

The percussion axis on the mirror point B' is thus

$$\boxed{ p = r_G \sin\theta = \frac{r_G^2}{c} }$$

I will leave it up to the reader to prove that a force through the percussion axis on a resting body will cause a rotation about the pivot without any reaction forces.