Angular momentum of a rigid body in 3d
In this blog post, I explore the concept of angular momentum for a rigid body undergoing three-dimensional motion. I derive the formula for angular momentum about an arbitrary point, starting from the basic definition involving the integral of the cross product between the position vector and velocity. I show how to incorporate the relative velocity equation to describe the velocity of mass elements within the rigid body. I then introduce the inertia tensor and show how it simplifies the expression for the angular momentum. I also cover specific cases where the reference point is the center of mass or a fixed point in the inertial frame, highlighting how the equation simplifies under these conditions.
The angular momentum \mathbf{L}_P of a rigid body about a point \mathbf{P} is given by the integral over the body:
\mathbf{L}_P = \int (\mathbf{r} \times \mathbf{v}) \, \mathrm{d}m
where \mathbf{r} is the position vector from \mathbf{P} to a mass element \mathrm{d}m and \mathbf{v} is its velocity.
Using relative velocity kinematics, I can express the velocity of the mass element \mathrm{d}m as:
\mathbf{v} = \mathbf{v}_P + \boldsymbol{\omega} \times \mathbf{r}
where \mathbf{v}_P is the velocity of point \mathbf{P} and \boldsymbol{\omega} is the angular velocity of the body. Substituting this into the integral, I have:
\mathbf{L}_P = \int \mathbf{r} \times (\mathbf{v}_P + \boldsymbol{\omega} \times \mathbf{r}) \, \mathrm{d}m
This equation shows that the angular momentum depends on both the translational motion of point P and the rotational motion of the body. By expanding this expression and using the concept of the center of mass and the inertia tensor I am able to show that:
\mathbf{L}_P = m \, \mathbf{r}_{PC} \times \mathbf{v}_P + \mathbf{I}_P \boldsymbol{\omega}
where \mathbf{r}_{PC} is the position vector from \mathbf{P} to the center of mass \mathbf{C}, m is the total mass, and \mathbf{I}_P is the inertia tensor about point \mathbf{P}.
If \mathbf P is the center of mass, \mathbf{r}_{PC} = 0 the term \mathbf{r}_p \times m \mathbf{v}_p vanishes, resulting in a simplified equation:
\mathbf L_C = \mathbf{I}_C \boldsymbol{\omega}
Similarly, if \mathbf{P} is a point \mathbf{A} fixed in an inertial frame, or if point \mathbf{P} has no velocity the term \mathbf{r}_p \times m \mathbf{v}_p vanishes, and I am left with:
\mathbf L_A = \mathbf{I}_A \boldsymbol{\omega}
These special cases illustrate how the angular momentum calculation simplifies when considering specific points of reference.
For more insights into this topic, you can find the details here.