What is the angular momentum of a particle rotating around an axis in 3D?

Question

What would be the angular momentum of the particle at position $r_i$ in the diagram above?

The vector from the axis of rotation is $R_i$ and the tangential velocity is $v_i$ so the magnitude of angular momentum should be $|R_i|m_i|v_i|$ which gives the vector $R_i \times (m_iv_i)$ since $R_i$ and $v_i$ are at right angles in the diagram.

However, if you look at most resources (for example or another example or another example) they all take each particle's angular momentum to be $r_i \times (m_iv_i)$ instead of $R_i \times (m_iv_i$). Which is correct and why?

Answer 1

The definition $\vec{L}_i = m \vec{r}_i \times \vec{v}_i$ is the correct one. Note that this vector will not lie along $\hat{n}$, and that's as it should be! Your derivation above makes the mistake of assuming the angular momentum vector of a rigid body will always be parallel to the axis of rotation. This is not the case. In general, an arbitrary rotating body will not have $\vec{L} \parallel \vec{\omega}$, which is why we need a tensor (the inertia tensor) to describe rigid-body motion in three dimensions. Only when you look at rotations around certain special axes of the body (called the principal axes) do you find that $\vec{L}$ is parallel to $\vec{\omega}$.

That said, the component of $\vec{r}_i \times \vec{v}_i$ along the axis of rotation $\hat{n}$ is the same as the component of $\vec{R}_i \times \vec{v}_i$ along that same axis. To see this, note that the component of any vector $\vec{A}$ along a unit vector $\hat{n}$ is $\vec{A} \cdot \hat{n}$. Applying this, we have $$ (\vec{r}_i \times \vec{v}_i) \cdot \hat{n} = \left[(\vec{R}_i + \alpha \hat{n} ) \times \vec{v}_i \right] \cdot \hat{n} = \left[\vec{R}_i \times \vec{v}_i+ \alpha \hat{n} \times \vec{v}_i \right] \cdot \hat{n} = \left[\vec{R}_i \times \vec{v}_i\right] \cdot \hat{n}. $$ (The factor of $\alpha$ introduced is the component of $\vec{r}$ along the axis $\hat{n}$, $\alpha = \vec{r} \cdot \hat{n}$; in the last step, the second term vanishes because $\hat{n} \times \vec{A} \perp \hat{n}$ for any vector $\vec{A}$.) So your method using $\vec{R}_i$ correctly calculates the component of $\vec{L}_i$ along the axis of rotation, but ignores the components along axes perpendicular to $\hat{n}$.

Answer 2

they all take each particle's angular momentum to be ri×(mivi) instead of Ri×(mivi). Which is correct and why?

I can understand your confusion since both terms do not seem to be the same. I think the main point can be extracted, for example, from the second ressource you cite

The total angular momentum of the body (about the origin) is written ...

Therefore, both can be correct, depending on the point you're interested in. The second ressource states out quite clearly that they consider the angular momentum wrt. the origin. Hence, their result is correct, although it is not what you might have "expected", namely something that matches with $I\omega = mr^2 \omega$. Here,, the angular momentum is described using the angular velocity $\omega$, which indeed requires some rotational axis. In more detail, when speaking about $\omega$ you implicitly define a rotational axes. Probably this caused the misconception that angular momentum is defined wrt. an axis(?)

Summing up:

• Angular momentum is defined with respect to a point, not an axis, which can be seen in the formula $\vec{L} = r \times m\vec{v}$ since both quantities are vectors (defined wrt. some origin). In this definition there does not occur any axis.
• However, you implicitly define an axis once you introduce the angular velocity $\omega$. So, once you describe the angular momentum using the angular velocity, there is some axis present, although the angular momentum is still defined wrt. a point.