How do I find the angular momentum of a body about any point? We know that $L=I\omega$ for a body rotating in space, where $L$ denotes the angular momentum, $I$ denotes the moment of inertia and $\omega$ denotes the angular velocity. However, this is only applicable for fixed axis of rotation, instantaneous axis of rotation and center of mass. Can somebody state and prove the value of the angular momentum of a body about any point? (if a formula for that exists)?
Answer
Let's suppose I have some system and I know the system's total mass $M$, the system's center of mass position $\newcommand{\v}[1]{\mathbf{#1}}\v{r}_{cm}$, and the systems angular velocity $\v{L}_{cm}$, in the frame where the center of mass is the origin. How do I find $\v{L}'$, the angular momentum with respect to some other origin, say $\v{r}_{0} $, which is moving at a velocity $\v{v}_0$? That is the question I will answer.
The answer is easy to understand intuitively. The total angular momentum in the new frame is the sum of two terms. The first term is the angular momentum in the center of mass frame $\v{L}_{cm}$. This piece is intrinsic to the motion in the sense that it does not depend on the frame. The second piece is frame dependent, but has a simple form which does not depend on the details of the system. The frame dependent piece is $ M \left(\v{r}_{cm}-\v{r}_0\right) \times \left(\v{v}_{cm}-\v{v}_0\right)$. Notice this is just the angular momentum of a point particle with position $\v{r}_{cm}$ and velocity $\v{v}_{cm}$. So the system can be modeled as a point particle for the purposes of calculating the frame dependent piece.
It is not hard to prove that the angular momentum decomposes this way. To do this, we will introduce the notation $\langle X \rangle = \int X dm$, so that when we write $\langle \v{r} \rangle$, we mean $\int \v{r} dm = M \v{r}_{cm}$. With this notation, the angular momentum in the frame with origin $\v{r}_0$ moving at velocity $\v{v}_0$ is $$\begin{equation} \begin{aligned} \v{L}'=&\langle \left(\v{r}-\v{r}_0\right) \times \left(\v{v}-\v{v}_0\right)\rangle\\ =&\langle \left(\left(\v{r}-\v{r}_{cm}\right) +\left(\v{r}_{cm} -\v{r}_0\right)\right) \times \left(\left(\v{v}-\v{v}_{cm}\right)+\left(\v{v}_{cm}-\v{v}_0\right)\right)\rangle\\ =&\overbrace{\langle \left(\v{r}-\v{r}_{cm}\right) \times \left(\v{v}-\v{v}_{cm}\right) \rangle}^{\v{L}_{cm}}\\ &+\langle \left(\v{r}-\v{r}_{cm}\right) \times\left(\v{v}_{cm}-\v{v}_0\right) \rangle\\ &+\langle \left(\v{r}_{cm} -\v{r}_0\right) \times \left(\v{v}-\v{v}_{cm}\right) \rangle\\ &+\langle \left(\v{r}_{cm} -\v{r}_0\right) \times \left(\v{v}_{cm}-\v{v}_0\right) \rangle\\ =&\v{L}_{cm}\\ &+ \overbrace{\langle\v{r}-\v{r}_{cm} \rangle}^{0} \times\left(\v{v}_{cm}-\v{v}_0\right)\\ &+\left(\v{r}_{cm} -\v{r}_0\right) \times \overbrace{\langle \v{v}-\v{v}_{cm}\rangle}^{0}\\ &+M \left(\v{r}_{cm} -\v{r}_0\right) \times \left(\v{v}_{cm}-\v{v}_0\right) \\ =&\v{L}_{cm} + M \left(\v{r}_{cm} -\v{r}_0\right) \times \left(\v{v}_{cm}-\v{v}_0\right) \end{aligned} \end{equation}$$
Above, in the third line, we find that $\v{L}'$ is the sum of four terms. The first is the angular momentum in the center of mass frame, $\v{L}_{cm}$, in the second and third terms, a constant can be factored out of the angle brackets and what remains in the brackets averages to zero. In the fourth term, the quantity in brackets is just a constant, so the brackets amount to multiplication by $M$. The two surviving terms are exactly the terms described in the previous paragraph.
You might think that you use the parallel axis theorem here. The parallel axis theorem is actually a special case of this where the displacement of the origin is perpendicular to the axis of rotation, and your new origin is some point embedded in the object (assumed to be rigid). By embedded in the object, I mean that the new origin is moving at the same velocity of the object at that point so that $\v{v}_0-\v{v}_{cm}= \boldsymbol{\omega} \times \left( \v{r}_0 - \v{r}_{cm}\right)$.
The equation we derived in this answer then predicts \begin{equation} \begin{aligned} \v{L}' &= \v{L}_{cm} + M \left(\v{r}_{cm}-\v{r}_0\right) \times \left(\boldsymbol{\omega} \times \left(\v{r}_{cm} - \v{r}_0\right)\right)\\ &= \v{L}_{cm} + M\left( \boldsymbol{\omega} \left(\v{r}_{cm}-\v{r}_0\right)^2 - \left(\v{r}_{cm}-\v{r}_0\right) \overbrace{\left(\v{r}_{cm}-\v{r}_0\right)\cdot \boldsymbol{\omega}}^{0} \right)\\ &=\v{L}_{cm} + M \boldsymbol{\omega} \left(\v{r}_{cm}-\v{r}_0\right)^2. \end{aligned} \end{equation}.
On the other hand, the parallel axis theorem would tell us to make the subsitution $I_{cm} \to I_{cm} + M \left(\v{r}_{cm}-\v{r}_0\right)^2$. Thus we would have $$I_{cm} \boldsymbol{\omega} \to \left(I_{cm} + M\left(\v{r}_{cm}-\v{r}_0\right)^2\right)\boldsymbol{\omega} = I_{cm} \boldsymbol{\omega} +M \left(\v{r}_{cm}-\v{r}_0\right)^2\boldsymbol{\omega}.$$ So that $\v{L}_{cm} \to \v{L}_{cm} + M\left(\v{r}_{cm}-\v{r}_0\right)^2\boldsymbol{\omega}$. I.e., $\v{L}' = \v{L}_{cm} + M\left(\v{r}_{cm}-\v{r}_0\right)^2\boldsymbol{\omega}$. Thus we see how the answer we get are the same in this special case, and the parallel axis theorem can be used. However your question concerns more general transformations.
No comments:
Post a Comment