I have seen an expression for the angular momentum of a rotating sphere calculated from outside the sphere as
$$L = I\omega + mvr,$$
where $v$ is the velocity of the center of mass, $m$ is the mass of the sphere, and $r$ is the distance of center of mass of the sphere from the point of calculation. My concern is if $v=0$, then $L = I \omega$, which is the angular momentum of the sphere when calculate through the center of mass. How can the angular momentum be the same when the point of calculation is changing?
[Physics] Angular momentum of a rotating sphere from a point outside the sphere
angular momentumnewtonian-mechanicsreference framesrotational-kinematics
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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.
Relation to parallel axis theorem
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.
The formula you mention is for general motion of an object that is both rotating and translating. Let $\mathbf{r}$ be the position of a point in the object and $\mathbf{r}_{\text{CM}}$ be the position of the center of mass of the object. We define the vector $$\mathbf{r}' \equiv \mathbf{r} - \mathbf{r}_{\text{CM}}$$ which represents the position of a point relative to the center of mass. The angular momentum $\mathbf{L}$ can be found by $$\mathbf{L} = \int \mathbf{r} \times \text{d}\mathbf{p} = \int \mathbf{r} \times \dot{\mathbf{r}} \text{d}m = \int \left(\mathbf{r}_{\text{CM}}+\mathbf{r}'\right) \times \left(\dot{\mathbf{r}}_{\text{CM}}+\dot{\mathbf{r}}'\right) \text{d}m \\ = \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} \text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}'\text{d}m + \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}'\text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}_{\text{CM}}\text{d}m$$
The last two terms vanish by the definition of the center of mass $\int \mathbf{r}' \text{d}m = 0$, so we have $$\mathbf{L} = \int \mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} \text{d}m + \int \mathbf{r}' \times \dot{\mathbf{r}}'\text{d}m \\ = m\mathbf{r}_{\text{CM}} \times \dot{\mathbf{r}}_{\text{CM}} + \int \mathbf{r}' \times \left(\boldsymbol{\omega} \times \mathbf{r}'\right)\text{d}m \\ = \mathbf{r}_{\text{CM}} \times \mathbf{p}_{\text{CM}}+ I\boldsymbol{\omega}$$
The first term is the $mvr$ that you mention which is commonly called the "orbital" angular momentum. The second term represents the rotation (spin) of the body.
The point is that the first term only makes sense when you consider an origin. In this case, the origin can be taken as any point fixed on the surface. The center of mass then moves in a straight line above the surface and parallel to it. You can check that $\mathbf{r}_{\text{CM}} \times \mathbf{p}_{\text{CM}}$ is constant.
The body does not have to actually orbit the origin. As long as the position and velocity of the center of mass are not parallel, the first term will be non-zero. It will only be zero if the body is heading directly towards or away from the origin. In other words, it can be thought of as how much the body is "going past" or "missing" the origin.
Best Answer
Don't confuse angular momentum and mass moment of inertia. The MMOI changes with location because of the assumption that the body rotates about that point (and hence the COM moves with $v \neq 0$).
But angular momentum depends on the position only if there is linear momentum present. This is entirely equivalent to a torque varying by position in the presence of forces, and of linear velocity varying by position in the presence of rotational velocity. The equations look the same too when considering an arbitrary point A and the center of mass point C.
$$\begin{array}{r|c} \mbox{Quantity} & \mbox{Transformation} \\ \hline \mbox{Linear Velocity} & {\boldsymbol v}_A = {\boldsymbol v}_C + {\boldsymbol r} \times {\boldsymbol \omega}\\ \hline \mbox{Angular Momentum} & {\boldsymbol L}_A = {\boldsymbol L}_C + {\boldsymbol r} \times {\boldsymbol p}\\ \hline \mbox{Torque} & {\boldsymbol \tau}_A = {\boldsymbol \tau}_C + {\boldsymbol r} \times {\boldsymbol F} \end{array}$$
Where ${\boldsymbol r}$ is the position vector of the center of mass C relative to the point of interest A.
So in the absence, of linear momentum ${\boldsymbol p} = m\, {\boldsymbol v}_C$ the angular momentum ${\boldsymbol L}_A$ at an arbitrary point A is the same as the angular momentum at the center of mass ${\boldsymbol L}_C = \mathrm{I}_C {\boldsymbol \omega}$.
Also, in the absence, of rotational velocity ${\boldsymbol \omega}$ linear velocity ${\boldsymbol v}_A$ at an arbitrary point A is the same as the linear velocity at the center of mass ${\boldsymbol v}_C$. (pure translation)
Finally, in the absence, any applied force ${\boldsymbol F}$ the equipollent torque ${\boldsymbol \tau}_A$ at an arbitrary point A is the same as the equipollent torque at the center of mass ${\boldsymbol \tau}_C$. (pure torque applied)
Now to consider the mass moment of inertia about a point A not the center of mass C.
The first part of this expression is the intrinsic angular momentum and the second part the parallel axis theorem.
Mathematically the rotational velocity vector if factored out of this expression and the 3×3 matrix between the momentum and velocity is the mass moment of inertia
$$ \mathrm{I}_A = \mathrm{I}_C - m \,[{\boldsymbol r} \times] [{\boldsymbol r} \times] $$
A little mathematical trick here is to convert the vector cross product into a matrix operation $ a \times b = [a \times] b = \left[\matrix{0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0}\right] \pmatrix{b_x\\b_y\\b_z} $