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I was reading about the angular momentum of rigid bodies, and I cam across the following problem.

Imagine a solid sphere is rolling down over an inclined plane without slipping, and I was trying to find the angular momentum of this body.

I found the following formula : $L = I\omega + mvr = \frac{7}{5}mvr$

It is obvious, that the first term is the angular momentum, due to the body spinning about its central axis. However, where does the second term come from ? It seems similar to the orbital angular momentum of a point particle about some barycenter. However, not only is the sphere here, a point particle, but it also isn't in any orbit. The motion comprises of rolling about its axis, and the linear motion of its center of mass. However, the center of mass isn't rotating about any point whatsoever, so why is the $mvr$ component coming from.

Moreover, $r$ refers to the radius of the sphere. Hence, $mvr$ must be the moment of inertia of a point particle of mass $m$ orbiting the center of the sphere while it rolls down. If that is the case, haven't we already taken care of that in $I\omega$ ?

Why are we considering, the rotation of the entire sphere about its axis, and the rotation of a point of mass $m$ about the center of the sphere separately, and then summing over them ?

Any intuitive explanation of what exactly does $mvr$ represent would be highly appreciated.

I did the derivation for angular momentum, and the two terms popped up. However, isn't angular momentum always related to the rotation? The term $mvr$ is written as translational angular momentum. However, this name seems to be a little contradictory to me. How can pure translation be related to rotation - unless we claim that rotation is always related to angular momentum, but angular momentum might not always signify rotation as in this case ? Is this claim valid ? If not, can someone explain to me, the physical intuition or help me visualize the idea of translational angular momentum.

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    $\begingroup$ Consult Kleppner and Kolenkow: An Introduction to mechanics. It is much better than having existential crisis every two weeks wondering how all the formulas came to be in mechanics. $\endgroup$ Aug 12, 2021 at 18:21
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    $\begingroup$ Asking "what is the angular momentum of the body?" doesn't make any sense. You need to ask "What is the angular momentum of the body about X point?" Remember that the angular momentum is just the moment of linear momentum of all the material points in the body added up. This moment of linear momentum is about some point, perhaps the CG, perhaps a fixed point, perhaps an accelerating point. Consider a box sliding across a table in pure translation. Did you know that this box has angular momentum about a point on the surface of the table? It does not have angular momentum about its CG. $\endgroup$
    – Evan
    Aug 13, 2021 at 23:46

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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.

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  • $\begingroup$ Hmmm how is $\int r' \times( \omega \times r') dm = I \omega$? The I in OP's sense, where inertia is a scalar, could only be found if $\omega \perp r'$ $\endgroup$ Aug 12, 2021 at 18:21
  • $\begingroup$ @Buraian $I$ is supposed to be the inertia tensor. But here in 2D it is a scalar (and so is $\omega$). $\endgroup$ Aug 12, 2021 at 22:05
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The confusion arises because the $r$ in $I\omega=\frac{2}{5}mvr$ is the distance to the center of the rolling sphere. The $r$ in the second term $mvr$ is the distance to some fixed origin about which the total angular momentum is defined.

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The linear angular momentum is defined as $$\vec{L}=m(\vec{v}\times\vec{r})=mvr_{\perp}$$ Now the $r_{\perp}$ doesn't change as the body moves, yes the angle changes but so does the radius from the origin, so they compensate for each other keeping the perpendicular radius constant, and in case of a sphere this radius is the radius of the sphere provided you keep your origin at the base of the surface.

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EDIT: The latter part of this answer(which was the original answer) is correct, although as I've noticed many people do not prefer another theorem I am just explaining briefly here. There are two special types of angular momentum of an object, the spin angular momentum is the angular momentum about the object's centre of mass, while the orbital angular momentum is the angular momentum about a chosen center of rotation. The total angular momentum is the sum of the spin and orbital angular momenta. That is the situation here. We are considering the origin to be a point on the sphere which touches the plane, therefore net angular momentum is $I_{\text{spin}}+I_{\text{orbital}}=I\omega+mvr$


Angular momentum is defined for any motion about a particular axis. The formula $L=I\omega$ basically comes from $L=mvr$. You can see $mv$ in this equation is the linear momentum of the object and $r$ is the distance to the centre of mass from that particular axis. Here you are referring to the angular momentum about the contact point between the plane and the sphere. If you find the moment of inertia of the sphere about that point, from the parallel axis theorem you can get, $$\begin{align} I&=I_0+mr^2 \\&=\frac 25mr^2+mr^2\\&=\frac 75mr^2\end{align}$$ From that you can find the angular momentum about that point $$L=\frac 75mr^2\omega=\frac 75mr^2\frac vr=\frac75mvr$$ You have directly found that using $L=I\omega+mvr$, so you can call it as parallel axis theorem of angular momentum.

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  • $\begingroup$ This has nothing to do with the parallel axis theorem. This is a common incorrect usage of it that just happens to produce the right answer in this case. $\endgroup$ Aug 12, 2021 at 10:25
  • $\begingroup$ @VincentThacker ,Did you go through the link? And do you agree this is the right answer(according to your comment)? And please read the 3rd paragraph on wikipedia article. $\endgroup$
    – ACB
    Aug 12, 2021 at 10:26
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    $\begingroup$ Okay, so this might not be technically incorrect, but it is still bad usage. In addition, the OP is not asking about the parallel-axis theorem or a solution to the problem. The OP is asking about the technical meaning of a formula. I don't see how this answers the question. $\endgroup$ Aug 12, 2021 at 10:49
  • $\begingroup$ It seems to me that you are only considering my answer is bad just because I mentioned about parallel axis theorem. But I basically used it to find MOI about that point. And I can't understand why it is bad, if it is technically correct(as you agree). Something cannot be considered bad always, although it is not in regular usage. I think OP is asking why it is not just $L=I\omega$ and I have explained it as it is due to the chosen origin. Angular momentum about the centre of the sphere is $I\omega$. $\endgroup$
    – ACB
    Aug 12, 2021 at 11:23
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    $\begingroup$ @ACB Fair enough, after your edit, I have removed my downvote. $\endgroup$ Aug 14, 2021 at 3:27

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