# I Total angular momentum is the sum of angular momentum of CM and that about CM

1. Dec 2, 2016

### Happiness

Consider a flat 2D rigid body rotating about an axis perpendicular to the body passing through a point P that is
(1) in the same plane as the body and
(2) different from the body's center of mass (CM).

In this case does Theorem 7.1 (eqn 7.9) still apply?

In the last step of the derivation of (7.9), $\omega'$ is taken out of the integral with respect to $m$. This is valid as long as each mass element $dm$ has the same angular velocity $\omega'$ with respect to the perpendicular axis passing through the CM. But this is not the case here, since here each mass element $dm$ has the same angular velocity with respect to the perpendicular axis passing through point P.

On the other hand, I'm thinking the theorem may still apply because any rotation about point P (with angular velocity $\omega$) can be viewed as a superposition of a rotation of the CM about point P (with angular velocity $\Omega$ in a reference frame where P is stationary) and a rotation of the rigid body about the CM (with angular velocity $\omega'$ in a reference frame where the CM is stationary). Is this true? Why? (A rotation about point P means a rotation of the rigid body about the axis perpendicular to the object and passing through point P. EDIT: When I look at this again, I find this useless because $\omega'=0$. And it only shows the theorem is true if we take the origin/pivot to be at point P but not at a different point Q for the calculation of angular momentum.)

Second question
Suppose a force is applied and then removed such that it causes a rigid body to rotate (with or without the translation of the CM). The object can only rotate about an axis passing through the CM and not any other point. Is it true? Why?

Last edited: Dec 2, 2016
2. Dec 6, 2016

### Simon Bridge

You can actually answer your questions yourself by dividing the body into lots of small masses and doing the math.

3. Dec 6, 2016

### Stephen Tashi

I think it does, but I don't regard the text you quoted as a complete proof of the theorem because it doesn't deal with the objection you raised. Theorem 7.1 is a more general statement that eq. 7.9. Theorem 7.1 speaks of the angular momentum "of a body". We can define the angular momentum "of a body" even when we cannot assign an angular velocity to the body.

To answer that, you'd have to state precisely what it means for one rotation to be a "superposition" of two other rotations. That might be a distraction to the question at hand. The computation of angular momentum is defined in terms of velocity (i.e. translational motion). The connection between angular momentum and angular velocity happens in the case of a rigid body because the parts of the body don't move independently of each other. If we have a "body" that is just a swarm of particles moving around independently of each other then we can still define a total angular momentum of the "body" about a point P, but there isn't a unique number $\omega$ that gives the angular velocity of each of the individual parts about P.

Perhaps the best approach is to prove theorem 7.1 in the general case first. That involves stating the precise definition of "center of mass", but doesn't involve the question of mathematically defining a "rigid body".
I think that involves two technicalities - the definition of "rigid body" and the definition of "angular velocity". When we speak of "the angular rotation of body B about point P", how can we define this quantity? Would our definition makes sense when B is a swarm of independently moving particles? If B is a rigid body and P is just an arbitrary point then can "the angular rotation of B about point P" be interpreted by using our definition ?

Angular momentum is a different issue than angular velocity Angular momemtum can be interpreted in either case by defining angular momentum to the sum of the individual (and possibly different) angular momenta of the pieces of the body. But angular velocity is a concept that does not depend on mass. So if we have collection of particles of different masses then summing their individual angular velocities (as vectors) defines a number, but it isn't going to be a useful number except in special situations.

4. Dec 8, 2016

### Happiness

Is the following explanation correct?

After the force is removed, there is no force acting on the body. So the CM is either stationary or moving at a constant velocity, by the conservation of linear momentum. We may pick the reference frame to be one where the CM is stationary without a loss in generality. The body can only rotate about the CM and not any other point in this reference frame. This is because if the body rotates about another point, that point would be stationary but the CM would NOT, resulting in a contradiction.

5. Dec 8, 2016

### Stephen Tashi

I think we have to define "rotate about a point" precisely.

It isn't clear why you say "if the body rotates about another point, that point would be stationary." Are you saying that rotation about a point that is not the C.M. is violation of conservation of (linear) momentum. How can that be shown in detail?

An even simpler question is "Can a rigid body rotate about two distinct points at the same time ?"

6. Dec 8, 2016

### Happiness

A 2D body rotates about a point P if there exists a reference frame in which point P is stationary and every point of the body rotates about (an axis passing through) point P with the same angular velocity.

It would be stationary by the above definition.

Yes, you may place the contradiction at the violation of conservation of linear momentum or at the violation of the fact that a rigid body cannot rotate about two distinct points at the same time.

I didn't work out in detail how conservation of linear momentum implies the CM must be stationary or moving at constant velocity. But I believe it can be done.

This is true, but we also need to establish that CM is stationary. And it is stationary, purportedly because of conservation of linear momentum.

7. Dec 9, 2016

### vanhees71

Let's make a simple model of a rigid body by supposing you have a collection of mass points that are held together with massless rigid rods. This implies that to describe the motion of this rigid body in an inertial frame you need to fix one point of the body (which doesn't need to be the location of one of the point masses) and an arbitrary body-fixed Cartesian basis system $\vec{e}_j'$ and the body-fixed point as origin. As the body-fixed point we choose the center of mass, i.e.,
$$\vec{x}_s=\sum_{i=1}^{N} \frac{m_i \vec{x}_i}{M},$$
where $M=\sum_i m_i$ is the total mass of the body, and $\vec{x}_i$ the position vector wrt. to the space-fixed origin of the inertial frame. Now it is convenient to introduce the position vectors relative to the center of mass,
$$\vec{x}_i=\vec{x}_s+\vec{r}_i.$$
Then the components of the $\vec{r}_j$ wrt. the body-fixed reference frame are time-independent, which defines the body to be rigid. The bases are both Cartesian by assumption, and we also assume that both are right-handed. Then they are connected by a rotation, i.e.,
$$\vec{e}_j'=D_{kj} \vec{e}_k.$$
The rotation matrix $\hat{D}=(D_{kj}) \in \mathrm{SO}(3)$ is time dependent but the space-fixed basis vectors are time independent, and we need the time derivative,
$$\dot{\vec{e}}_j'=\dot{D}_{kj} \vec{e}_k = \dot{D}_{kj} D_{kl} \vec{e}_l',$$
where in the final step we have used the orthogonalty of the rotation matrix. Since $D_{kj} D_{kl}=\delta_{jl}=\text{const}$ the matrix $\omega_{jl}=\dot{D}_{kj} D_{kl}$ is antisymmetric, i.e., we can write
$$\omega_{jl}=\epsilon_{ijl} \omega_i.$$
We note that $\vec{\omega}$ is the momentary angular velocity of the rotation of the rigid body around its center of mass.

Thus we have
$$\dot{\vec{x}}_j=\epsilon_{ijk} \omega_i \vec{e}_k'.$$
thus we have
$$\dot{\vec{x}}_i=\dot{\vec{x}}_s + \vec{r}_j = \dot{x}_{sj} \vec{e}_j + r_{ij}' \dot{\vec{e}}_j' = \dot{x}_{sj} \vec{e}_j + r_{ij}' \omega_l \epsilon_{ljk} \vec{e}_k'.$$
Now we can rewrite this again as vector equations
$$\dot{\vec{x}}_i = \dot{\vec{x}}_s + \vec{\omega} \times \vec{r}_i.$$
For the total angular momentum this implies
$$\vec{J}=\sum_i m_i \vec{x}_i \times \dot{\vec{x}}_i = \sum_i m_i (\vec{x}_s + \vec{r}_i) \times (\dot{\vec{x}}_s + \vec{\omega} \times \vec{r}_i).$$
Now we multiply out the vector product and use that
$$\sum_i m_i \vec{r}_i=0,$$
$$\vec{J}=\sum_i m_i [\vec{x}_s \times \dot{\vec{x}}_s + \vec{r}_i \times (\vec{\omega} \times \vec{r}_i)].$$
So obviously we have
$$\vec{J}=M \vec{x}_s \times \dot{\vec{x}}_s + \sum_i m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i)].$$
The first term is the angular momentum due to the motion of the center of mass around the space-fixed origin of the reference frame, and the second term is due to the rotation of the mass points around the body-fixed center of mass, i.e., the spin of the rigid body. We can rewrite this term a bit to get a more familiar form,
$$\vec{s}=\sum_i m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i) = \sum_i m_i [\vec{\omega} \vec{r}_i^2-\vec{r}_i (\vec{r}_i \cdot \vec{\omega})].$$
Introducing the tensor of inertia we can write
$$\vec{s}=\hat{\Theta} \vec{\omega},$$
where $\hat{\Theta}=(\Theta_{jk})$ with
$$\Theta_{jk}=\sum_{i} m_i (\vec{r}_i^2 \delta_{jk}-r_{ij} r_{jk}).$$
Finally we get the familiar form for the total angular momentum
$$\vec{J}=\vec{L}+\vec{s}=M \vec{x}_s \times \dot{\vec{x}}_s + \hat{\Theta} \vec{\omega}.$$

Last edited: Dec 11, 2016
8. Dec 15, 2016

### Stephen Tashi

That sounds reasonable, although I'm not sure what properties you assume when you say "reference frame".

Let's consider the case of a N "point masses". Let the origin of our coordinate system be point $O$. We won't assume the masses form a rigid body. Assume mass $m_i$ has position vector $r_i$ relative to the origin and velocity $v_i$, but both $r_i$ and $v_i$ are functions of time and there is no other particular relationship between $O$ and the masses.

Let $P$ be an arbitrary (fixed) point in 2D space. (i.e. it's coordinates are constant with respect to time).

We can write each $r_i$ as the sum of two vectors $\overrightarrow{OP} + p_i$ where $p_i$ is the vector from $P$ to mass $m_i$.

The total angular momentum of the masses about $O$ is:

$A_{total} = \sum_{i=1}^N ( r_i \times m_i v_i) = \sum_{i=1}^N ( (\overrightarrow{OP} + p_i) \times m_i v_i)$

$= \sum_{i=1}^N ( \overrightarrow{OP} \times m_i v_i + p_i \times m_i v_i )$

$= \sum_{i=1}^N (\overrightarrow{OP} \times m_i v_i) + \sum_{i=1}^N ( p_i \times m_i v_i)$

The contrast between the two sums above is interesting. The first sum involves the common factor $\overrightarrow{OP}$. It looks like an angular momentum computation where the position of each mass has been translated so it lies on the end of vector $\overrightarrow{OP}$. The other sum is the usual calculation for the total angular momentum of the masses about point $P$.

Now let's add the assumption that the masses are rotating about point $P$ at the current time. I'll specify the meaning of this assumption without insisting the masses are rotating at an angular velocity that is constant with respect to time. I also won't insist that $P$ is the center of mass or that the masses continue to rotate about point $P$ as time passes. All I will assume is that (at the current instant of time): $v_i = |p_i| \omega \hat{s_i}$ where $\hat{s_i}$ is the unit vector perpendicular to $p_i$ chosen to have the same "sense" (right hand vs left hand) for all the masses.

The second sum becomes $\sum_{i=1}^N (p_i \times m_i v_i) = \sum_{i=1}^N( p_i \times m_i |p_i|\omega \hat{s_i})$

$= \sum_{i=1}^N ( m_i |p_i|^2 \omega \hat{z})$ where $\hat{z}$ is the unit vector that points "into the page" or "out of the page" depending on the sense of the $\hat{s_i}$.

$= \omega ( \sum_{i=1}^N ( |p_i|^2 m_i) ) \hat{z}$

The factor with the summation suggests the "moment of inertia" about point $P$ , i.e. $\int |p(m)|^2 m\ dm$. From the way physicists reason, we can proceed from a system of "point masses" to the continuous case with a wave of the hand. I would hate to have to formulate a rigorous mathematical argument that makes that leap, but the physics way of thinking seems reasonably reliable.

The sum $\sum_{i=1}^N (\overrightarrow{OP} \times m_i v_i)$ can be expressed as $\overrightarrow{OP} \times (\sum_{i=1}^N m_i v_i)$ using the fact that the cross product distributes over addition.

That cross product is an angular momentum computation involving one position, $\overrightarrow{OP}$ and different linear momenta $m_i v_i$. Can we express it as a sum $\overrightarrow{OP} \times MV$ involving only one linear momentum $MV$? The algebraic way to do it is $\sum_{i=1}^N m_i v_i = M \frac{\sum_{i=1}^N m_i v_i}{M}$ where $M$ is the total mass of all the particles. We can regard the vector $\frac{\sum_{i=1}^N m_i v_i}{M}$ as a single linear velocity $V$ (even though it contains terms involving masses).

$V = \frac{\sum_{i=1}^N m_i v_i}{M} = \frac{d}{dt} \frac{\sum_{i=1}^N m_i r_i}{M}$. This introduces the expression $\frac{\sum_{i=1}^N m_i r_i}{M}$, which is , by definition, the "center of mass". So $V$ can be interpreted as the velocity of the center of mass.