How Does Angular Momentum Apply to Earth and Spinning Bike Wheels?

[alkaspeltzar]

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same.

Another example is the spinning bike wheel/person holding it in a chair. It has angular momentum about its axis which is equal to the angular momentum of the center of the chair say if I were holding it, but myself not moving.

I am trying to see a concrete/more physical understanding of how this.

Like an object moving in a straight line relative to a point still 'has some rotation'. I understand that. Trying to see how an object spinning about its axis has rotation to an axis in parallel that is not moving around it in a circle.

Here is a breakdown example. As you can see the spin angular momentum about Q is the same as it is for the ball center of mass.
https://scripts.mit.edu/~srayyan/PE...of_a_Rigid_Body_both_Rotating_and_Translating

 

[Dale]

Summary:: Why is angular momentum of object about its axis equal to that of a non-moving parallel axis?

if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same.

Yes, if you ignore the change then it doesn’t change.

The angular momentum does change. The total is for convenience split into a changing (orbital) and a non-changing (spin) part. If you ignore the change then it doesn’t change, but I don’t think that is particularly important.

 

[alkaspeltzar]

Yes, if you ignore the change then it doesn’t change.

The angular momentum does change. The total is for convenience split into a changing (orbital) and a non-changing (spin) part. If you ignore the change then it doesn’t change, but I don’t think that is particularly important.

Dale, can you explain though why it is the same?

Basically the ang. Momentum of an objects rotation of cof mass is same as any other axis assuming not moving? Why

Does it not matter? Despite it is still rotating about its cofm is there still rotation and Ang momentum about Q

 

[kuruman]

Start with a rigid body spinning about an axis passing through its center of mass. Let $\vec R_0$ be the position of the object's center of mass relative to the origin. Say you want to find the angular momentum of the object about the origin. You already know that the angular momentum about the spin axis is $\vec L_s=I_{cm}~\vec \omega.$

Now consider mass element $dm_i$ on the object. Let its position relative to the origin be $\vec r_i=\vec R_0+\vec r'_i$ where $\vec r'$ is the position vector of the element relative to the CM. The angular momentum of $dm_i$ relative to the origin is $$d\vec L_i=\vec r_i\times \vec p_i=dm_i\vec r_i\times \vec v_i=dm_i(\vec R_0+\vec r'_i)\times \vec v_i.$$ Since the axis is not moving, the linear velocity of the mass element relative to the origin is related to the angular velocity by $\vec v_i=\vec \omega \times \vec r'_i.$ Then $$d\vec L_i=dm_i(\vec R_0+\vec r_i')\times ( \vec \omega \times\vec r'_i).$$The triple cross product has two terms. The second term is $$\vec r'_i\times (\vec \omega \times \vec r_i)=\vec \omega( \vec r'_i \cdot \vec r'_i)-\vec r'_i(\vec\omega \cdot \vec r'_i)=\vec \omega {r'}_i^2.$$ The total angular momentum is the sum, $$\vec L=\sum_i d\vec L_i= \vec R_0\times \left( \vec \omega \times \sum_i dm_i\vec r'_i\right)+\vec \omega\sum_i dm_i {r'}_i^2.$$By the definitions of the center of mass and of the moment of inertia about the center of mass, $$ \sum_i dm_i \vec r'_i=0~;~~\sum_i dm_i {r'}_i^2=I_{cm}$$ Thus, $\vec L=\vec L_s=I_{cm}~\vec\omega.$

 

[alkaspeltzar]

Start with a rigid body spinning about an axis passing through its center of mass. Let $\vec R_0$ be the position of the object's center of mass relative to the origin. Say you want to find the angular momentum of the object about the origin. You already know that the angular momentum about the spin axis is $\vec L_s=I_{cm}~\vec \omega.$

Now consider mass element $dm_i$ on the object. Let its position relative to the origin be $\vec r_i=\vec R_0+\vec r'_i$ where $\vec r'$ is the position vector of the element relative to the CM. The angular momentum of $dm_i$ relative to the origin is $$d\vec L_i=\vec r_i\times \vec p_i=dm_i\vec r_i\times \vec v_i=dm_i(\vec R_0+\vec r'_i)\times \vec v_i.$$ Since the axis is not moving, the linear velocity of the mass element relative to the origin is related to the angular velocity by $\vec v_i=\vec \omega \times \vec r'_i.$ Then $$d\vec L_i=dm_i(\vec R_0+\vec r_i')\times ( \vec \omega \times\vec r'_i).$$The triple cross product has two terms. The second term is $$\vec r'_i\times (\vec \omega \times \vec r_i)=\vec \omega( \vec r'_i \cdot \vec r'_i)-\vec r'_i(\vec\omega \cdot \vec r'_i)=\vec \omega {r'}_i^2.$$ The total angular momentum is the sum, $$\vec L=\sum_i d\vec L_i= \vec R_0\times \left( \vec \omega \times \sum_i dm_i\vec r'_i\right)+\vec \omega\sum_i dm_i {r'}_i^2.$$By the definitions of the center of mass and of the moment of inertia about the center of mass, $$ \sum_i dm_i \vec r'_i=0~;~~\sum_i dm_i {r'}_i^2=I_{cm}$$ Thus, $\vec L=\vec L_s=I_{cm}~\vec\omega.$

So by mere break down of the math and by the definition of ang. momentum, relative to the origin, if one were to look at the mass element there still is rotation. Therefore upon working it out, they are the same.

That makes more sense now. Took me awhile to see how an object moving straight has ang. momentum to a point. The is similar. When we think about how the mass element changes positions relative to the origin, there is still a rotational components that must be conserved, hence the ang. momentum.

 

[kuruman]

To make another connection, consider the Moon that is orbiting the Earth with the same $\omega$ as its spin. The total angular momentum of the Moon relative to the Earth is $$L_{\text{Earth}}=L_{\text{orbit}}+L_{\text{spin}}=M_{\text{Moon}}R^2\omega+I_{\text{cm,Moon}}\omega$$where $R$ is the Earth-Moon distance. But you can also write the angular momentum about the Earth as $L_{\text{Earth}}=I_{\text{Earth}}\omega.$ With the parallel axis theorem, $I_{\text{Earth}}=I_{\text{cm,Moon}}+M_{\text{Moon}}R^2$, you get the same expression.

 

[alkaspeltzar]

To make another connection, consider the Moon that is orbiting the Earth with the same $\omega$ as its spin. The total angular momentum of the Moon relative to the Earth is $$L_{\text{Earth}}=L_{\text{orbit}}+L_{\text{spin}}=M_{\text{Moon}}R^2\omega+I_{\text{cm,Moon}}\omega$$where $R$ is the Earth-Moon distance. But you can also write the angular momentum about the Earth as $L_{\text{Earth}}=I_{\text{Earth}}\omega.$ With the parallel axis theorem, $I_{\text{Earth}}=I_{\text{cm,Moon}}+M_{\text{Moon}}R^2$, you get the same expression.

This is very helpful but I am curious is my comment above made sense. On a basic level, the angular momentum doesn't change as rotation about an objects cofm still dictates there is rotation relative to a fix axis R away, assuming it's not moving.

For angular momentum to be conserved, there has to be. And these examples shoe that to be true. Right?

 

[kuruman]

This is very helpful but I am curious is my comment above made sense. On a basic level, the angular momentum doesn't change as rotation about an objects cofm still dictates there is rotation relative to a fix axis R away, assuming it's not moving.

For angular momentum to be conserved, there has to be. And these examples shoe that to be true. Right?

Which comment "above" are you referring to?
The idea is simple: Relative to an arbitrary point, the angular momentum of an object can be decomposed into (a) angular momentum of the object's CM, a.k.a. orbital part; (b) angular momentum about the center of mass, a.k.a. spin part. An moving object may have one, the other, both or none depending on how it is moving and where the reference point is.

Example
Consider a sphere on a horizontal surface. The point of contact P is the reference point for the angular momentum in all cases shown below.
Case I: Ball is moving with constant velocity straight up such that point P is always below its center.
L_orbit = 0; L_spin = 0 ⇒ L_Total = 0.

Case II: Ball is sliding to the right with constant velocity v on frictionless surface .
L_orbit = MvR; L_spin = 0 ⇒ L_Total = MvR.

Case III: Ball is spinning in place with constant angular velocity ω on frictionless surface.
L_orbit = 0; L_spin = I_cmω ⇒ L_Total = I_cmω.

Case IV: Ball is rolling without slipping so that its center moves to the right with constant speed v.
L_orbit = MvR; L_spin = I_cmω ⇒ L_Total = MvR + I_cmω.
In this case because v = ω R, we can rewrite
L_Total = MωR² + I_cmω = ( I_cm + MωR² ) ω = I_Pω, where I_P is the moment of inertia about point P.

See how it works?

 

[alkaspeltzar]

Which comment "above" are you referring to?
The idea is simple: Relative to an arbitrary point, the angular momentum of an object can be decomposed into (a) angular momentum of the object's CM, a.k.a. orbital part; (b) angular momentum about the center of mass, a.k.a. spin part. An moving object may have one, the other, both or none depending on how it is moving and where the reference point is.

Example
Consider a sphere on a horizontal surface. The point of contact P is the reference point for the angular momentum in all cases shown below.
Case I: Ball is moving with constant velocity straight up such that point P is always below its center.
L_orbit = 0; L_spin = 0 ⇒ L_Total = 0.

Case II: Ball is sliding to the right with constant velocity v on frictionless surface .
L_orbit = MvR; L_spin = 0 ⇒ L_Total = MvR.

Case III: Ball is spinning in place with constant angular velocity ω on frictionless surface.
L_orbit = 0; L_spin = I_cmω ⇒ L_Total = I_cmω.

Case IV: Ball is rolling without slipping so that its center moves to the right with constant speed v.
L_orbit = MvR; L_spin = I_cmω ⇒ L_Total = MvR + I_cmω.
In this case because v = ω R, we can rewrite
L_Total = MωR² + I_cmω = ( I_cm + MωR² ) ω = I_Pω, where I_P is the moment of inertia about point P.

See how it works?

I was curious if based on the math and my comments on post #5 did you agree?

I get how it works. Just trying to conceptually understand how a ball spinning in place has angular momentum relative to a non moving parallel axis, mentioned in the example

If I understand the math right it doesn't need to have direct circular motion to have Ang momentum.

@Dale made a comment once to me, that torque is about change of ang momentum and rotation but not necessarily traveling in a circle. From the math you presented there is a rotation of the mass element as it moves via position vector ri=r0+ri'. It isn't necessarily circular but neither is an object as it passes a point traveling straight. Still an object traveling straight to a point has Ang momentum and this is similar. Doesn't need to be a circular movements to have Ang momentum. Agree?

 

[kuruman]

Doesn't need to be a circular movements to have Ang momentum. Agree?

Agree. There needs to be an angle changing with respect to time which means there is a non-zero ω. Imagine sitting on street bench. To keep looking at a man who passes in front of you left to right, you have to turn your head left to right. The angle of your head relative to your body changes. The man has orbital angular momentum relative to your neck even though the man is moving in a straight line. If the man crosses the street in front of you in a straight line so that you don't have to turn your head to keep looking at him, he has zero angular momentum relative to your neck.

 

[alkaspeltzar]

Agree. There needs to be an angle changing with respect to time which means there is a non-zero ω. Imagine sitting on street bench. To keep looking at a man who passes in front of you left to right, you have to turn your head left to right. The angle of your head relative to your body changes. The man has orbital angular momentum relative to your neck even though the man is moving in a straight line. If the man crosses the street in front of you in a straight line so that you don't have to turn your head to keep looking at him, he has zero angular momentum relative to your neck.

and that's what I was trying to understand with the ball and it's angular momentum in my examole. If you examine the ang. Momentum of the ball relative to an axis that's not at the center of mass, even though it spins about its center of mass, there is rotation to that point

Thanks to the math it shows as the position vector changes there is an angular velocity. So not circular rotation, but still rotation.

Thanks

PS, like your footnote

 

[hutchphd]

And all of this is recapitulated in the Parallel Axis Theorem which has to do directly with the moments of inertia. The physics is the same and I think someone needed to say those words.

 

[jartsa]

Summary:: Why is angular momentum of object about its axis equal to that of a non-moving parallel axis?
See link or examples below

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same.

Earth has rotational inertia $I$, according to everyone
Earth has angular velocity $w$, according to everyone
Earth has angular momentum $L=I*w$, according to everyone

We can move the Earth and its angular momentum anywhere we want, and then make the Earth to exert a torque on some axle there. Everywhere the same amount of angular momentum can be extracted.

I don't know what "angular momentum about on axis" means, except just angular momentum.

Hmm, oh yes, angular momentum of a linearly moving abject is observer dependent, so when we are talking about that kind of angular momentum, then it's necessary to say "angular momentum about an axis".

 

[Dale]

The angular momentum depends on the axis, full stop. The moment of inertia also depends on the axis.

There is a minimum angular momentum, which is the angular momentum about the center of mass, which is called the spin angular momentum. As you choose other axes the angular momentum will become larger than the spin angular momentum. You can write the total angular momentum as the sum of the spin angular momentum and the rest of the angular momentum. The rest of the angular momentum is called the orbital angular momentum.

 

[jartsa]

There is a minimum angular momentum, which is the angular momentum about the center of mass, which is called the spin angular momentum. As you choose other axes the angular momentum will become larger than the spin angular momentum.

Well that's wrong. Orbital angular momentum may cancel out spin angular momentum.

Disclaimer: I didn't actually understand anything about what you said.

 

[kuruman]

There is a minimum angular momentum, which is the angular momentum about the center of mass, which is called the spin angular momentum. As you choose other axes the angular momentum will become larger than the spin angular momentum.

This is a bit misleading in the sense that it is not necessarily true. In post #4 it is shown that if you choose another axis that is at rest w.r.t. the spin axis, the total angular momentum remains the same.

 

[vanhees71]

I'm not sure, what you are discussing about. From #14 I assume you discuss rigid bodies. So let's check this with math (only math can help to sort out all this mess).

To describe the motion of the rigid body we use as a model a discrete set of point particles bound together in such a way that we can assume that all applied external forces (usually gravity of the Earth) don't deform the body. Then the mathematical description of the motion is given in an arbitrary inertial reference frame.

The body's degrees of freedom are determined by the components of the position vector $\vec{R}$ of the origin $O'$ of a body-fixed reference frame (which is of course non-inertial in general) and the orientation of body-fixed right-handed Cartesian basis system $\vec{e}_j'$ relative to the right-handed Cartesian coordinate system. Then any point of the body has a position vector wrt. the inertial system given by
$$\vec{x}_i=\vec{R}+\vec{r}_i=R_j \vec{e}_j + r_{ij}' \vec{e}_j'.$$
The $r_{ij}'$ are the components of the position vectors in the body-fixed frame wrt. the body-fixed basis and thus time-independent.

The bases are related by a time-dependent rotation matrix $D_{kj}$ such that
$$\vec{e}_j'=D_{kj} \vec{e}_k.$$
and thus (Becase $D_{kj} D_{kl}=\delta_{jk}$)
$$r_{ik}=D_{kj} r_{ij}' \Leftrightarrow \; r_{ij}'=D_{kj} r_{ik}$$
So you have
$$\vec{x}_i = R_j \vec{e}_j + r_{ij}' D_{kj} \vec{e}_k.$$
To evaluate the total angular momentum you need the velocities
$$\dot{\vec{x}}_i=\dot{R}_j \vec{e}_j + r_{ij}' \dot{D}_{kj} \vec{e}_k=\dot{R}_j \vec{e}_j + r_{il} D_{lj} \dot{D}_{kj} \vec{e}_k=\dot{\vec{R}}+\vec{\omega} \times \vec{r}_i.$$
This we can write as
$$\dot{\vec{x}}=\vec{e}_k (\dot{R}_k + \epsilon_{kml} \omega_m r_{il}).$$
Here
$$\Omega_{kl}=\dot{D}_{kj} D_{lj}=-\epsilon_{klm} \omega_m=\epsilon_{kml} \omega_m.$$
We have used that
$$\dot{\hat{D}} \hat{D}^{\text{T}}=-(\dot{\hat{D}} \hat{D}^{\text{T}})^{\text{T}},$$
which follows from the orthogonality of $\hat{D}$, i.e., $\hat{D} \hat{D}^{\text{T}}=\hat{1}=\text{const}$ by taking the time derivative.

Finally we have for the total angular momentum
$$\vec{J}=\sum_{i} m_i \vec{x}_i \times \dot{\vec{x}}_i = \sum_i m_i (\vec{R}+\vec{r}_i) \times (\dot{\vec{R}} + \vec{\omega} \times \vec{r}_i).$$
This gives
$$\vec{J}=M \vec{R} \times \dot{\vec{R}} + M \vec{s} \times \dot{\vec{R}} + M \vec{R} \times (\vec{\omega} \times \vec{s}) + \sum_i m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i)$$
with
$$M=\sum_i m_i, \quad \vec{s}=\frac{1}{M} \sum_{i} m_i \vec{r}_i.$$
Of course $\vec{s}$ is the center of mass position vector relative to the origin of the body-fixed reference frame.

So the angular momentum consists of the first three parts dependent on the choice of the body-fixed origin:
$$\vec{L}=M \vec{R} \times \dot{\vec{R}}$$
is the orbital angular momentum of the entire body wrt. the space-fixed origin of the inertial reference frame,
$$\vec{J}_{LS}=M \vec{s} \times \dot{\vec{R}} + M \vec{R} \times (\vec{\omega} \times \vec{s})$$
is a mixed "spin-orbit" contribution, and the "spin",
$$\vec{S}=\sum_i m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i).$$
[EDIT: corrected]
One should note that the body-fixed basis doesn't depend on the choice of the body-fixed reference point, and thus also the rotation matrix $\hat{D}(t)$ and the angular velocity $\vec{\omega}$ are independent of this choice. While $\vec{\omega}$ is independent of the choice of the body-fixed reference point, the three parts of the angular momentum depend on it.

It's also clear that the entire calculation becomes very much simplified, when the center of mass of the body is chosen as the reference point, i.e., to make
$$\vec{R}=\frac{1}{M} \sum_{i} m_i \vec{x}_i,$$
because then $\vec{s}=0$. With this choice of the body-fixed reference point you simply have
$$\vec{J}=\vec{L}+\vec{S}.$$

 

[Dale]

Well that's wrong. Orbital angular momentum may cancel out spin angular momentum.

Disclaimer: I didn't actually understand anything about what you said.

This is a bit misleading in the sense that it is not necessarily true. In post #4 it is shown that if you choose another axis that is at rest w.r.t. the spin axis, the total angular momentum remains the same.

I think that we must be talking past each other.

I am describing a single rigid object and considering its angular momentum about a variety of axes (all parallel), some of which are stationary wrt the object (including the axis through the CoM) and others are moving with respect to the object at a constant speed. The angular momentum through the CoM axis is the spin angular momentum and it is a minimum. The angular momentum wrt other axes will include the spin plus the orbital angular momentum.

 

[jartsa]

I think that we must be talking past each other.

I am describing a single rigid object and considering its angular momentum about a variety of axes (all parallel), some of which are stationary wrt the object (including the axis through the CoM) and others are moving with respect to the object at a constant speed. The angular momentum through the CoM axis is the spin angular momentum and it is a minimum. The angular momentum wrt other axes will include the spin plus the orbital angular momentum.

Now I understand. And agree, except for the one unimportant detail.

The OP wanted to know about the spin angular momentum of the Earth about the axis of the sun, I think.

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same.

 

[kuruman]

I think that we must be talking past each other.
I am describing a single rigid object and considering its angular momentum about a variety of axes (all parallel), some of which are stationary wrt the object (including the axis through the CoM) and others are moving with respect to the object at a constant speed. The angular momentum through the CoM axis is the spin angular momentum and it is a minimum. The angular momentum wrt other axes will include the spin plus the orbital angular momentum.

Maybe we are talking past each other. I agree with your last statement above, "The angular momentum wrt other axes will include the spin plus the orbital angular momentum", but I also think that it needs a footnote of sorts saying something like "the orbital angular momentum to be thus included will be zero if the other axis is at rest w.r.t. the spin axis."

My rationale for including this qualification to your statement is this. We spend a lot of time trying to establish the general rule that angular momentum depends on the reference point. It behooves us then to bring forth the exception, that the spin-only angular momentum of a rigid body remains spin-only about any reference point that is at rest w.r.t. the spin axis.

 

[kuruman]

The OP wanted to know about the spin angular momentum of the Earth about the axis of the sun, I think.

Yes, my understanding is that the OP was essentially asking "I know that angular momentum is position-dependent. However, if one ignores the orbital angular momentum of the Earth, how come that the spin angular momentum is not position-dependent?" That is why I think pointing out the exception to the rule is important.

 

[Dale]

I also think that it needs a footnote of sorts saying something like "the orbital angular momentum to be thus included will be zero if the other axis is at rest w.r.t. the spin axis."

My rationale for including this qualification to your statement is this. We spend a lot of time trying to make establish the general rule that angular momentum depends on the reference point. It behooves us then to bring forth the exception, that the spin-only angular momentum of a rigid body remains spin-only about any reference point that is at rest w.r.t. the spin axis.

Yes, I agree, that is a good pedagogical approach.

 

[A.T.]

The angular momentum through the CoM axis is the spin angular momentum and it is a minimum. The angular momentum wrt other axes will include the spin plus the orbital angular momentum.

I don't think it's a minimum of the total angular momentum, because spin and orbital angular momentum can have opposite directions, cancelling each other.

But for other parallel axes at rest relative to the CoM the orbital angular momentum is zero, so the total angular momentum is the same.

This is analogous to a pure torque from a force couple (zero net force), which is also independent of of the reference point.

 

[alkaspeltzar]

But for other parallel axes at rest relative to the CoM the orbital angular momentum is zero, so the total angular momentum is the same.

What A.T. said was my original question. If there is no orbital ang. momentum, then was does the spin angular momentum stay the same relative to other parallel axes?

Looking for a general conceptual answer. from what i can gather, it has to be to conserve angular momentum. And as the math from kuruman shows, even as an object spins about its CofM, there is still rotation relative to other points as there is ang. velocity.

 

[Dale]

I don't think it's a minimum of the total angular momentum, because spin and orbital angular momentum can have opposite directions, cancelling each other.

D’oh! Yes, you are right. I was implicitly thinking of actual orbits where it is usually in the same direction, but even then there are planets with retrograde rotation.

I see also that @jartsa made this same point to me

 

[alkaspeltzar]

Yes, if you ignore the change then it doesn’t change.

The angular momentum does change. The total is for convenience split into a changing (orbital) and a non-changing (spin) part. If you ignore the change then it doesn’t change, but I don’t think that is particularly important.

Can you explain why spin angular momentum doesn change with choice of axis?

Seem people are talking but missing the point of this topic. Thks

 

[hutchphd]

D’oh! Yes, you are right. I was implicitly thinking of actual orbits where it is usually in the same direction, but even then there are planets with retrograde rotation.

OK now I am confused. I tacitly assumed you were talking about rigid bodies where all of what you said was absolutely true. Absent that constraint, I don't really know what you were intending to say..but I may well be missing something

 

[kuruman]

OK now I am confused. I tacitly assumed you were talking about rigid bodies where all of what you said was absolutely true. Absent that constraint, I don't really know what you were intending to say..but I may well be missing something

I think @Dale in post #25 retreated from the position that the total angular momentum is at a minimum when it is spin-only because of the condition $$|\vec L_{\text{orb}}-\vec L_{\text{spin}}|\leq J\leq | \vec L_{\text{orb}}+\vec L_{\text{spin}} |.$$ He is still talking about rigid bodies.

 

[kuruman]

Can you explain why spin angular momentum doesn change with choice of axis?

Spin angular momentum doesn't change when the center of mass is not moving relative to the new axis. That was explained mathematically in post #4. If the math daunts you, here is an explanation without math.

The orbital angular momentum is $\vec L_{\text{orb}}=\vec r \times \vec p$ where $\vec p$ is the momentum of the CM relative to the new axis. If $\vec p$ is zero because the relative motion is zero, then the orbital part is zero and spin is all you have left.

 

[hutchphd]

Still muddled sorry. Are we saying anything more than $\vec J=\vec L+\vec S$ ?

 

[alkaspeltzar]

Spin angular momentum doesn't change when the center of mass is not moving relative to the new axis. That was explained mathematically in post #4. If the math daunts you, here is an explanation without math.

The orbital angular momentum is $\vec L_{\text{orb}}=\vec r \times \vec p$ where $\vec p$ is the momentum of the CM relative to the new axis. If $\vec p$ is zero because the relative motion is zero, then the orbital part is zero and spin is all you have left.

Thanks

 

[kuruman]

Still muddled sorry. Are we saying anything more than $\vec J=\vec L+\vec S$ ?

It's about what we are doing with that equaton. The initial question was

Summary:: Why is angular momentum of object about its axis equal to that of a non-moving parallel axis?

So the task is to convince the OP that when $\vec L=0$ (as in the case of the object's CM not moving with respect to an arbitrary parallel axis) it follows from the equation that $\vec J = \vec S$. This needs to be done in a way OP will understand and accept.

 

[jartsa]

Summary:: Why is angular momentum of object about its axis equal to that of a non-moving parallel axis?
See link or examples below

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same.

Let's say there is a straight 20 m wide street, on the left side of which a 100 kg guy is walking at at velocity 1m/s northwards, and on the right side of which another 100 kg guy is walking at at velocity 1m/s southwards.

The angular momentum of the pair of guys about the center line of the street is:

angular momentum of guy1 about the center line + angular momentum of guy2 about the center line =
$(100kg * 1m/s * -10m) + (100kg * -1m/s * 10m) = -2000 kgm^2/s$

That is $-2000 kgm^2/s$ of spin angular momentum.
Now let's calculate angular momentum about a different axis:

The angular momentum of the pair of guys about the left side line of the street is:

angular momentum of guy1 about the left side line + angular momentum of guy2 about the left side line =
$(100kg * 1m/s * 0m) + ( 100kg * -1m/s * -20m) = -2000 kgm^2/s$

That is $-2000 kgm^2/s$ of orbital angular momentum.Well I'm surprised now, I never thought that spin angular momentum can change to orbital angular momentum when we change the axis about which we calculate the angular momentum.

But the amount of angular momentum stayed the same when we changed the axis.
Correction:Actually the orbital angular momentum of a system is the orbital angular momentum of the center of mass of the system about some axis. In my scenario the velocity of the center of mass of the system is zero. So there is no orbital angular momentum about any axis.

Therefore my system only has spin angular momentum, about the center of mass axis. Which by definition is the only axis about which spin angular momentum can be defined.https://en.wikipedia.org/wiki/Angular_momentum

 

[vanhees71]

As you can see from the general calculation in my posting #17, for a general body-fixed reference point there are three terms for the total angular momentum,
$$\vec{J}=\vec{L}+\vec{J}_{LS}+\vec{S},$$
where $\vec{L}$ is the orbital angular momentum describing the orbiting of the body-fixed reference point around the space-fixed origin, $\vec{J}_{LS}$ is the angular momentum from a coupled spin-orbit motion (which I have no good intuitive picture I must admit) and the spin $\vec{S}$. The latter is completely independent of the choice of the body-fixed reference point, while the two other contributions are dependent on this choice. [EDIT: The know crossed-out sentence is of course wrong, what's independent of the choice of the body-fixed reference point is of course $\vec{\omega}$ but not $\vec{S}=\hat{\Theta} \vec{\omega$, because $\hat{\Theta}$ of course changes when you change the body-fixed reference point according to "Steiner's Rule" (also known as "parallel-axis law").]

If you have a body in free fall in the homogeneous gravitational field of the Earth it's of course advantageous to choose the center of mass of the body as the body-fixed reference point and then $\vec{J}_{\text{LS}}=0$, and the equations of motion for the center of mass is simply a free fall and the rotation of the body around the center of mass is just the free spinning top equations (Euler equations).

 

[A.T.]

Can you explain why spin angular momentum doesn change with choice of axis?

The explanation is in the vector math that was already posted.

To understand it intuitively make sure you understand the geometrical interpretation of the cross product as an area spanned by two vectors:


Then consider the simple case: Two connected point masses spinning with a fixed distance around their common CoM. Compute their total AM around different reference points, as the sum of their orbital AMs around those points. Draw yourself some diagrams on how the areas representing the AM of the individual points change.

When moving the reference points around you will see, than any change of the orbital AM for one point mass is exactly canceled by the change of the orbital AM for the other point mass. So the total AM remains the same. The same applies for the total torque by a force couple (zero net force).