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

[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).

 

[jbriggs444]

"the orbital angular momentum to be thus included will be zero if the other axis is at rest w.r.t. the spin axis."

If you clarify that by "spin axis" you mean "center of mass" then I will agree.

 

[kuruman]

If you clarify that by "spin axis" you mean "center of mass" then I will agree.

Yes, of course. I specified as much in post #4 and in other posts where I wrote $I_{\text{cm}}~\vec\omega$ for the spin angular momentum in equations. I confess I got lazy and abbreviated "spin angular momentum about an axis that goes through the CM" to just "spin angular momentum" in sentences.