I am trying to understand how a gyroscope works, which in the broad strokes is due to conservation of angular momentum. I understand the case when the angular momentum passes through the origin of the field, but what I'm having some trouble understanding is when it doesn't. For example, enter image description here

Where O is the origin of a central field, $\textbf{a}$ is the position of the centre mass for the rotating particle, $\textbf{p}$ is the instantaneous momentum of the particle at a given point, $\textbf{r}$ is the position of the rotating particle measured from the O frame, $\textbf{r'}$ is the position of the particle measured from it's own centre of rotation, and $\textbf{M'}$ is the angular momentum of the particle measured from the centre of rotation. All points are coplanar, and $\textbf{M'}$ is perpendicular to the plane.

From this, to transform back into the frame of the origin of the field (O), $$\textbf{M}=\textbf{M'}+\textbf{a}\times\textbf{p}$$ but since $\textbf{a}$ is constant and $\textbf{p}$ is changing, wouldn't the angular momentum change as well when measured from the rest frame of O? While the angular momentum would still pass through the origin, wouldn't the fact that it is oscillating mean that it is not conserved? Where is the flaw in m reasoning?

  • $\begingroup$ The Lagrangian of a particle in central field potential is independent of the angular coordinates, thus the momentum related to those coordinates - the angular momentum - is conserved. It is essentially Newton's laws in polar coordinates. $\endgroup$ – Alexander Mar 3 at 10:25

If $\vec r$ is the location of the particle, it will not execute circular motion about the point located at $\vec a$. Rather it will execute elliptical motion with the origin as one focus of the ellipse. Also the CoM is not at $\vec a$ since you have a point particle: the CoM would be at the particle itself.

In other words, your circular orbit does not describe the motion in a central field so cannot be used to understand conservation of angular momentum. Alternatively the orbit you drew doesn’t conserve angular momentum, but that’s not surprising as it’s not a possible orbit.

  • $\begingroup$ I've edited my question a bit but I'm trying to understand how a gyroscope works, which from what I understand is from conservation of angular momentum. Since the angular momentum is conserved, the angular momentum of the gyroscope remains constant, wherever it is, and I'm trying to figure out why. $\endgroup$ – Redcrazyguy Mar 2 at 22:57
  • $\begingroup$ @Redcrazyguy then I fail to understand your diagram and the concept of an orbit as you describe in your diagram. $\endgroup$ – ZeroTheHero Mar 3 at 0:41
  • $\begingroup$ it's not an orbit. $\endgroup$ – Redcrazyguy Mar 3 at 14:01

Your reasoning about the angular momentums is corrrect. What is needed is an interpretation: note that $\frac{d\vec{L}}{dt} = \vec{\tau}$, and both $\vec{L}$ and $\vec{\tau}$ are dependent on the choice of the origin.

Point O should not be the center of a central field, Since the particle is moving as a circular motion with a constane speed $v = |\vec{p}| / m$, there is a centripedal force toward the center of rotation, say point A:

$$ \vec{F} = -\frac{m v^2}{r'} \hat{r}' =-\frac{p^2}{m r'} \hat{r}' . $$

where $\hat{r}'$ is the unit vector of $\vec{r}'$.

In frame A

In the frame A (the center of rotation), the angular momentum $$ \vec{M}' = \vec{r}' \times \vec{p} = \{ r' p \} \hat{z} = \text{a constant}. $$ Why $ \vec{M}'$ is a constant in frame A? because the torque $$ \vec{\tau}' = \vec{r}' \times \vec{F} = 0 = \frac{d\vec{M}'}{dt}. $$ The zero torque renders a constant angular momentum in Frame A.

In Frame O

In the frame O, the angular momentum $$ \tag{1} \vec{M} = \vec{r} \times \vec{p} = \{ r' p + a p \cos\theta \} \hat{z}. $$ where $\theta$ is the angle between $\vec{a}$ and $\vec{r}'$, $-\pi \lt \theta \lt \pi$. The angular momentum $\vec{M}$ is not a constant, but varies with time, $\theta = \omega t$. Let examine the torque $$ \tag{2} \vec{\tau} = \vec{r} \times \vec{F} = \vec{r}' \times \vec{F} + \vec{a} \times \vec{F} = a F (-\sin\theta )\hat{z}. $$ The angular momentum is not a constant, because the torque is not vanishes.

We can furthermore show that indeed: $\frac{d\vec{M}}{dt} = \vec{\tau}$

$$ \frac{d\vec{M}}{dt} = \frac{d}{dt}\{ r' p + a p \cos\theta \} \hat{z} \\ =-a p \sin\theta \omega \hat{z} = - a p \sin\theta \frac{v}{r'} \hat{z} = -a \frac{p^2}{m r'}\sin\theta\hat{z} = -a F \sin\theta\hat{z} = \vec{\tau} $$


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