Physical meaning of the angular momentum

I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position?

1) - Angular momentum($L = rmv $)

It is very simple: in the other question you have understood the concept of linear momentum, now you have only to join it to the concept of the lever.

Imagine that ball B is the same ball which was in the linear-momentum question $(m = 2 Kg)$ which was traveling at a velocity of three meters per second and had a momentum of six kilograms meters per second

We can think that it has a line in the direction in which it is moving and a hook hanging. This hook gets caught by a peg $F$. What will happen? $B$ will start to rotate around the fulcrum $F$ (sketch on the left). The direction of motion will be perpendicular to the radius (line), therefore the angle will be $90°$, and it's $\sin$ will be $1$.

Consider another scenario (sketch on the right; same as in a lever) the torque exerted depends also on the radius, the distance of the body from the fulcrum which is the arm of the lever. The magnitude of the torque depends on the value of $r$. A weight of $6 kg$ will exert a torque of $12$ $Nm$ at the distance of $2$ $m$, and you will have balance only if you put (on the other arm) a weight of $6 Kg$ at $2 m$ or weight of $12$ $Kg$ at $1 m$.

If you understand the concept of the lever, you can easily understand the physical explanation of the formula of the angular momentum. In the same way, if B ($m = 2$) is rotating anticlockwise at $v$ = $3 m/s$ (linear $momentum$ = $6$) at distance $2 m$ from the fulcrum it will have angular momentum (6 * 2 =) 12 Kg * m²/s). If the line hanging from B had been only $1 m$ long, the magnitude of $L$ would have been (6 * 1) = 6.

Likewise, if another body A ($m$ = $2$, $v$ = $3$, $p$ = $6$) is rotating clockwise on the other arm, there will not be equilibrium, even though mass, speed and linear momentum are the same; the same would happen if a force of $6N$ is applied at $r$ = $2m$ and another opposite force of $6N$ is applied at $r$ = $1m$. Note that B had angular momentum with reference to F even before it started to rotate around it all along its trajectory and it always was (p * r) = $12 Kg * m^2/s$.

2) - Definition of L

A body B with velocity (and linear momentum) has a potential rotational momentum L with reference to/around any point/body O which does not lie on its trajectory.

The magnitude of L can be found multiplying its linear momentum (p = m*v) by the distance of point O from the trajectory: $r$.

In the full formula: $L = m * [v * sinλ * d]$, L is obtained multiplying mass by tangential velocity $V_t = v * sinλ$ times distance $d$, but $d * sinλ$ is always equal to $r$

3) - Conservation of angular momentum

Angular momentum L is conserved if no external torque is exerted on the system and this property helps you understand the importance of radius. When body B is bound to O by a line/rod or by a non-contact force (like g) it starts rotating around it and acquires actual rotational momentum L.

If, while rotating around O, B impacts with a similar ball A ($m$ =2, $v$ = 0), B stops dead and A acquires same v/p/E, and potential L with reference to point F, if it collides with the bob of a pendulum A ($m$ = 2, $r$ = 2) it will acquire same v/p/L/E. If the line/rod of the pendulum $r_p = k$, p will be conserved, but $L_p$ will become $L \times \frac{k}{r}$.

This is a simple example in which the body is considered a point mass rotating on the circumference, if the mass is distributed along the radius, then we must apply a different formula $ L = I * \omega$, where $ω = v/r$ and $I = m *r^2$. P is not conserved, but KE and L are, in this way we can work out the outcome of the collision. You can find a simple example of conservation of L here

Ultimately, what's special about angular momentum is this:

• Look up in the sky. A certain set of physical laws pertain in that direction.
• Look to the north. A certain set of physical laws pertain in that direction.
• Look to the west. A certain set of physical laws pertain in that direction.

Those physical laws: They're the same in all directions. There's an underlying conserved quantity whenever you find a symmetry like this. A similar concept applies if you take a trip to China, Proxima Centauri, the Andromeda galaxy, or even further. Here the laws of physics are the same regardless of translation. What about time? Blink your eyes and the laws of physics don't change. Age 80 years and the laws of physics don't change. The laws of physics are timeless.

The timelessness of the laws of physics means that energy is a conserved quantity. The translational independence of the laws of physics means that linear momentum is a conserved quantity. Finally, the rotational independence of the laws of physics means that angular momentum is a conserved quantity. These are all consequences of Noether's theorem. There are a number of other conserved quantities that result from Noether's theorem, and this turned out to be very important for quantum mechanics. This is obviously critical to classical mechanics as well. Noether's theorem explains exactly why those conserved quantities are conserved.

Angular momentum would be a rather useless concept if angular momentum was not a conserved quantity in the absence of external torques. It is a conserved quantity thanks to the rotational symmetry of space.

Why does the angular momentum depends on the position?

Angular momentum is always defined relative to a reference point, say $\mathbf r_0$, (which is often, but not necessarily the origin).

If the system is invariant under rotation around this reference point the quantity that we call "angular momentum with respect to $\mathbf r_0$" is conserved. (Note, that if only a rotation around a certain axis leaves the system unchanged, only this component of the angular momentum is conserved).

So since the angular momentum depends on a point of reference it is not a surprise that the angular momentum explicitly depends on the position.

How does this conservation of angular momentum work?

The abstract answer deals with the Noether theorem and the Lagrangian of the system you are looking at. For simplicity, let's just look at a single point particle moving on a straight line.

Note that, even a free particle moving on a straight line has a non-zero angular momentum with respect to certain points of reference. In fact the angular momentum is only zero, if the momentum and the connection between the point of reference are parallel (i.e. the point of reference is on the path of the particle).

Noether Theorem for a free particle under rotation

Let's use this free particle to see where this conservation of $\mathbf x\times \mathbf p$ comes from. The Lagrangian is here just the kinetic energy. If we rotate the coordinates around the origin and along the fixed axis $\mathbf n$ by the angle $\varphi$. The kinetic energy (the Lagrangian) should not depend on the angle of rotation.

Let the rotated positions be given by $\mathbf x'$. The kinetic energy is $\frac 12 m \dot {\mathbf x'}^2(\varphi)$, so our condition that the kinetic energy is independent of $\varphi$ can be written as:

$$\frac {\mathrm d(m \dot {\mathbf x'}^2(\varphi))}{\mathrm d\varphi}= \mathbf p \frac{\mathrm d \dot {\mathbf x'}(\varphi)}{\mathrm d \varphi} =0 \,,$$

since there are no forces acting on the free particle ($\dot{\mathbf p}=0$), we can write this as: $$\frac{\mathrm d}{\mathrm dt}\left(\mathbf p \frac{\mathrm d {\mathbf x'}(\varphi)}{\mathrm d \varphi}\right) = 0\,.$$

How does $\mathbf x'$ change with the angle? Take a look at the chapter on infinitesimal rotations and you should find something like $$\mathbf x' = \mathbf x + \varphi (\mathbf n \times \mathbf x)$$

Computing the derivative and plugging it in the equation above leads to:

$$\frac {d}{dt}\left(\mathbf p\frac {d({\mathbf x + \varphi (\mathbf n \times \mathbf x)})}{d \mathbf \varphi}\right) = \frac {d}{dt}\mathbf p \cdot (\mathbf n \times \mathbf x) = \frac {d}{dt}\mathbf n \cdot (\mathbf x \times \mathbf p)=0$$

Which tells us, that the $\mathbf n$-component of $\mathbf x \times \mathbf p$ does no change over time (i.e. it is conserved). In this case this is true for arbitrary axes,which means that the angular momentum $\mathbf L = \mathbf x \times \mathbf p$ is conserved.