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Diagram of the system

Description of the system

Assume two point masses one at the point $C$ and the other on the circumference of the circle with radius $R$. They are attracting one another gravitationally and no external forces are acting on them. The point mass at C has very large mass such that it is at the COM of the system consisting of both the particles and the COM is at rest. The point mass is revolving about the COM with an angular velocity $\omega$. We fix our coordinate axis at one point on the circumference and find the angular momentum of both the particles by the following formula:

$$ \boldsymbol {\ell} = \mathbf r \times \mathbf p$$

Question

Clearly the magnitude of the angular momentum $\ell _{\alpha}$ of the particle at the circumference is given as follows: $$\ell _{\alpha} = Rp\sin (\omega t) \tag 1 $$ Whereas that of the one at centre is $0$.

This means the total angular momentum of the system is: $$L = Rp\sin (\omega t)$$

This is a time dependent equation meaning that angular momentum is variable which should not be the case as this system is isolated one.

I think that since law of conservation of angular momentum cannot be violated therefore there is something wrong with the method / conclusion.

So

  • What am I doing wrong here that I am getting this conclusion?

  • Is it possible to show mathematically that angular momentum is conserved?


Please don't skip this section and then later tell me that angular momentum here is $\ell = rp$

Derivation of the Eq. (1)

enter image description here

Magnitude of angular momentum at point $O'$ if given by $$\ell _{\alpha} = r(t) p \sin \phi$$

Here $$\theta (t) = \omega t $$

Clearly ( via the sum of Internal angel of triangle)

$$ \alpha = \frac {\pi}{2} - \frac {\theta (t)}{2}$$

Therefore $$ \phi = \frac {\pi}{2} + \alpha = \pi - \frac {\theta (t)}{2}$$

Therefore

$$\boxed {\begin {align} \ell _{\alpha} & = r(t) p \sin \left (\pi - \frac {\theta (t)}{2} \right ) \\ & = r(t) p \sin \left ( \frac {\theta (t)}{2}\right)\end {align}}$$

Now (using law of sines)

$$\frac {R}{\sin \alpha} = \frac {r(t)}{\sin \theta (t)}$$

Therefore

$$r(t) = 2R \cos \left ( \frac {\theta (t) }{2} \right) $$

Now substituting this into the equation for $\ell$ we get

$$\Rightarrow \ell _{\alpha} = Rp \left (2 \cos \left ( \frac {\theta (t) }{2} \right) \sin \left ( \frac {\theta (t)}{2}\right) \right) $$

$$ \ell _{\alpha} = Rp \sin \theta (t) $$

Substituting $\theta (t) = \omega t $

$$\boxed { \ell _{\alpha} = Rp \sin {\omega t}}$$

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    $\begingroup$ Yes the angular momentum transformation law is $$ \boldsymbol{\ell}_0 = \boldsymbol{\ell}_A + \boldsymbol{r}_A \times \boldsymbol{p} $$ where 0 is the origin and A is the point of interest (like the center of mass). $\endgroup$ – John Alexiou Jan 29 at 13:35
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    $\begingroup$ Why $Rp \sin\omega t$ instead of $r(t)p\sin\gamma(t)$ where $\gamma$ is the angle between p and r? $\endgroup$ – Wolphram jonny Jan 29 at 14:06
  • $\begingroup$ Have you considered the torque applied by the (non) central force? $\endgroup$ – JEB Jan 29 at 14:21
  • $\begingroup$ @wolphram I have added the derivation. $\endgroup$ – Kenzo Tenma Jan 29 at 15:18
  • $\begingroup$ Thanks, you are right! and I agree with the other answers saying that you cannot neglect the mass at the center $\endgroup$ – Wolphram jonny Jan 29 at 15:29
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I think that this is similar to something we learn in gravitation quite commonly called a double star system.

In that 2 bodies of some mass rotate about their stationary COM. This is just the same case except one of the masses is extremely large.

It is known that $L_{p} = L_{com,p} + L_{system,com}$. Using this we can first prove that the angular momentum is conserved about COM, since the velocities of both the masses are constant. Also, we can say that $L_{com,p} = 0$ since the velocity of the centre of mass is zero in this case. Thus $L_{p} = L_{system,com}=$ contant.

Thus the angular momentum about P should be conserved.

I think the mistake made by you was that you did not consider the mass at the centre. Even though it has a small velocity, since it has a large mass it can make a difference.

One more thing I would like to add is that, since the force of gravity acts in a pair, the torque is $0$ about any axis. This further confirms conservation of angular momentum.

Hope this helps!

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What am I doing wrong here that I am getting this conclusion?

The problem is that even if the inner mass is quite large then too it would have angular velocity and hence angular momentum.

Is it possible to show mathematically that angular momentum is conserved?

Yes it is. The following picture says most of it. (A description is coming ahead):

Diagram-1

Description

The outer mass is at a distance $R_{out}$ from the com and has a linear momentum as $\mathbf p_{out}$. The inner mass has is a distance $R_{in}$ from the center and has linear momentum $\mathbf p_{in}$. Two things to note here:

  1. The momentum of each particle isn't going to change because

    • the distance between the two objects is always the same (i.e., $R+R'$).

    • the force (here gravitational) is centripetal.

  2. the centre of mass is always inbetween these two on the line joining these. Therefore the angular velocity, $\omega$, of both the particles is the same.

Now in the centre of mass reference frame:

$$\mathbf L_{total} = \mathbf R_{in} \times \mathbf p_{in} +\mathbf R_{out} \ \mathbf p_{out}$$

$$\mathbf L_{total} = (p_{in}R_{in}+ p_{out}R_{out}) \hat k$$

Here you can see that angular momentum of the system is constant with time so it is conserved. Note that $\hat k$ is vector pointing out of your screen perpendicular to the plane. Also using vectors would make things easier.

Also since position of centre of mass in this reference frame is at origin:

$$\Rightarrow R_{in}m_{in} + R_{out}m_{out}= 0$$

[This equation is going to be quite useful in the last part of derivation]

Now let's shift the axis to a point on the orbit of outer particle. The labels diagram is as follows:

Labeled Diagram

Inner Particle

The position vector $\mathbf r_{in}(t)$ of the inner particle with shifted axis is:

$$\begin {align} \mathbf r _{in}(t) & = \mathbf {OB} - \mathbf {OA} \\ & = R _{in} \hat r - R_{out} \hat i \end {align}$$

And $\mathbf p = p_{in} \hat {\theta}$

$$\begin {align} \boldsymbol {\ell} _{in} & = \mathbf r _{in}(t) \times \mathbf p \\ & = (R_{in} \hat r - R_{out} \hat i) \times p_{in} \hat {\theta} \\ & = R_{in}p_{in} \hat k- R_{out}p_{in} (\hat i \times \hat {\theta}) \end {align}$$

$$\mathbf {\ell}_{in} = R_{in}p_{in} \hat k- R_{out}p_{in} (\hat i \times \hat {\theta}) \tag 1$$

Now carrying this process similarly for outer particle we get:

$$\boldsymbol {\ell}_{out} = R_{out}p_{out} \hat k- R_{out}p_{out} (\hat i \times \hat {\theta} ) \tag 2$$

Total Angular momentum $L_{total}'$ is given by:

$$\mathbf L_{total}' = \boldsymbol {\ell}_{out} + \boldsymbol {\ell}_{in}$$

$$\mathbf L_{total}' = (R_{out}p_{out} + R_{in}p_{in}) \hat k - R_out (p_{out} +p_{in} )(\hat i \times \hat {\theta})$$

Now

$$p_{out} +p_{in} = \omega R_{in}m_{in} + \omega R_{out}m_{out}$$ And using equation $(1)$ we get:

$$p_{out} +p_{in}=0$$

Therefore

$$\mathbf L_{total}' = (R_{out}p_{out} + R_{in}p_{in}) \hat k$$

Clearly this equation is constant therefore this implies that the angular momentum is conserved.

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  • $\begingroup$ Have you taken the angular velocities of both masses to be equal? $\endgroup$ – Manvendra Somvanshi Jan 30 at 11:08
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    $\begingroup$ @Manvendra I didn't take the value of both to be equal (it wasn't my intention) rather I got that as a result. As you can see that the centre of mass of both the particles should remain inbetween therefore you can see by taking two separate instances that angular velocity is indeed the same. $\endgroup$ – Kenzo Tenma Jan 30 at 11:25
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This problem is seriously over-contrived. Gravity is not necessary, the COM stuff is not required to get at the main concept.

The main concept is that angular momentum is not a vector, is a pseudo-vector. True vectors are independent of choice of coordinate origin, pseudo vectors: not so much.

Consider a bead on a hoop moving around at constant angular velocity: it at has an oscillating angular momentum about any fixed point on the hoop. There is a "central" (for the hoop) force at all times keeping the bead on the hoop, but that force is not central in the wonky coordinate system about the fixed point on the hoop: it applies a torque (also a pseudo-vector) that satisfies:

$$ {\bf{\tau}}=\frac{d{\bf{L}}}{dt}$$

Note that angular momentum and torque are also axial-vectors, but that has to do with their positive parity and the fact that, while they rotate as vectors, they are really antisymmetric rank-2 tensors.

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    $\begingroup$ Small note, there are certain examples in mechanics of a bead on a vertical hoop. After reading the entire answer I know that you are talking about ahorizontal hoop, but maybe an explicit mention would be helpful. Also, the order of your words makes it seem like you are saying the hoop has the constant angular velocity (also what happens in some vertical hoop examples). $\endgroup$ – BioPhysicist Jan 29 at 15:16
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    $\begingroup$ Most sources I'm familiar with use the word "pseudovector" use to mean the same thing you're calling an "axial vector". I've honestly never heard word "pseudovector" to mean a quantity that is origin-dependent. Is, for example, the position vector of a particle a pseudovector? $\endgroup$ – Michael Seifert Jan 30 at 0:59
  • $\begingroup$ @MichaelSeifert I thought that for 40 years, and then youtube....(science asylum, I think?). $\endgroup$ – JEB Jan 30 at 1:47
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    $\begingroup$ Magnetic field is another counterexample: It doesn't depend on the coordinate system's origin, but it does depend on its chirality. $\endgroup$ – mr_e_man Jan 30 at 1:58
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    $\begingroup$ In the set-up you describe here, the hoop is an external agent exerting a torque on the system, the "system" here being just the bead. In the OP's example, there is no external torque - the system consists of both masses, and the only force is an internal force between the two masses. Your explanation doesn't explain why the total angular momentum wouldn't be conserved in that case (and in fact, it is conserved - as the OP discovered in their self-answer, the problem was that they were incorrectly assuming the angular momentum of the heavier particle would be negligible). $\endgroup$ – Carmeister Jan 30 at 2:49
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While reading the question’s starting part I remembered my first lecture on modern physics

See what you have stated in the start of your question is a very good approximation for calculating motion parameters of the small body but not for calculating angular momentum of the system

See the larger body will not remain at a fixed point (if it isn’t pivoted to a point in that case the angular momentum can’t be conserved as there would be a external torque about the point you are conserving angular momentum)

The larger body will infact move with however small velocity it might have but in that case its angular momentum would not be a significant quantity due to its advanced mass

You might see how will it move in picture below      motion

However in this case the bodies will get closer

About my class

this case is specially valid in movement of electron around proton and the reason why in a atom electron and proton doesn’t get closer is the system has zero net angular momentum i.e. both proton and electron move in a circular orbit about their c.o.m. and the electrostatic pull on both of them is used in keeping both of them in circular orbits and doesn’t bring them closer.

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  • $\begingroup$ @Wolphramjonny If you didn't choose your reference point to be the center of rotation then it wouldn't be conserved though $\endgroup$ – BioPhysicist Jan 29 at 15:04
  • $\begingroup$ @AaronStevens right, it is because the object rotating with the cord is not an isolated system, correct? $\endgroup$ – Wolphram jonny Jan 29 at 15:05
  • $\begingroup$ @Wolphramjonny Yes. But you can get around that by putting your reference point right at the constraint, because then the external force has no torque. It all depends on the reference point. $\endgroup$ – BioPhysicist Jan 29 at 15:13
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The easiest way that I can see to show that angular momentum is not conserved in this problem is to look at the torque in the system. If angular momentum is conserved, then the torque (the time derivative of the angular momentum) must be identically zero.

Torque is $\mathbf{\tau} = \mathbf{r}\times\mathbf{F}$, using the vectors on your diagram. It is clear, in both of your diagrams,that in the given position, the force and position vectors are not parallel and not zero. Therefore, there is torque, and angular momentum can't be conserved.

As others have pointed out, the main problem is your assumption that the more massive body is so massive that it doesn't move. Both masses move, so its necessary to keep track of the angular momentum and torque on both. The force on both is the same, but in opposite directions. The total torque is $\mathbf{\tau} = \mathbf{r}\times\mathbf{F} - \mathbf{r'}\times\mathbf{F} = (\mathbf{r}-\mathbf{r'})\times\mathbf{F} = \mathbf{0}$ because the line between the masses is parallel to the forces. So torque is identically zero, and thus angular momentum is conserved.

Note that when you include the movement of both masses, the total torque is zero, and the actual location of the coordinate axis doesn't matter. That's one of the characteristics of the conservation laws. While the value of the angular momentum may be different when measured in different coordinate systems, it doesn't change in any of them.

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I read all the answers above and decided to write my own, since most of them are missing a complete treatment.

We know that when a point mass is subject to a radially symmetric potential $V(|\mathbf{r}|)$ (as is the case for the usual gravitational interaction) angular momentum is conserved, and that's a fact. What happens if I move the origin by a vector $\mathbf{a}$? From the perspective of the new origin, the potential is not central anymore, it is clearly centered on a different point, and thus the angular momentum measured in that frame of reference must depend on time. How does it depend on time? Measuring everything in new frame of reference(all quantities in the shifted origin are denoted by a prime)

$$\mathbf{L}'=\mathbf{r}'\times\mathbf{p}=\mathbf{r-a}\times\mathbf{p}=\mathbf{L}-\mathbf{a}\times\mathbf{p}$$

In the equation above the angular momentum of the point mass in the original frame, but the angular momentum in the shifted frame is modified by a term proportional to the momentum of the particle.

So what gives? Is there any context in which conservation of angular momentum is independent of shifts in the reference frame? The answer is certainly yes, but that only happens when the total momentum of the system is conserved. Imagine there are some point masses in the system, then their TOTAL angular momentum transforms under shifts as follows:

$$\mathbf{L}'_{tot}=\mathbf{r}'\times\mathbf{p}_{tot}=\mathbf{r-a}\times\mathbf{p}_{tot}=\mathbf{L}_{tot}-\mathbf{a}\times\mathbf{p}_{tot}$$

which implies that if the total momentum of the system is conserved, then total angular momentum is conserved in any frame of reference, regardless of origin.

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For circular motion ${\bf r} \times {\bf p} = rp$ as these are perpendicular vectors and $\sin 90=1$. So angular momentum is constant hence conserved. There is no torque with respect to the centre of the circle. Of course angular momentum is not conserved with respect to any other position. The reason is that there is a torque ${\bf r}\times {\bf f} $ with respect to any other point than the centre, as r is then no longer parallel to f. This torque is opposite and equal to the torque exerted by the small mass on the big one. The _total _ AM is conserved, regardless of the choice of origin.

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