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B Something I really don't get about Angular momentum

  1. Aug 20, 2016 #1
    Hi there guys.

    I'm new here, and I'm learning physics by myself. Please, if the answer to my question is too obvious, be kind hee hee, I'm a 13 and trying to learn physics by myself.

    So, I know that we have

    ## \textbf{L}= \textbf{r} \times \textbf{p} ##

    Then, we can write

    ## \frac{d}{dt} \textbf{L} = \dot{\textbf{L}} = \frac{d}{dt} \left( \textbf{r} \times \textbf{p} \right) = \frac{d}{dt} \left( \textbf{r} \times m \textbf{v} \right) ##

    So, this is

    ## \dot{\textbf{L}} = \frac{d}{dt} \textbf{r} \times m \textbf{v} + \textbf{r} \times \frac{d}{dt} \left( m \textbf{v} \right) ##

    This yields

    ## \dot{\textbf{L}} = \textbf{v} \times m \textbf{v} + \textbf{r} \times \textbf{F} ##

    The first term on the RHS vanishes, since

    ## \textbf {v} \times \textbf{v} = (v)(v) \sin \theta = (v)(v) \sin 0 = (v)(v) 0 = 0##

    So, we have, then

    ## \dot{\textbf{L}} = \textbf{r} \times \textbf{F} = \textbf{N} ##

    Where ## \textbf{N} ## is the torque (a pseudovector by the way)

    Then, it follows that if the total torque is zero, then angular momentum is conserved.

    So far so good.

    My questions is, then, if a particle is moving about a point, the linear momentum of the particle is changing indeed, so is its position vector. Which then made me wonder certain things, then I realized that the only way for the angular momentum to be conserved is that the force must be pointing in the direction of the position vector, that way ## \textbf{r} \times \textbf{F} = 0 ##

    Now, here's my question, can I state that "If the angular momentum is conserved, it implies that the sum of all torques is zero, but, also, conservation of angular momentum do implies the presence of a force directed along the same line of the radius vector ## \textbf{F} ## ", in other words, "In order to have angular momentum conservation THERE MUST EXIST A FORCE acting on the particle"?

    Am I right?

    Thanks in advance for your help!
     
  2. jcsd
  3. Aug 20, 2016 #2
    I am also a newbie. So, feel free to correct me, if I am wrong. :)

    I think, we consider about angular momentum when an object is moving in a curved path, and for an object to move in a curved path, there must be a force acting along the radius vector (to turn the body from its linear path, i guess), ex : centripetal force. So, conserved or not conserved, there will be force along the radius vector.

    Yes, conservation means sum of all torques acting is zero, and also, no external torque is acting on the objects.

    Again this is what I understood, if I am wrong correct me. :)
    Thanks.
     
  4. Aug 20, 2016 #3
    Nope you can't say that a force must be acting on the particle since if ## \vec F = \vec 0 ## then the cross product will be ## \vec 0 ## as well, otherwise if a force is acting on the particle then it must be parallel to ## \vec r ##
     
  5. Aug 20, 2016 #4
    Thanks for your responses.

    I agree with Molar.

    And to nashed, I will respond that in my mind, in such a case where ## \textbf{F} = 0 ##, the linear momentum will keep its direction, then, it would not follow a circular path.
     
  6. Aug 20, 2016 #5
    I'm perfectly aware that in such a case there is momentum conservation, the thing is that the definition of angular momentum ## \mathbf L = \mathbf r \times \mathbf P ## doesn't really care which kind of movement the particle is doing, as long as you provide a reference point you can talk about angular momentum and as such you can't discard linear paths...
     
  7. Aug 20, 2016 #6

    PeroK

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    Angular momentum (about any point) is conserved for a particle moving with constant velocity in a straight line.
     
  8. Aug 20, 2016 #7
    Thank you very much. I just did not know angular momentum could be defined for linear motion. That just never came to my mind. Thanks.
     
  9. Aug 20, 2016 #8
    Ahh yes, I too didn't think of it. Thanks.
    Now I have a little confusion, can you please help ?

    From
    dL/dt = r x F ,
    for straight line motions, if F = 0, so we get L is conserved. Okay.
    What happens when F is constant ?

    Also from
    L
    = r x mv ,
    for straight line motions, r is changing constantly from that certain point. So, L will be changing in each second. Then how angular momentum is conserved from this view point ?
     
    Last edited: Aug 20, 2016
  10. Aug 20, 2016 #9
    The angle is also changing.
     
  11. Aug 20, 2016 #10
    In such a case, it is nothing more than a particle moving in a straight line. And now we know that angular momentum is conserved, indeed.

    In this case, as stated before, if you have that ## \textbf{L} = \textbf{r} \times m \textbf{v} = (r)(mv) \sin \theta ##, and we see from here that, letting ## v ## constant, and just varying ## r ##, then, in order to the result to be the same (i.e., conservation of angular momentum), we must vary the only remaining variable, namely, the angle ## \theta ##
     
  12. Aug 21, 2016 #11

    vanhees71

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    Well, let's calculate it. For ##\vec{F}=0## your motion is uniform along a straight line. Now you have to remember that for the definition of the angular momentum you have to choose a reference point. This reference point can be chosen as the origin of the reference frame. Then you have
    $$\vec{L}=\vec{r} \times \vec{p}=m \vec{r} \times \dot{\vec{r}}=m \vec{r} \times \vec{v}.$$
    Now you have
    $$\vec{r}(t)=\vec{r}_0+\vec{v} t, \quad \vec{v}=\text{const}$$
    and thus
    $$\vec{L} = m (\vec{r}_0+\vec{v} t) \times \vec{v}=m \vec{r}_0 \times \vec{v}=\text{const}.$$
    So indeed, angular momentum is conserved, as is also clear from the previous postings by taking the time derivative of ##\vec{L}##.

    In #1 the OP however made a much more profound discovery. From
    $$\dot{\vec{L}}=\vec{r} \times \vec{F}$$
    you get that angular momentum is also conserved, if ##\vec{F} \propto \vec{r}##, i.e., if you have what's called a central force.

    Usually the forces in Newtonian mechanics are conservative, i.e., derivable from a potential,
    $$\vec{F}=-\vec{\nabla} V.$$
    In order to have $$\vec{F} \propto \vec{r}$$ you must have $$V(\vec{r})=V(r)$$, i.e., the potential must depend on the distance from the center only, not on its direction. This means the potential doesn't change under rotations of ##\vec{r}##.

    There is a much more powerful way to describe mechanics than in its original Newtonian form, i.e., with the socalled action principle. This doesn't lead to anything new concerning the physics of motion but it provides powerful mathematical tools to analyze this motion. As it turns out all conservation laws are due to symmetries and any symmetry implies a conservation law (there are some technical details about how the symmetry is described mathematically, known as a Lie group, but that's not so important here).
     
  13. Aug 21, 2016 #12
    That sounds very interesting! I mean, once again, I'm learning physics by myself, and that comment about the potential doesn't changing under rotations made me think there must be more to it, a more profound concept deep inside. I have heard about symmetries and how important they are, but for now I have not studied more about them.

    Thanks for your help and comment everyone :)
     
  14. Aug 21, 2016 #13

    vanhees71

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    Sure, my comment was meant to be a motivation to study into these symmetry concepts. It's a lot of fun, and really at the heart of all of physics.
     
  15. Aug 21, 2016 #14

    jbriggs444

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    Assuming that the moment arm associated with F is fixed so that F imparts a fixed torque then angular momentum changes at a fixed rate.

    r is a vector. It can change without affecting r x mv. Work out the math. Any change of r in the direction of v leaves r x mv unchanged.
     
  16. Aug 21, 2016 #15
    Yeah, I understood. It's much clearer now. Thanks to all. :)
     
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