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Conservation of angular momentum derivation

  1. Oct 1, 2012 #1
    Confusion over derivation of angular momentum equation

    Hello, I'm a little confused over the relation between torque and angular momentum.



    When [tex]L=r×mv[/tex]

    [tex]\frac{dL}{dt}=r×m\frac{dv}{dt}+mv×\frac{dr}{dt}[/tex]


    According to Wikipedia,

    [tex]v=\frac{dr}{dt}[/tex]
    [tex]mv×\frac{dr}{dt}=mv×v=0[/tex]

    So [tex]\frac{dL}{dt}=r×m\frac{dv}{dt}=r×F=τ[/tex]


    But isn't [tex]v=r\frac{dθ}{dt}≠\frac{dr}{dt}[/tex]


    This is the part which bugs me. I hope someone can clarify it.
     
    Last edited: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2
    actually [tex]v_{\bot}=r\frac{dθ}{dt}≠v[/tex]

    but i still can't see how this leads to [tex]v=\frac{dr}{dt}[/tex]
     
  4. Oct 1, 2012 #3

    vanhees71

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    I think there is, first of all, some confusion in your notation. You should clearly distinguish vector and scalar quantities and define you quantities.

    In coordinate-free vector notation for the angular momentum of a point particle
    [tex]\vec{L}=m \vec{r} \times \dot{\vec{r}}.[/tex]
    Here, [itex]\vec{r}[/itex] is the position vector with respect to an (arbitrary) fixed point. Taking the time derivative you get from Newton's Equation of motion
    [tex]\dot{\vec{L}}=m \dot{\vec{r}} \times \dot{\vec{r}}+m \vec{r} \times \ddot{\vec{r}}=\vec{r} \times \vec{F}=\vec{\tau}.[/tex]
    Here, [itex]\vec{\tau}[/itex] is the torque of the total force on the particle measured with respect to the arbitrary fixed origin of your reference frame.

    I do not know how you define [itex]\theta[/itex]. So I can't say anything about it.
     
  5. Oct 1, 2012 #4
    sorry, that's the notation used in Wikipedia.

    afraid i don't understand your notation with all those dots.

    θ is used in the derivation of angular velocity.
     
  6. Oct 1, 2012 #5

    vanhees71

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    The dots are time derivative, i.e., for any quantity [itex]f[/itex] I use the usual notation
    [tex]\dot{f}:=\frac{\mathrm{d} f}{\mathrm{d} t}.[/tex]

    The section on the definition of the quantities in Wikipedia is quite confusing. Forget it. Use the definitions in terms of vectors. Then there shouldn't be any problems. The derivation, I've given is identical to the one in Wikipedia in the Section titled "Proof of the equivalence of definitions".
     
  7. Oct 1, 2012 #6
    L=r×p, now to avoid cross only that component of p should be chosen which is perpendicular to r so
    L=mr2 dθ/dt
    now dL/dt can be obtained just by differentiation of above.
    now torque=r×F ,again F perpendicular should be chosen .now it is possible to show that acceleration along θ direction is 1/r d/dt( r2 dθ/dt) which you can prove by yourself or see somewhere else( transverse acceleration)
    so torque=m d/dt( r2 dθ/dt) which is of course equal to dL/dt.( this is all two dimension)
    to avoid confusion one should use cartesian coordinates rather polar in which it is much simple.
     
  8. Oct 1, 2012 #7

    vanhees71

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    I also don't understand, which one should use this angle. If you want to describe physics in non-cartesian coordinates, better use Hamilton's principle of least action, where such issues become much less complicated.

    Of course even there you need to calculate time derivatives of the position vector. Let's look at the case of two-dimensional motion in cylinder coordinates. First of all one must realize that also the basis vectors become position dependent. So let's first derive this dependence by using a cartesian basis first. The position vector reads
    [tex]\vec{r}=r (\vec{e}_x \cos \varphi + \vec{e}_y \sin \varphi).[/tex]
    The basis of the cylinder coordinates [itex](r,\varphi)[/itex] is given by the derivatives with respect to these coordinates (and then normalized). First the non-normalized basis vectors are
    [tex]\vec{b}_r=\partial_r \vec{r}=\vec{e}_x \cos \varphi + \vec{e}_y \sin \varphi, \quad \vec{b}_{\varphi}=r (-\vec{e}_x \sin \varphi + \vec{e}_y \cos \varphi).[/tex]
    One finds that both vectors are orthogonal to each other. Thus usually one also normalizes the basis vectors, using
    [tex]\vec{e}_r=\vec{b}_r=\vec{e}_x \cos \varphi + \vec{e}_y \sin \varphi \quad \vec{e}_{\varphi}=(-\vec{e}_x \sin \varphi + \vec{e}_y \cos \varphi).[/tex]
    As you see, the new basis vectors depend on the position, and you have
    [tex]\partial_r \vec{e}_r=\partial_r \vec{e}_{\varphi}=0, \quad \partial_{\varphi} \vec{e}_r=\vec{e}_{\varphi}, \quad \partial_{\varphi} \vec{e}_{\varphi}=-\vec{e}_r.[/tex]
    Now you can calculate the time derivative of the position vector directly in cylinder coordinates. You have by definition [itex]\vec{r}=r \vec{e}_r[/itex] and thus the velocity is given by
    [tex]\dot{\vec{r}}=\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}=\dot{r} \vec{e}_r + r \dot{\vec{e}}_r=\dot{r} \vec{e}_r + r \dot{\varphi} \vec{e}_{\varphi}.[/tex]
    In the last step, I've used the chain rule for deriving the basis vector [itex]\vec{e}_r[/itex], using the above given partial derivatives. Of course you can also use the cartesian coordinates to come to the same result.

    The second derivative with respect to time, i.e., the acceleration gives
    [tex]\vec{a}=\dot{\vec{v}} = \ddot{r} \vec{e}_r + \dot{r} \dot{\vec{e}}_r + \frac{\mathrm{d} (r \dot{\varphi})}{\mathrm{d} t} \vec{e}_\varphi + r \dot{\varphi} \dot{\vec{e}}_{\varphi}.[/tex]
    Again using the chain rule you get
    [tex]\vec{a}=(\ddot{r} - r \dot{\varphi}^2) \vec{e}_r + (2 \dot{r} \dot{\varphi}+r \ddot{\varphi}) \vec{e}_{\varphi}.[/tex]
     
    Last edited: Oct 1, 2012
  9. Oct 1, 2012 #8
    Re: Confusion over derivation of angular momentum equation

    [tex]v=\frac{dr}{dt}[/tex] This is the definition of the vector on the left via the quantity on the right. Velocity is the rate of change of position.

    [tex]v=r\frac{dθ}{dt}≠\frac{dr}{dt}[/tex] This makes no sense, because velocity is not equal to the quantity in the middle. The quantity in the middle, via an abuse of notation, might be meaningful only in the particular case of circular motion.
     
  10. Oct 1, 2012 #9
    Re: Confusion over derivation of angular momentum equation


    But for an object undergoing uniform circular motion, its angular momentum is constant, its radius is also constant, so [tex]\frac{dr}{dt}=0[/tex]

    Its tangential velocity would be [tex]v=r\frac{dθ}{dt}[/tex]

    Also its angular momentum is still [tex]m(v×r)=mvr[/tex]since v is the tangential velocity.


    I don't understand how the velocity can be construed as the rate of change of radius.
     
  11. Oct 2, 2012 #10
    You are confusing vectors and scalars, speed with velocity and and radius with position. You should start from defining what all those letters you use are.
     
  12. Oct 2, 2012 #11
    if you want to treat the components of velocity then use what I have written above otherwise you should better stick to the definition of velocity for it.
     
  13. Oct 2, 2012 #12

    vanhees71

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    Re: Confusion over derivation of angular momentum equation

    Again, in this way you never will get a clear understanding what's going on. You have to clearly distinguish between vector and scalar quantities. That's what the little arrows are good for. In Wikipedia they print vectors bold face. I prefer the arrow notation, because it's better visible.

    Look at my last posting. There everything is evaluated in polar coordinates for the motion in a plane. The special case of uniform circular motion around a fixed axis (here the [itex]z[/itex]) is given by [itex]\dot{r}=0[/itex] (note, here it's the magnitude of the position vector! That's why there's no arrow). For the velocity from this you get, using the general formulas, derived in my posting
    [tex]\vec{v}=r \dot{\varphi} \vec{e}_{\varphi}.[/tex]
    BTW: You should draw all these vectors, appearing in my derivation. Then a lot becomes much more clear and you get a better intuition of what's going on!
     
  14. Oct 4, 2012 #13
    unfortunately, your explanation has made it very unclear and counter-intuitive for me.

    not to criticize you, its probably because im still a noob when it comes to angular momentum.

    maybe next time.
     
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