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Angular Momentum In Polar Coordinates

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the xy-plane, with the sun at the origin, and label the planet's position by polar coordinates [itex](r, \theta)[/itex]. (a) Show that the planet's angular momentum has magnitude [itex]L = mr^2 \omega[/itex], where [itex]\omega = \dot{\theta}[/itex] is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" is [itex]\frac{dA}{dt} = \frac{1}{2} r^2 \omega[/itex], and hence [itex]\frac{dA}{dt} = \frac{L}{2m}[/itex],

    2. Relevant equations

    3. The attempt at a solution
    If we consider the sun and the planet as one system, then the gravitational force will be an internal force, which means there are no external force, and consequently [itex]\sum \vec{\tau} = \vec{0}[/itex], which means angular momentum is constant.

    [itex]\vec{L} = \vec{r} \times \vec{p}[/itex]

    [itex]L = rp \sin \theta[/itex], where the radial distance, r, and the angle, theta, both vary as a function of time.

    Substituting in the expression for linear momentum,

    [itex]L = rmv_{tan} \sin \theta[/itex], where [itex]v_{tan} = r \omega[/itex]

    [itex]L = rm(r \omega) \sin \theta[/itex]

    [itex]L=mr^2 \omega \sin \theta[/itex]. This is where I get stuck. What am I to do with the sin theta, which varies as time progresses? I would appreciate a hint, but not the answer.
  2. jcsd
  3. Oct 7, 2013 #2


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    ##mv_{tan}## is not equal to ##p## in general. If you think about a general ##\vec{p}## then how is ##p \ \sin(\theta )## related to ##v_{tan}## ? You could try drawing it out to see what makes sense.
  4. Oct 7, 2013 #3

    George Jones

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    [itex]p \ne mv_{tan}[/itex]

    Express [itex]\vec{v}[/itex] in terms of [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex].
  5. Oct 7, 2013 #4
    Why doesn't p = mvtan?
  6. Oct 7, 2013 #5

    George Jones

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    Actually, defined appropriately, [itex]p = mv_{tan}[/itex], i.e., [itex]\vec{v}[/itex] is always tangent to the path of the planet, and thus [itex]v = v_{tan}[/itex], but, in general, [itex]v \ge r \omega [/itex], because the planet is not necessarily moving in a circular orbit.

    Also, I think that you are confusing two angles that are different, the angle between [itex]\vec{r}[/itex] and [itex]\vec{p}[/itex], and the angle that is a coordinate for polar coordinates.


  7. Oct 7, 2013 #6
    Oh, so should I be using [itex]\phi[/itex] as my angular position, and [itex]\theta[/itex] as the angle between [itex]\vec{r}[/itex] and [itex]\vec{p}[/itex]?

    So, I should would considering the centripetal acceleration be at all helpful?
  8. Oct 7, 2013 #7

    George Jones

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    In your original post, you wrote

    so I think that we should stick with [itex]\theta[/itex] as the polar coordinate. I think that we do not need to consider the angle between [itex]\vec{r}[/itex] and [itex]\vec{p}[/itex].

    Use the hint that I gave, and calculate [itex]\vec{r} \times \vec{p}[/itex] as a vector.
  9. Oct 7, 2013 #8
    I believe [itex]\vec{r} = r \hat{r}[/itex]. So, would the velocity vector be the derivative of this?
  10. Oct 7, 2013 #9

    George Jones

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    Yes. Use the product rule, since both [itex]r[/itex] and [itex]\hat{r}[/itex] change with time.

    I thought you might have seen, either in your notes or text, the velocity vector expressed in polar coordinates. You are going to need a relationship between [itex]d\hat{r}/dt[/itex] and [itex]\hat{\theta}[/itex] that is simple, but that is difficult to motivate without diagrams.

    If you haven't seen any of this, then maybe thsi isn't the approach you should take to solve the problem.
  11. Oct 7, 2013 #10
    I actually just remembered that my textbook provides a derivation of Newton's 2nd Law in terms of polar coordinates. The velocity vector is [itex]\vec{v} =\dot{r} \hat{r} + r \dot{\theta} \hat{\theta}[/itex]. I have a question, suppose our planet is following an elliptical path, does this mean there is a centripetal and tangential acceleration, and that is why the tangential velocity is changing at every point on the path; or is there only a centripetal acceleration, and is the velocity changing because radial distance is changing?
  12. Oct 7, 2013 #11
    Okay, I was able to figure it out. Now I am working on part b).

    I drew an arbitrary elliptic/circular path, that the planet follows around the sun. I considered two arbitrary points on the path, [itex]P_1 = (r_1, \theta_1)[/itex] and [itex]P_2 = (r_2, \theta_2)[/itex]. I the supposed that the two positions where sufficiently close to each other so that the path followed is approximately a straight line, rather than curved. The position vector for P1 is [itex]\vec{r_1}[/itex], and for P2, [itex]\vec{r_2}[/itex]; then [itex]d\vec{r} = \vec{r_2} -\vec{r_1}[/itex] is the vector connecting these two points two points, and the magnitude of the vector is the length of the path that the planet travels during some time [itex]dt[/itex]. Let [itex]dA[/itex] be the area enclosed by the three vectors [itex]d\vec{r},~\vec{r_1}~,~and~\vec{r_2}[/itex].

    This is sort of where I am suck. I am not sure if the assumptions I have made are correct, and of how to proceed.
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