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Show that dA/dt is constant in time

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle is traveling with a constant velocity v along a line that is a distance b from the origin O. Let dA be the are swept out by the position vector from O to the particle in time dt. Show that dA/dt is constant in time and equal to (1/2)*L/m, where L is the angular momentum of the particle about the origin.

    2. Relevant equations

    L= rmv

    3. The attempt at a solution

    Any hint on how to do this problem ???
  2. jcsd
  3. Feb 12, 2009 #2


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    Homework Helper

    You'll need a nice diagram showing O, the distance b straight down from O to B. Off to the side a bit you need a bit of the path of the particle heading straight for B. Draw lines from O to each end of this bit of path, to form the triangle the question talks of. Label the area dA.

    You must get an expression for the area of the triangle. You'll need the base and height, one of which is easy. For the other, think about how far the particle goes in time dt at its uniform speed. Just a bit of playing to get that dA/dt = L/(2m).
  4. Feb 12, 2009 #3


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    Homework Helper

    Trying to wrap my mind around this. Why one-third of the area of a circle???
  5. Feb 12, 2009 #4

    Doc Al

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    Staff: Mentor

    It's just nonsense--ignore it.
  6. Feb 13, 2009 #5
    What you're working on seems to be Kepler's second law. It's used in the derivation of the Volume of a sphere, since this has to be determined to yield mass.

    It's not the only way to approach, as there's no absolute. It's just the way I did it.

    You might find it easier this way, though, since the derivations will come in handy when you start playing around with gravity and the constant yield vectors.
  7. Feb 13, 2009 #6
    The BOMB!!!
  8. Feb 13, 2009 #7
    Ok. Here is what I have been trying to do.

    so L=mrv. Thus, r= L/(m*v)

    The area of triangle is (1/2)*base*height. Height will be "b" so the base will just be r then ?

    the I substitute as: (1/2)*b*(L/(m*v)) and get b*L/(2*m*v).

    Am I going the right direction ?

    Any further suggestion ??
  9. Feb 13, 2009 #8

    Doc Al

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    Staff: Mentor

    In general, L = r X mv = rmv sinθ. But r sinθ = b in this case.

    Imagine the particle sweeping out a right triangle that has height b. The position vector forms the hypotenuse, but what about the other side? Express that other side as a function of time. Then find the area of the triangle as a function of time.

    (Draw yourself a diagram.)
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