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Closed trajectory in a central field of force; find the mass

  1. Oct 30, 2015 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    A particle moves along a closed trajectory in a central field of force where the particle's potential energy is U=kr2 (k is a positive constant, r is the distance of the particle from the center O of the field). Find the mass of the particle if it's minimum distance from O equals r1 and it's velocity at the point farthest from O equals v2.

    2. Relevant equations
    ##rv=r^2\dot \theta=\text{constant}\equiv c_a##
    ##0.5mv^2+kr^2=0.5m(\dot r^2+r^2\dot \theta^2)+kr^2=\text{constant}\equiv c_b##
    ##\dot r = \dot \theta \frac{dr}{d\theta}##
    ##\vec F=-\nabla U=-2k\vec r=m(\ddot r-r\dot \theta^2)\hat r##

    3. The attempt at a solution
    My attempt ignored the force equation (the last of my "relevant equations").
    What I did was to eliminate the time dependency in the energy equation to get a differential equation between r and θ, like this:
    ##c_b=kr^2+\frac{mc_a^2}{2r^4}\big ( \frac{dr}{d\theta} \big )^2+\frac{mc_a^2}{2r^2}##
    which is a separable equation:
    ##d\theta=\frac{c_adr}{r^2\sqrt{\frac{2}{m}(c_b-kr^2)-\frac{c_a^2}{r^2}}}##
    I haven't tried, but I don't think I can solve that integral. At any rate, there must be a simpler way. What I am doing will lead to the arbitrary path of an object in this field (the constants ca and cb are defined by the initial condition). I don't think it is necessary to solve for the entire trajectory like this... but I don't know what else to do.

    [edited to include that k is a positive constant]
     
    Last edited: Oct 30, 2015
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  3. Oct 30, 2015 #2

    TSny

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    Use each of these two equations to relate ##r_1, r_2, v_1## and ##v_2##. Combine the equations and simplify.
     
  4. Oct 30, 2015 #3

    Nathanael

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    That's 3 unknowns with only two equations. We want to find the mass. r2 and v1 are unknown.
     
  5. Oct 31, 2015 #4

    andrewkirk

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    It seems to me that more information is needed. There are three unknowns: ##r_2,\ v_1## and ##m##. But there are only two equations - one from conservation of angular momentum and one from conservation of energy. Can we get another, independent, equation, or is the problem under-specified?

    Edit: I see that Nathaneal has pointed this out. For some reason it didn't appear on my screen.
     
  6. Oct 31, 2015 #5

    Nathanael

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    I was thinking the same, but I think the final constraint may come from the fact that it is a "closed trajectory," because surely not all initial conditions will lead to a closed trajectory (right?). I'm just not sure how to use this constraint. (This is why I attempted to solve for the entire trajectory r(θ) in the OP by eliminating time.)
     
  7. Oct 31, 2015 #6

    Nathanael

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    I just checked and the answer is supposed to be ##m=2k(\frac{r_1}{v_2})^2##

    This is what you get if you assume the path is circular, but I see no reason why this should be the answer in general. Perhaps the problem creator made a mistake with this one? (The problem statement is word for word.)
     
  8. Oct 31, 2015 #7

    haruspex

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    No, it does not depend on the shape of the trajectory, and you do not need to solve the differential equation.
    Just follow TSny's lead to obtain a differential equation that does not involve theta. From this you can obtain expressions for the min and max radii. You can use your energy equation to relate the maximum radius to the velocity at that point.
     
  9. Oct 31, 2015 #8

    TSny

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    Just use the two equations
    ##rv=\text{constant}##
    ##0.5mv^2+kr^2=\text{constant}##.

    The first equation tells you that ##r_1v_1 = r_2v_2##. Similarly, you can get a relation involving ##r_1, v_1, r_2, v_2##, and ##m## from the second equation. Due to the form of the potential energy, you will be able to solve for ##m## in terms of just ## r_1## and ##v_2##.
     
  10. Oct 31, 2015 #9

    Nathanael

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    ##0.5m(\frac{r_2v_2}{r_1})^2+kr_1^2=0.5mv_2^2+kr_2^2##
    ##m\big(\frac{v_2}{r_1}\big)^2(r_2^2-r_1^2)=2k(r_2^2-r_1^2)##
    ##m=2k\big (\frac{r_1}{v_2}\big )^2##

    @TSny How did you know it would cancel like this?
     
  11. Oct 31, 2015 #10

    TSny

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    I didn't! I just wanted to see what you could conclude from the two equations. Bingo, the answer fell out unexpectedly:wideeyed:.
     
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