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Motion of a system

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Particles of mass m, 2m, and 3m are arranged as shown in (Figure 1) , far from any other objects. These three particles interact only gravitationally, so that each particle experiences a vector sum of forces due to the other two.
    Mazur1e.ch13.p08.jpg
    Is the analysis of the motion of this system straightforward?
    Is the analysis of the motion of this system straightforward?
    a. No. If released from rest, the system will rotate clockwise around the common center of mass. The accelerations are not constant, so the motion is difficult to describe completely.
    b. No. If released from rest, each particle moves in a straight line toward the common center of mass. The accelerations are not constant, so the motion is difficult to describe completely.
    c. Yes. If released from rest, the system will rotate clockwise around the common center of mass. The accelerations are constant, so the motion is easy to describe completely.
    d. Yes. If released from rest, each particle moves in a straight line toward the common center of mass. The accelerations are constant, so the motion is easy to describe completely.


    3. The attempt at a solution
    is a the right answer?
     
  2. jcsd
  3. Oct 25, 2015 #2

    gneill

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    Staff: Mentor

    We won't answer such a question without seeing your work/reasoning.
     
  4. Oct 25, 2015 #3
    I calculated the forces between the three particles. m&2m: F=2Gm2/5d2 m&3m: F=3Gm2/d2 2m&3m: F=6Gm2/4d2 then compare the magnitude of the three forces to get the answer. They are not equal and they can't cancel out so they must be accelerating.
     
  5. Oct 25, 2015 #4

    gneill

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    True, but all of the answers provided include acceleration, so acceleration alone is not a distinguishing factor. How did you settle on choice a?
     
  6. Oct 25, 2015 #5
    the forces are in different directions, so they can't be accelerating on a straight line or can they? and then I am struggled in picking a or c
     
  7. Oct 25, 2015 #6

    gneill

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    Perhaps you're approaching the problem in a more complex way than necessary, trying to analyze the forces and motions. What do the various conservation laws suggest to you about how the system must behave?
     
  8. Oct 25, 2015 #7
    There is no external force, so there is energy conservation, momentum conservation? should the answer be d then?
     
  9. Oct 25, 2015 #8

    gneill

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    Again, you'll have to make your argument for all the features of the chosen answer.

    It's a straightforward isolated system so you can expect all the conservation laws to apply. Try them all against each of the proposed solutions. Hint: don't forget angular momentum! Hint: if you pare them down to a couple of choices that the conservation laws don't seem to be able to distinguish between, then take a look at the forces (accelerations) as the system evolves.
     
  10. Oct 25, 2015 #9
    Since the forces on each particle point inward to the triangle. But the change in acceleration is inversely proportional to the square of the distance, so the acceleration is not constant?
     
  11. Oct 25, 2015 #10

    gneill

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    Is that a question or a statement? Seems like a reasonable argument rendered weak by adding a question mark :wink:

    Don't be afraid to draw conclusions from a sound use of the laws of physics :smile:
     
  12. Oct 25, 2015 #11
    that is a statement:smile: seems like answer b is most reasonable now?
     
  13. Oct 25, 2015 #12

    gneill

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    It could be! Why don't you list the reasons in support of each of the characteristics of answer b?
     
  14. Oct 25, 2015 #13
    The forces point inward to the triangle, so the particles move to the center of mass. As the distance between them decreases, the acceleration increases. Only b seems to be suitable.
     
  15. Oct 25, 2015 #14

    gneill

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    How can you be sure that the accelerations must be directly toward the center of mass? How are answers with rotating systems excluded?

    What I'm looking for are arguments like, "By the conservation of momentum, ..." and "By the conservation of angular momentum,...". If you can fill in those blanks then you will have a rock solid answer.
     
  16. Oct 25, 2015 #15
    by the conservation of angular momentum, there is no external torque, so the angular velocity is constant, and the angular acceleration is thus constant. By the conservation of momentum, the center of mass must be at rest the whole time. But for each of the particle, the momentum is not conserved because of the forces exerted by the other two particle, its velocity must increase because of this force. is this correct? I have no clue...
     
  17. Oct 25, 2015 #16

    gneill

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    It's a good attempt, and covers the major ideas. I might have used the conservation of angular momentum to first exclude answers that suggest rotation (a & c), then argue that while both momentum laws are in favor of b & d (stationary center of gravity and no rotations about it so straight trajectories are suggested) that ultimately the law of gravitation implies that the accelerations will not be constant (must increase as particle separations decrease), making the final choice between b & d on that basis.
     
  18. Oct 25, 2015 #17
    The acceleration will not be constant because the distance is changing.
     
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