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Very basic mechanics

  1. Jul 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Two equal masses hang on either side of a pulley at the same height from the ground. The mass on the right is given a horizontal speed, then after some time--
    (A) The mass on the left will be nearer to ground.
    (B) The mass on the right will be nearer to ground.
    (C) Both the masses will be at equal distance from the ground.
    (D) Nothing can be said regarding their positions.

    2. Relevant equations


    3. The attempt at a solution
    Just thinking visually, I feel that as the mass on the right moves with a horizontal velocity, the left mass will lift up.
    So, answer is B, which is in fact the right answer.
    But the masses also feel tension and I just think there must be a more concrete explanation.
     
  2. jcsd
  3. Jul 11, 2015 #2
    Both masses have a net force of zero i.e. the tension = weight.for each a condition of equilibrium and thus there is no acceleration but given a shove ( or impulse) they will move at constant speed depending on the magnitude of that impulse. The masses are out of equilibrium only during the time of the impulse.
     
  4. Jul 11, 2015 #3
    look at the scetch
    upload_2015-7-11_20-48-11.png

    Let's say that the length of the swinging part of the rope is ##r## and we give the right mass a ##{v_0}## speed. Then when it's in the ##\theta## angle position. it will have a speed ##v## such that

    ##mgr\left( {1 - \cos \theta } \right) + {1 \over 2}mv_0^2 = {1 \over 2}m{v^2}##.

    The radial forces on the right body will give

    ##T - \cos \theta mg = m{{{v^2}} \over r}##.

    Now you need to show that ##T > mg##
     
  5. Jul 11, 2015 #4
    OOPSS.

    The first equation should be

    ##{1 \over 2}mv_0^2 = {1 \over 2}m{v^2} + mgr\left( {1 - \cos \theta } \right)##
     
  6. Jul 11, 2015 #5
    After that corection I will say that it's depand on ##{v_0}##.
     
  7. Jul 12, 2015 #6

    haruspex

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    I'm not sure how to interpret that. Does it mean, for all time after some point?
    If so, consider what the eventual motion will look like.
    That would certainly show that the left hand mass rises initially, but it does not directly follow that it will at some point be higher than the right hand mass. When a pendulum is at one extreme of its swing, the tension is less than mg.
     
  8. Jul 12, 2015 #7
    Not necessarily, it's depands on ##{v_o}##.
     
  9. Jul 12, 2015 #8

    haruspex

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    What depends on v0? That the left hand mass rises initially? That at some point it will be higher than the right hand mass? Whether the tension in a pendulum is less than mg at some point in its swing?
     
  10. Jul 12, 2015 #9
    Whether the tension in a pendulum is less than mg at some point in its swing.
     
  11. Jul 12, 2015 #10

    haruspex

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    Consider a pendulum at one extreme of its swing.
     
  12. Jul 13, 2015 #11
    I don't know if this changes anything, but the initial figure is supposed to look like this
    http://www.drcruzan.com/Images/Physics/Atwoods/EqualMassesNoAcceleration.png
     
  13. Jul 13, 2015 #12
    When considering the mass on the right, it does show any vertical displacement. But it shows a horizontal displacement.
    Since the string is not considered to be elastic, let's take its length to be a constant.
    At the starting point the masses are at equal height so when the mass on the right moves in the horizontal direction, the mass on the left should move up. Again pointing out that the positiin of the mass doesn't change vertically.
     
  14. Jul 13, 2015 #13

    haruspex

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    In the early stages, both masses will accelerate upwards. The question is (at that point), which acceleration is greater?
    Consider the FBDs of the two masses when the right hand string is at some small angle theta to the vertical. What are the vertical forces on each?
     
  15. Jul 14, 2015 #14
    I don't understand why the mass on the right side has an acceleration. Is it because theta varies with time and therefore the tension also varies, and the vertical forces are not in equillibrium?
     
  16. Jul 14, 2015 #15
    Would the question be valid for any reasonable value of any of the variables ?
     
    Last edited: Jul 14, 2015
  17. Jul 14, 2015 #16

    haruspex

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    The left hand mass rises, so the tension must exceed mg. The right hand mass experiences the same tension, but, except right at the start, it will not be vertical.
     
  18. Jul 14, 2015 #17
    Why does the mass rise ?
     
  19. Jul 14, 2015 #18

    Nathanael

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    This assumes r is constant, which assumes the other mass isn't moving up or down.
    Your conservation of energy equation also assumes the other mass isn't moving up or down (you have no term for the other mass's potential energy).

    The equations of motion would be:
    ##F_r=mg\cos\theta - T =m( \ddot r - r\dot \theta ^2)##
    ##F_{\theta} = -mg\sin\theta = m(r\ddot \theta +2\dot r \dot \theta)##
    We can eliminate T by considering the forces on the other mass: ##T-mg=m\ddot r##

    These equations can tell you the upwards acceleration of each mass at the initial time, but beyond that they're not very useful unless you know how to solve them (which I certainly don't).


    If we make the approximation V0<<r, then cosθ ≈ 1 and sinθ ≈ θ which would turn the equations into:
    ##2\ddot r \approx r\dot\theta ^2##
    ##-g\theta \approx r\ddot \theta +2\dot r \dot \theta##
    @haruspex or anyone:
    Do you think these equations have a simple-ish solution? Or are they not practically solvable?
     
  20. Jul 14, 2015 #19

    haruspex

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    I don't think the student is expected to solve, or even write down, those equations.
    Did you consider the line I suggested in post #13?
     
  21. Jul 14, 2015 #20

    Nathanael

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    I don't either, I am personally curious what the solution would look like.
    And yes I understood your post #13.
     
    Last edited: Jul 14, 2015
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