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Can you deduce the coefficient of restitution from impace force plot?

  1. Nov 15, 2012 #1
    If you have the impact force on the y axis (in Newtons) and the time on the x axis (in milliseconds), can you get the coefficient of restitution from just this? The plot is for the force of impact exchanged between two balls as a function of time.

    I know the coefficient of restitution is the change of velocities of the balls just after impact divided by the change just before.

    Thanks
     
  2. jcsd
  3. Nov 15, 2012 #2

    tiny-tim

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    welcome to pf!

    hi mokwana1221! welcome to pf! :smile:

    hint: what does the area under the curve represent? :wink:

    (in the centre-of-mass reference frame, the coefficient of restitution is also the square-root of the ratio of kinetic energy after to kinetic energy before)
     
  4. Nov 15, 2012 #3
    hmm it gives us the impulse (Newton*Seconds) but can the coeff. of restitution be deduced from that? i am still kind of lost unfortunately
     
  5. Nov 15, 2012 #4
    also the area under the graph is triangular
     
  6. Nov 15, 2012 #5
    hopefully somebody can help me out before my final tomorrow, thanks guys
     
  7. Nov 15, 2012 #6

    haruspex

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    I gather that the graph shows the force increasing linearly to a max then decreasing linearly to zero, the two slopes being different.
    Intuitively, with restitution = 1 the two slopes would be the same. For less than 1, the second slope would be steeper. So it seems reasonable that the restitution is related to the ratio of the two areas.
    Try considering the impact in those two phases. Let the two slopes be s1, s2, lasting t1, t2. The compression would peak, would it not, when the two bodies are at the same velocity? That should enable you to get enough equations.
     
  8. Nov 15, 2012 #7

    tiny-tim

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    have you done integrals? area represents an integral
     
  9. Nov 15, 2012 #8

    haruspex

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    Hi tiny-tim.
    I believe mokwana1221 established at post #3 that this much is understood. But that only gives the momentum exchange. It will not in itself say anything about the energy loss. See my previous post.
     
  10. Nov 15, 2012 #9
    i see now i am kind of getting it. yes the two slopes are different (and naturally the two areas under the graph as well, divided by the peak).

    1) can we get a definitive quantitative answer using this graph for the coefficient of restitution?

    2) if yes, would it be the ratio of the two slopes, or the ratio of the two areas under the peak?

    this was under the professor's 'advance topics' for the exam as we did not cover it but i am putting in the extra mile just in case he throws it on the exam

    thanks
     
  11. Nov 15, 2012 #10
    the area of the first triangle is 10 and the area of the second is 5, the ratio is .5.

    the slope of the first triangle is 5 and the slope of the second triangle is 10. the ratio is also .5.

    so my questions is, is the coefficient of restitution .5? or am i missing something? or is it not possible to get the coefficient of restitution from just this plot (if this is the case, what other data is needed to get the coefficient of restitution)?

    thanks guys i feel like this is a simple problem and that i am just missing something.
     
  12. Nov 15, 2012 #11

    haruspex

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    Perhaps - or it could be a more complicated relationship involving a square or square root. It cannot be guessed at. Try what I suggested: consider the compression and decompression phases separately so that the speeds are the same at peak compression. Write out equations relating speed, momentum, time, whatever for start and end of the two phases. See where it leads.
     
  13. Nov 15, 2012 #12
    i am truly at a dead end. i do not see how one can get energy or velocity from this graph since multiplying force (N) times time (milliseconds) equals impulse.

    i have been thinking and googling this question since i first posted and joined the forum hours ago. i am stuck.
     
  14. Nov 15, 2012 #13
    can anyone help me with a last effort?

    thanks, kwan
     
  15. Nov 16, 2012 #14

    haruspex

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    Have you tried what I suggested? Consider the compression and decompression phases separately so that the speeds are the same at peak compression.
    Let the masses m1, m2, start at speeds u1, u2 (same straight line), reach a common speed of v at peak compression, and finish at speeds w1, w2.
    Let the impulse during compression (i.e. area under first triangle) be I and that during decompression be J.
    Let the coefficient of restitution be R.
    Write out equations relating the above:
    - four for momentum
    - one from the definition of coefficient of restitution
    If you do all that you'll be surprised how easy it is to get the answer.
     
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