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How to find the acceleration of a ball rolling on ground?

  1. Feb 26, 2016 #1
    1. The problem statement, all variables and given/known data
    "A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?"

    2. Relevant equations
    ##x_0=15m##
    ##x=0m##
    ##t_2=0.02s##
    ##v^2=v_0+2a(x-x_0)##
    ##v=v_0+at##
    Answer as given by book: (a) 857 m/s2; (b) up

    3. The attempt at a solution
    ##v^2=2(9.8\frac{m}{s^2})(15m)##
    ##v_0=0\frac{m}{s}##
    ##v=17.15\frac{m}{s}##

    Next, I treat this as my next initial velocity when the clay rolls on the ground for 0.02 s.

    ##v_0=17.15\frac{m}{s}##
    ##v=0\frac{m}{s}##
    ##t=0.02s##

    ##0\frac{m}{s}=|17.15\frac{m}{s}|+a(0.02s)##
    ##a=|\frac{-17.15\frac{m}{s}}{0.02s}|=|-875\frac{m}{s^2}|≠857\frac{m}{s^2}##

    Aside from not matching the answer in the book, the sign of the acceleration is negative, which would mean it is going down. Unless I'm mistaken, and the negative acceleration is going up from -875 m\s2 to 0 m\s2... I'm not sure if I'm correct or not, in this assumption, though. Could someone confirm or deny my answer and my assumption?
     
  2. jcsd
  3. Feb 26, 2016 #2

    haruspex

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    When you solve v2= some number, what are the possible values of v?
    I agree with the 17.15, as a magnitude. Check the next step. (Looks like you dropped a digit.)
     
  4. Feb 26, 2016 #3
    Oh, I made a miscalculation.

    ##a=\frac{-17.15\frac{m}{s}}{0.02s}=-857.3\frac{m}{s^2}##
    ##|a|=857.3\frac{m}{s^2}##

    So does that mean that the average acceleration is up, when it rises from -857 m/s2 to 0 m/s2?
     
  5. Feb 26, 2016 #4

    haruspex

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    The acceleration does not rise from -857 m/s2 to 0 m/s2.
    You did not answer my question about solving for v2.

    Edit: by the way, the ball does not roll on the ground, it squishes on the ground.
     
  6. Feb 26, 2016 #5
    It's going to be ±17.15 m/s. I also saw that I typed in 17.5 on my calculator; not 17.15.
     
  7. Feb 26, 2016 #6

    haruspex

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    Right. Which is it here?
     
  8. Feb 26, 2016 #7
    Negative, because it's going down?
     
  9. Feb 26, 2016 #8

    haruspex

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    Right. So what do you get for the acceleration?
     
  10. Feb 26, 2016 #9
    -857 m/s2?
     
  11. Feb 26, 2016 #10

    haruspex

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    The velocity has gone from negative to zero. Does that make the acceleration positive or negative?
     
  12. Feb 26, 2016 #11
    ##a=\frac{v_f-v_i}{t}##
    ##a=\frac{0-(-17.15\frac{m}{s})}{0.02s}##
    ##a=857\frac{m}{s^2}##

    So positive.
     
    Last edited: Feb 26, 2016
  13. Feb 26, 2016 #12
    But that's only for the initial fall. What about after it touches the ground and squishes?
     
  14. Feb 26, 2016 #13

    haruspex

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    No, you took the final velocity as zero to get that acceleration. That is the average acceleration while squishing.
     
  15. Feb 26, 2016 #14
    Oh, so...

    ##v_0=-17.15\frac{m}{s}##
    ##v=0##
    ##t=0.02s##

    ##v=v_0+at##
    ##a=\frac{17.15\frac{m}{s}}{0.02s}=857\frac{m}{s^2}##
     
  16. Feb 26, 2016 #15

    haruspex

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    Yes.
     
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