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Ball and tray on a Spring

  1. Jun 25, 2011 #1
    A 1.2 kg horizontaal uniform tray is attached to a vertical ideal spring of force constant 195N/m and a 275 g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down 13 cm below its equilibrium point called point A and released from rest.

    how long does it take the ball to leave the tray from point A?

    So the time it took should equal T/4+ the time it takes to get to the spot it leaves at

    T=.5464

    x=Acos([itex]\omega[/itex]t)

    x is .07420
    A is .13

    [itex]\omega[/itex]=sqrt(k/m)=11.4979

    solving for t at dsiplacement 0 I get t=.0837

    so my total time is .22. Some reason that is not the right answer and i cant figure out where i went wrong
     
  2. jcsd
  3. Jun 25, 2011 #2

    gneill

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    How did you decide that x is .07420 ? The period, T= .5464 seconds looks okay for the tray+ball on the spring.

    Where do you suppose the "launch" point is?
     
  4. Jun 25, 2011 #3
    the launch point is at point A so -.13m

    for x I did -kx=mg

    x=mg/k

    1.475*9.81/195

    =.0742m
     
  5. Jun 26, 2011 #4

    gneill

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    Okay, so your x is the distance above point A where the spring+tray alone are at equilibrium (spring force balances gravitational force on tray). Let's call this point B.

    When the tray+ball just pass point B (on the way upwards) the tray is just beginning to experience a deceleration due to the sum of spring and gravitational forces on the tray being zero at point B. This deceleration is still much less than g , so the ball will remain in contact with the tray. It's not until the downward acceleration of the tray begins to exceed g that the ball and tray will part.

    So, where will the downward acceleration of the tray be g? Hint: acceleration at g sounds like free-fall...
     
  6. Jun 26, 2011 #5
    but if you use x as .20 you cant take arccos
     
  7. Jun 26, 2011 #6

    gneill

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    Why would you be taking the arccos of a distance? Can't have (net) units in the arccos function argument...

    Where did your 0.20 (meters?) come from?
     
  8. Jun 26, 2011 #7
    because


    x=Acos([itex]\omega[/itex]t)



    im solving for t to add it to 1/4 the period.

    is that not the right way to go about this.
     
  9. Jun 26, 2011 #8

    gneill

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    In this case x must be less than the magnitude of the harmonic motion, A. So x/A is less than one.
     
  10. Jun 26, 2011 #9
    Yes and I did that, but my answer is wrong,so im wondering were I went wrong.
     
  11. Jun 26, 2011 #10

    gneill

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    I'm wondering, too. Perhaps you should share your methodology and calculations so that we can puzzle it out.
     
  12. Jun 26, 2011 #11
    (arccos(.0742/.13))/sqrt(k/m)=t=.08378

    .0837+ .1366=.22sec
     
  13. Jun 26, 2011 #12

    gneill

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    The question asks for the time from point A to the launch of the ball. You shouldn't have to include the time it takes to go from the spring-compressed position to point A.

    Also, if you take the point A to be the zero reference for the motion, and if you set t=0 at that point, the displacement will correspond to a sine function rather than a cosine function.

    A*sin(ω*0) = 0
    A*cos(ω*0) = A
     
  14. Jun 26, 2011 #13
    So the answer should be .189430 secs
    Thank you!
     
  15. Jun 26, 2011 #14

    gneill

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    That value looks a bit large to me. What was the calculation?
     
  16. Jun 26, 2011 #15
    .1366+

    arcsin(.0742/.13)/sqrt(195/1.75)=.05754489

    if you round its .19secs
     
  17. Jun 26, 2011 #16

    gneill

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    Your total mass value looks high; 1.2kg + 0.275kg = 1.475kg

    Why are you adding 0.1366? You only want the time from point A to the launch.
     
  18. Jun 26, 2011 #17
    because point A is at 13cm below the equilibrium spot and that is 1/4 of the period

    and isnt 1.2+.275=1.475
     
  19. Jun 26, 2011 #18

    gneill

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    You had written: arcsin(.0742/.13)/sqrt(195/1.75)=.05754489
    So the mass you used was 1.75(kg). Presumably this was a typo?

    Point A is the equilibrium level:
     
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