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Vertical Spring System Problem

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1.60 kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 195 N/m and a 280 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 15.0 cm below its equilibrium point (call this point A) and released from rest.

    a) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.)
    b) How much time elapses between releasing the system at point A and the ball leaving the tray?
    c) How fast is the ball moving just as it leaves the tray?

    2. Relevant equations
    amax=Aω2
    x=Acos(ωt+[itex]\phi[/itex])
    v=-ωAsin(ωt+[itex]\phi[/itex])

    3. The attempt at a solution

    So I already solved part (a) by equating the accelerations of the ball and the spring system, g=xω2 , and got an answer for x. But I can't seem to figure out how to get the time at this instant for part (b): I tried using x=Acos(ωt+[itex]\phi[/itex]) and solving for t, but it doesn't make too much sense to me. And I believe I can solve for part (c) with the time t from part (b) using v=-ωAsin(ωt+[itex]\phi[/itex]), but this I'm also not 100 percent on.

    Could anyone nudge me in the right direction? I seem to be stuck after part (a) :confused:
    Thanks!
     
  2. jcsd
  3. Jan 15, 2012 #2
    You need to consider acceleration.
    If the tray loses contact with the ball it tells you something about the acceleration of the tray.
    Presumably this occurs when the tray is moving down so you need the time to get to this point
    What did you get for x?
     
    Last edited: Jan 15, 2012
  4. Jan 15, 2012 #3
    For part (a), I got 0.244m, which is the distance from the compression A to the point of separation; so the distance from equilibrium would be x=0.094m.

    As for the time at that point, I'm really iffy about how to solve for it.. not sure if I should be using a=-Aω2cos(ωt+[itex]\phi[/itex]) or not.. and since A is not at equilibrium, I think phi would be something other than zero..
     
  5. Jan 15, 2012 #4
    Using k = 195 N/m and total mass of tray+ball = 1.88kg I got ω^2
    to be 104
    The max acceleration is ω^2 A where A is the amplitude which I think is 0.15m
    so the max acceleration is 15.5 at the top of the oscillation and at this point the table will be accelerating down at 15.5 m/s^2. ie will leave the ball behind.
    I think this is a strange question because if the table leaves the ball behind at all it is bound to be at the top of the oscillation where the acceleration is max.
    This means that I get the height to be 0.15 +0.15 = 0.3m
    What do you think?
    I don't think I am misreading anything strange in the question wording.:confused:

    This analysis is not correct an I have modified it in post Below
     
    Last edited: Jan 16, 2012
  6. Jan 15, 2012 #5
    That approach was my initial approach, but I have verified that it is not correct :eek:
    It would be tempting to believe that the ball leaves at the top of the oscillation, but it actually leaves somewhere between the equilibrium and the top. So instead of using a max acceleration, Aω2, I used an arbitrary acceleration, xω2, where x is the displacement from equilibrium. And then, observing that the ball's normal force has to disappear as a condition for leaving the tray, I equated that acceleration xω2 to g, the acceleration due to gravity, and solved for x.

    But now the problem is how to solve for that time t :cry:
     
  7. Jan 16, 2012 #6
    I solved it!
    Here's the solution for part (b):

    We know that the ball-tray-spring system is released under the equilibrium at point A, and we also know that the ball leaves the tray at some point between the equilibrium and the top of the oscillation (this is because as soon as the oscillation passes equilibrium, its acceleration points downward).
    I first found the period of oscillation to be T=2[itex]\pi[/itex]sqrt(Ʃm/k)=0.616 seconds, and took T/4=0.154 seconds as the time it took to get from point A to equilibrium.
    Then I used the acceleration function:
    a=ω2Acos(ωt+[itex]\phi[/itex]),
    and took x=0 when t=0 to solve for [itex]\phi[/itex]:
    0=Acos([itex]\phi[/itex])
    [itex]\phi[/itex]=[itex]\pi[/itex]/2

    So, at the point of separation:
    -g=ω2Acos(ωt+[itex]\pi[/itex]/2)
    where ω=sqrt(k/Ʃm)

    . . . after some math and calculations, I get t=0.067 seconds, which I add to T/4 to get:
    t'=0.221 seconds
    (the instructor-website verified answer :surprised )

    And for the sake of completion, part (c) is solved in a similar way:
    v=ωAsin(ωt+[itex]\phi[/itex]),
    where [itex]\phi[/itex]=[itex]\pi[/itex]/2,
    using t=t'=0.221 seconds from part (b),
    v=1.186 m/s
    (correct me if I am wrong anywhere in my solutions)

    Sorry that I posted and answered my own question! But thank you so so much for the discussion! That's where it really counts. I hope this stuff helps other weary souls looking for an answer.
     
  8. Jan 16, 2012 #7
    I now think that my analysis is wrong. I believe the ball will leave the table before it gets to the top of the oscillation.
    I am going to find the displacement when the acceleration becomes 9.81 and I will get back to you
    Have just woken up !! And want to solve this one
    Cheers
    Just done that and got acc = 9.81 when displacement = 0.094m
    So the height is 0.15+0.094 = 0.244m
    This is what you got in your first post so you were correct in your approach all along
    I will go through your other numbers and will get back to you.
    Cheers
     
    Last edited: Jan 16, 2012
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