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Conservation of Energy - Swing Problem

  1. Oct 25, 2006 #1

    AVD

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    Conservation of Energy - "Swing Problem"

    Hi everyone I'm having some trouble solving this problem because I'm not sure what method to use. It seems like it could involve Centripetal acceleration, conservation of energy, and/or conservation of momentum.

    Problem
    A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the seat and begins to swing with zero velocity at a position at which the cord makes a 60° angle with the vertical is shown in Figure I. The swing continues down until the cord is exactly vertical at which time the child jumps off in a horizontal direction. The swing continues in the same direction until its cord makes a 45° angle with the vertical as shown in Figure II: at that point it begins to swing in the reverse direction. With what velocity relative to the ground did the child leave the swing? (cos 45° = sin 45° = , sin 30° = cos 60° = 1/2, cos 30° = sin 60° = /2)

    Diagram

    [​IMG]

    So far, I figured you could use the conservation of energy formula
    Pe1+Ke1=Pe2+Ke2 to determine velocity in terms of L
    Ke1 cancels out when the swing is at rest and Pe2 cancels out when the swing is vertical.

    mgh = 1/2mv^2
    ( I wasn't sure if Pe2 should include the mass of the second person or not)
    This is assuming the person jumped off
    (2m * 9.8m/s^2 * L-Lcos60) = (1/2 * m * v^2)
    18.6L m/s = v^2

    This is if the person is still on the swing
    (2m * 9.8m/s^2 * L-Lcos60) = (1/2 * 2m * v^2)
    9.8L m/s = v^2

    I tried a few things from here - using centripetal acceleration and angular momentum formulas but nothing worked out. I'm not sure if I'm even starting it right.

    Thanks for any help & I'm still a high school student so I may not understand some terminology or if I'm even doing the math right (sorry for that) :smile:
     
  2. jcsd
  3. Oct 25, 2006 #2

    OlderDan

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    You are on the right track with energy conservation. You can use that to find the velocity of the child and swing at the vertical (yes, both masses are needed). You can also use it to find the velocity of the swing seat after the child leaves. The instant of separation at the vertical can be examined without resorting to any angular or rotational considerations. The child and swing are both moving horizontally just before and just after separation. What principle applies?
     
  4. Oct 25, 2006 #3

    AVD

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    Do you mean I should apply conservation of energy to the moment before and the moment after the child leaves the swing?
     
  5. Oct 25, 2006 #4

    OlderDan

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    I mean you should apply conservation of energy to the whole system of child and swing during the time between the start and reaching vertical, and then you should apply it to the swing after the child has jumped. This will tell you the velocity of the swing just before and just after the jump, and the velocity of the child before the jump.
     
  6. Oct 25, 2006 #5

    AVD

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    i actually tried this but didn't bother to mention it because I kept hitting a dead end. For both velocities, my answer has the term L. Is there any way for this to cancel out? Also is my math right for the velocity of the swing and the child together? I got v = \/(9.8L) <-- p.s. that's my best attempt at square root of 9.8 L =)

    Anyway, I'm not sure if this even makes sense / if my math is right and how to get a value for v without the variable L
     
  7. Oct 25, 2006 #6

    OlderDan

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    You answer will be in terms of L. Everything else is numerical.
     
  8. Oct 25, 2006 #7

    AVD

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    ok thanks - well here is the rest of what i tried:

    After child leaves the swing

    1/2mv^2=mgh
    1/2(m)(v^2) = (m)(9.8)(L-Lsin45)
    v^2 = 4.9(L-Lsin45)
    v = 2.4 * (square root of L)

    Before the child leaves the swing

    I already established that
    v = 3.2 * (square root of L)


    Am I on the right track?

    -edit-

    from here should i use conservation of momentum since this is like the reverse of an inelastic collision ( is that an "explosion" i forget). Anyway i got an answer of approx 4(radical L)

    (m+m)Vi = MVf1 + MVf2
    6.4radicalL = 2.4rad L + v ( i canceled out M1 and M2 since they're both the same)
    (6.4 - 2.4)rad L = v
    4radL m/s = v (approx)


    I know i degenerated from saying the square root of L to radical L to rad L but I was getting a bit lazy & don't know how else to write it :)
     
    Last edited: Oct 25, 2006
  9. Oct 25, 2006 #8

    OlderDan

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    You don't mean 4.9, but you must have used 2*9.8 in your calculation to get the result you got. I also suspect you used (L-Lcos60) to get the 3.2sqrt(L) before the separation, so I'm a bit surprise by the sin45. Of course the result is not affected since sin45 = cos45, and for all I know you used (L-Lsin30) for the first velocity calculation. The units of you answer should be sqrt(meters)/second, since you have to multiply by the square root of a length to get velocity. The method is right on.
     
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