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Homework Help: Circular motion?

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A child's playground swing is supported by chains that are 4.0m long. If the swing is 0.50m above the ground and moving at 6.0 m/s when the chains are vertical, what is the maximum height of the swing?

    2. Relevant equations

    3. The attempt at a solution
    Back again.. this time I am LOST. I feel like this should be a circular motion type of problem, but that doesn't entirely make sense since I'm working on energy conservation, linear momentum, torque, and angular momentum. This is what I DO understand:

    radius = 4.0m
    vi = 6.0 m/s
    vf = 0 m/s
    The total height is whatever it is plus 0.50m.

    I'm lost because I have no clue how to do the problem without a time component or a mass component or an angle. I understand that gravity plays a vertical role, slowing down the speed of the swing.
  2. jcsd
  3. Jul 12, 2008 #2


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    Staff Emeritus
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    Gold Member

    It may be easier in this case to consider conservation of energy rather than focusing on circular motion.
  4. Jul 12, 2008 #3
    The conservation of energy should be sufficient to answer this problem.

    You don't need mass or time. I am pretty sure it is assumed that the ground is flat, otherwise the problem wouldn't be solvable. In the case the ground is flat then you are given an angle which is 90 degrees when the swing as at the bottom of its swing.

    The swing is at is maximum velocity when it is at the bottom of its swing (when its not moving upwards). Like you said the swing has no velocity at vf of this problem. This means that at this time it has transferred all of its kinetic energy into ....
  5. Jul 12, 2008 #4
    Sorry, I had a dr. appointment....

    The kinetic energy was transferred into potential energy. This helps a bit, but I feel like I'm still missing something.

    I know the change in potential energy is U = mgΔy; I don't know m, but I can figure Δy from the equation: Δy = vi^2 / 2g = (6.0 m/s)^2 / (2)(9.8 m/s) = 1.84 m. So am I just adding 1.84m + 0.50m = 2.34m ??

    I feel like I'm missing something, otherwise why would the length of the swing chain (4.0m) be included?
  6. Jul 12, 2008 #5

    Doc Al

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    Staff: Mentor

    That's all there is to it. Realize that the equation you used is just an application of energy conservation.
    I can think of several reasons: (1) Just to see if you know what matters and what doesn't, or (2) The problem might have another part that will require the length of the swing. :smile:
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