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Father pushing child on swing problem

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data

    A father is about to push his child on the swing. He pulls the swing back by 1 radian, and start pushing the child with a constant force perpendicular to the swing's chain and with a magnitude equal to mg, where m is the mass of the child and the swing seat. The father stops pushing when the swing's chain is again vertical. In the approximation that sinα ≅ α (where α is an angle up to 1 radian), calculate the time the father is pushing. Neglect the weight of swing's chain and the retardation by the air drag.

    2. Relevant equations

    Umm, some second order differential equation that I have no clue how to derive?

    3. The attempt at a solution

    I really have no idea where to start with this one and I only have like 4 hours to get this done :frown: . Any help would be greatly appreciated.
  2. jcsd
  3. Apr 23, 2007 #2


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    First, please do not use "special symbols". I do not have a Font that allows me to see what you have between [itex]sin \alpha[/itex] and [itex]\alpha[/itex]. I am assuming that you are saying that [itex]\alpha[/itex] is small enough that [itex]sin\alpha[/itex] is approximately equal to [itex]\alpha[/itex] (which is NOT very accurate for [itex]\alpha[/itex] up to 1 radian- sin(1) is only about .84- so I may be wrong about that).

    Okay, your basic equation is F= ma, force equals mass times acceleration. At any point, there is a force, -mg, straight down, but the swing chain offsets part of that- the force parallel to motion is [itex]-mg sin\alpha[/itex] so you have [itex]m d^2s/dt^2= -mg sin\alpha[/itex] or approximately [itex]m d^2s/dt^2= -mg sin\alpha[/itex] where s is measured along the arc of the swing. If [itex]sin(\alpha)[/itex] is approximately [itex]\alpha[/itex], then [itex]m d^2s/dt^2= -mg\alpha[/itex] approximately.
    If the swing chain has length L, then [itex]s= L\alpha[/itex] so the equation is [itex]d^2\alpha/dt^2= (-g/L) \alpha[/itex] or [itex]md^2\alpha /dt^2+ (mg/L) \alpha= 0[/itex].

    Now put in the father's push. He is pushing parallel to the arc of motion with magnitude mg so we have [itex]md^2\alpha/dt^2+ (mg/L) \alpha= mg[/itex]. As always with gravity problems, the "m"s cancel and we have [itex]d^2\alpha/dt^2+ (g/L) \alpha= g[/itex]. That's the equation you want to solve.
  4. Apr 23, 2007 #3
    You're a lifesaver, thank you so much for your help!
  5. Apr 23, 2007 #4
    Just when I thought I had this problem, turns out the 2.5 years of no differential equations has caught up to me. Can anyone help me on how to solve [itex]d^2\alpha/dt^2+ (g/L) \alpha= g[/itex]? I've been searching online and through books for the past 2 hours and can't get it :confused:. Thanks.
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