1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook