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!

Period of a child on a swing

  1. Dec 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Question: If a child gets up from sitting position to standing while swinging, how does the period change?

    2. Relevant equations
    Period of a physical pendulum: T = 2π√(I/mgL), where I is the moment of inertia and L is the distance between the pivot and center of mass
    Period of a simple (mathematical) pendulum: T = 2π√(L/g), where L is the distance between the (point) mass and the pivot

    3. The attempt at a solution
    The suggested answer I have seen is that a child on a swing is a physical pendulum. When the a child gets up, his center of mass moves up closer to the pivot point, so L (see the equation above) decreases, and the period therefore increases.
    The problem I have with this answer is that when child gets up, his/her moment of inertia changes as well - how this can be taken into consideration?

    Another possible answer is to consider the child a simple pendulum, in which case, when he gets up, L decreases and the period also decreases. But, in a real world, a child on a swing cannot be approximated by a simple pendulum!

    How should this question be approached?
  2. jcsd
  3. Dec 9, 2014 #2


    Staff: Mentor

    I would treat the child as a simple pendulum, an ideal case.

    In the real world a child on a swing is something to enjoy because it gives you some time to relax.
  4. Dec 10, 2014 #3
    This is probably how it was meant to be done. I just did not think it would be a reasonable approximation.

    That's only if the child is old enough. Otherwise, it is a way to get exercise (by pushing), contemplating forced oscillations.
  5. Dec 10, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Key is the word "up". With ##I = \int r^2 dm## and ##L = \int r dm## it becomes clear that ## I/(mL) ## decreases when changing from sitting to standing.

    Might need the parallel axis theorem to finish this off: worst case is changing from point mass at ##L## (swing length), so ##I = mL^2##, to a rod of length ##l## at ##L - l/2##: $$(L-l/2)^2 + l^2/12 < L^2 \ \ ?$$ leads to ## l(l-3L) < 0 ## which we can assume true.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Period of a child on a swing
  1. 400N child on swing (Replies: 12)