# Period of a child on a swing

## Homework Statement

Question: If a child gets up from sitting position to standing while swinging, how does the period change?

## Homework 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

## 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?

Related Introductory Physics Homework Help News on Phys.org
jedishrfu
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.

I would treat the child as a simple pendulum, an ideal case.
This is probably how it was meant to be done. I just did not think it would be a reasonable approximation.

In the real world a child on a swing is something to enjoy because it gives you some time to relax.
That's only if the child is old enough. Otherwise, it is a way to get exercise (by pushing), contemplating forced oscillations.

BvU
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.