# Homework Help: Father pushing child on swing problem

1. Apr 23, 2007

### rcw110131

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 . Any help would be greatly appreciated.

2. Apr 23, 2007

### HallsofIvy

First, please do not use "special symbols". I do not have a Font that allows me to see what you have between $sin \alpha$ and $\alpha$. I am assuming that you are saying that $\alpha$ is small enough that $sin\alpha$ is approximately equal to $\alpha$ (which is NOT very accurate for $\alpha$ 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 $-mg sin\alpha$ so you have $m d^2s/dt^2= -mg sin\alpha$ or approximately $m d^2s/dt^2= -mg sin\alpha$ where s is measured along the arc of the swing. If $sin(\alpha)$ is approximately $\alpha$, then $m d^2s/dt^2= -mg\alpha$ approximately.
If the swing chain has length L, then $s= L\alpha$ so the equation is $d^2\alpha/dt^2= (-g/L) \alpha$ or $md^2\alpha /dt^2+ (mg/L) \alpha= 0$.

Now put in the father's push. He is pushing parallel to the arc of motion with magnitude mg so we have $md^2\alpha/dt^2+ (mg/L) \alpha= mg$. As always with gravity problems, the "m"s cancel and we have $d^2\alpha/dt^2+ (g/L) \alpha= g$. That's the equation you want to solve.

3. Apr 23, 2007

### rcw110131

You're a lifesaver, thank you so much for your help!

4. Apr 23, 2007

### rcw110131

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 $d^2\alpha/dt^2+ (g/L) \alpha= g$? I've been searching online and through books for the past 2 hours and can't get it . Thanks.