# Swing Problem

• Apr 8th 2008, 03:50 PM
vesperka
Swing Problem
A 16 kg child on a playground swing moves
with a speed of 9.5 m/s when the 2 m long swing is at its lowest point.

The acceleration of gravity is 9.81 m/s^2

What is the angle that the swing makes
with the vertical when the swing is at its
highest point? Answer in units of degrees.

For this problem, i solved for KE first. KE = 1/2mv^2, so plugging in some numbers I get 76 joules. Then I plug this into the PE equation (PE = mgh), so when 76 = (16kg)(9.81)(h) the height I get is .4841997m.

From this point on, I think I get stuck. I understand that energy is conserved, but how do I find out the angle that the swim makes? I tried using pythagorean's theorem, but that doesn't seem to work because the swing is always 2m long (which gives it a round opposite).
• Apr 8th 2008, 04:13 PM
abender
Think this way: sin opposite hypotenuse, cos adjacent hypotenuse,

SOH CAH TOA

You have enough information to find theta (the angle).

-Andy
• Apr 8th 2008, 04:40 PM
vesperka
http://img221.imageshack.us/img221/1769/pic2xi5.jpg
The opposite side is curved, not straight, so when I tried using SOH CAH TOA the answer I got was wrong :(
• Apr 8th 2008, 06:15 PM
vesperka
I'm not really sure where to go with this one. I tried subtracting the height I got from 2m to create a "new" adjacent angle, and then tried using Cos^-1(1.52m/2m) and I get about 40 degrees or something (i forget) but that answer is incorrect.
• Apr 8th 2008, 06:31 PM
topsquark
Quote:

Originally Posted by vesperka
I'm not really sure where to go with this one. I tried subtracting the height I got from 2m to create a "new" adjacent angle, and then tried using Cos^-1(1.52m/2m) and I get about 40 degrees or something (i forget) but that answer is incorrect.

You have the right method. (I could have sworn I posted my own diagram here. Ah well.)

Check your height again. I'm getting h = 4.60 m.

-Dan

Edit: I see what went wrong. You didn't square the 9.5 in your kinetic energy equation.
• Apr 8th 2008, 06:37 PM
vesperka
Oh yeah I screwed that up big time. Still stuck though. So now with the new height, how do I go about setting it up? The hypotenuse is still 2m, but if the adjacent side's length is negative it messes up my calculator (maybe I'm typing it wrong :\ )
• Apr 8th 2008, 06:46 PM
topsquark
Quote:

Originally Posted by vesperka
Oh yeah I screwed that up big time. Still stuck though. So now with the new height, how do I go about setting it up? The hypotenuse is still 2m, but if the adjacent side's length is negative it messes up my calculator (maybe I'm typing it wrong :\ )

I didn't notice that. It does kind of screw the problem up, doesn't it. I'd say this one is FUBARed. Talk to your instructor.

-Dan
• Apr 8th 2008, 06:49 PM
vesperka
Heh thanks anyways, I'll be sure to give you an update tomorrow once I figure out how to get this solved :D