# Homework Help: Pendulum Problem

1. Jan 14, 2008

### bfr

1. The problem statement, all variables and given/known data

The bob of a pendulum 1.2m long is pulled aside so the string is 40 degrees from the vertical. When the bob is released, with what speed will it pass through the bottom of its path?

2. Relevant equations

PE=mgh
KE=(mv^2)/2

3. The attempt at a solution

Well, I started out with cos 40=x/1.2 and found x to be approximately .92, but I don't exactly know where to go from there.

2. Jan 14, 2008

### Tom Mattson

Staff Emeritus
Well, you wrote down the expressions for KE and PE. What do you think you should do with them? Think conservation.

3. Jan 14, 2008

### bfr

Wait...PE+KE is always a constant, right? So when the bob is first released at 40 degrees, it as zero kinetic energy, and PE=mgh=9.8(1.2-.92)~=2.75, where .92 is the solution to "cos 40=x/1.2". So, at the bottom of its path, it's height will be zero...right? Which leaves me with m(9.8)(0)+.5(m)(v^2)=2.75. Can I just eliminate "m" from the equation and solve from there?

4. Jan 14, 2008

### Tom Mattson

Staff Emeritus
You're close.

There should be an m on the right hand side. You've only accounted for the gh.

You can eliminate the m, but only after you make the correction on the right side of the equation. Do you see what I'm talking about?

5. Jan 14, 2008

### bfr

Oh, yeah...thanks!

So m(9.8)(0)+.5(m)(v^2)=2.75m -> .5v^2=2.75 -> v~=2.35 ?

6. Jan 15, 2008

### Tom Mattson

Staff Emeritus
Yes, that's it.