Solving the Pendulum Problem: Find Speed at Bottom of Path

[bfr]

Homework Statement

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?

Homework Equations

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

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.

 

[quantumdude]

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

 

[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?

 

[quantumdude]

You're close.

PE=mgh=9.8(1.2-.92)~=2.75

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

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?

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?

 

[bfr]

Oh, yeah...thanks!

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

 

[quantumdude]

Yes, that's it.