00:02
All right, so for this problem, we're going to start by just thinking about how the trapeze is moving.
00:14
And so it's starting at an angle of the theta 1.
00:17
And so the potential energy it's going to have at this point due to gravity will be due to this height right here, which we can find in terms of the length of the trapeze is l, which is the maximum or the minimum height.
00:34
Minus l cosine theta, which is the projection of this triangle right here.
00:40
As we swing down, this height's going to decrease, and we're going to get a new height, which is going to be minus l cosine theta as this theta approaches zero.
00:54
And so we can write a conservation of energy equation starting from here for any point along this path, where each, the initial kinetic energy is going to be zero.
01:07
Initial potential energy is mgh1, which is this one right here, in terms of theta 1.
01:14
The final kinetic energy is 1⁄2mb squared, and the final potential energy is mgh, where h is in terms of theta.
01:25
Now we can plug in these relations right here and solve for v squared.
01:31
If you solve for b squared first, you get 2g times h1 minus h.
01:37
And then you plug in the l's and minus cosines.
01:42
We can pull the l out, factor that up.
01:45
We get 1 minus cosine theta minus l minus, or 1 minus cosine theta.
01:52
So cosine theta 1 minus cosine theta, which minus the minus the minus becomes positive.
01:57
The ones cancel each other out.
02:00
And we end up with this.
02:03
Is equal to 2gl, cosine theta, minus cosine theta 1.
02:08
From here, we can look at the sum of the forces on the trapeze artist as she swings down...