00:01
Okay, in this question we have this bar with sliders at point b and a here.
00:09
They're moving along tracks that are in these directions that i'm going to highlight in red here.
00:18
This bar of length l is a fixed length, and this bar will move, these objects will move in given slots here, one of them at this angle of 40 degrees with the horizontal, one of them horizontally.
00:38
And we want to know what the angular velocity of the bar will be in terms of the displacement.
00:48
So to do that, we'll have to work out a little bit of geometry first.
00:53
So first we're going to use the law of signs to compare that horizontal position xa over the sign of theta.
01:05
We'll have to equal that length l over the sign of 40 degrees.
01:17
So next we can differentiate, well, we can solve for xa and then we can differentiate with respect to time to get the velocity of a.
01:31
So the velocity of a will equal l is not changing.
01:37
So that's just a constant.
01:39
I'm going to pull that out of our derivative here.
01:51
So we have the derivative respect to time, sine of theta, d theta, not a d theta there yet, just a sine of theta.
02:04
So when we take that derivative with respect to time, we're left with l over the sign of 40 degrees.
02:12
Times the cosine of theta, and then theta dot, which is just the time derivative of theta.
02:21
So this then gives us the rate at which that angle theta is changing, and that's just that theta dot.
02:31
So we can solve for that.
02:36
We'll call that omega here, and that will just equal the velocity of that bar, or velocity a times the sign of 40 degrees, divided by l times the cosine of theta.
03:02
However, we're not done yet.
03:04
We still need to find the value of the cosine of theta.
03:11
So we can do a few things to get there...