00:01
And this problem is asked to analyze this physical pendulum, which consists of a light rod here that has a smaller mass up here and a larger mass up here and it pivoted about its center.
00:21
So the forces that act are obviously the weight of this mass and the weight of this mass.
00:27
There are forces in this rod that hold the mass on here.
00:32
And then there are reaction forces here that then are at the pivot point that react to these forces here.
00:39
Now, what we're going to do in this problem is actually just look at this whole thing as the system.
00:45
And because and leave these forces here as well, these forces here as internal and not consider those.
00:58
Basically just that there would be a reaction.
01:02
These forces here would be offsetting these so again you just want to want to leave these ones out if you wanted to figure out what the tension was or the forces were in the in the rod here obviously you could you could then you know look at this body and then this body separately again get these things but we're going to look at them all together so that um this whole thing together and take moments about this point.
01:36
And you can also see that these don't generate any moments about that point either.
01:41
So if we take moments about this point, that is the moment of inertia times the angular acceleration.
01:49
And i've said positive is, you know, counterclockwise.
01:53
And that equals the moment generated from this mass up here, which is the weight of the mass times l over 2, this distance here, times sine of theta.
02:05
So the moment arm, that is, this is the moment arm here.
02:10
So here's the moment arm, and that is l over 2, sign of theta.
02:14
So this is the moment arm.
02:16
And then for this one, we get a torque that is in the negative direction.
02:21
So we have the weight, and again, its moment arm is this length here, which is l over 2, sine of theta also.
02:31
Now, the moment of inertia about this point, assuming that these are just point masses, is m1 times l2 squared over 4 or l over 2 squared, and then plus m2 times l squared over 4, which is for this one.
02:52
So in the end, we can see that when we plug this into here and then move everything to one side, we get the sum of the mass.
03:04
The total mass times l squared over 4 equals the difference in the masses m1 minus m2 times gl over 2 times sine of theta.
03:12
Now we see an l cancels out and we can cancel out a 2 here.
03:18
This becomes a 2.
03:19
And then we're going to assume that this is small motion.
03:22
So sine of theta is going to be approximately theta for small oscillations about the the vertical position here.
03:32
And so, and then, dividing through by m1 plus m2 and l.
03:40
And let's see here.
03:41
Did i forget? i think i forgot a two in here.
03:47
Hmm.
03:48
I think i did.
03:49
Um, there should be a two out in front of this guy.
03:56
It means there should be a two, um, in front of this guy.
04:01
And that means there should be a two out in front of this guy.
04:02
And that means there should be a two out in front of this guy.
04:03
A one half in here.
04:08
I forgot to keep this two around when i rewrote the equation down here.
04:16
So what we have then is that writing it this way, we can see this is the natural frequency squared.
04:24
So the natural frequency is this.
04:27
And then the period is 2 pi over the natural frequency, which gives us this.
04:32
And so an interesting thing here to note is obviously if m2 is zero, so m2 was zero, then m1 cancels out, just like in a regular pendulum.
04:45
Remember that in the regular pendulum, it's just g over l.
04:50
And in this case, you know, we have l over 2 is the length.
04:53
So it's kind of, that's why we have this factor of two here.
04:57
This is, this is l over 2, not l.
04:59
So the total length is l...