0:00
A very known question.
00:01
The solution is a bit known as well as to be patient.
00:05
So we basically look at a bar, right? of mass, above a mass.
00:11
What's happening is that you have a kind of a pole, right? kind of a pole around or whatever.
00:24
And this pool, you know, you have a mass around somewhere here.
00:31
Okay? some mass here, and you have a route which is massless and attached to a mass which is m here.
00:45
Free to slide along a line in the x direction.
00:58
Let me read it more clearly.
01:03
Now here we, in this question we have basically a something, a pop of mass, right? and you basically have a kind of a rod and this rod, suppose this is a rod is something like this.
01:19
You have a rod something like this.
01:21
Now this rod has a mass, right? it has a mass m, i guess.
01:26
And it can freely move along this x direction, right? suppose this x direction.
01:32
It can freely move slide along the x direction.
01:38
And it's attached to a mass right, m.
01:46
I think it's little m, right? at end of this, at end of, this is a support, right? this supporter can actually move, can actually slide about, right? so this is a support, support.
02:03
And this supporter can move about another x line.
02:07
And so this mass is attached to the support by this route, by this road, a cord, which actually is massless.
02:17
And then this point can is free, the pivot freely.
02:21
Okay, so this mass can actually do some oscillations, free oscillations about this pivot, right? and so this lens is air, right? and the angle we can call it sita, and the position of this support, the center of the support is x, okay? so it's very clear that that's a diagram we have, right? and then the tudokinand, energy of the system is to write down.
02:50
The total kinetic energy of the system, of course, is just the kinetic energy of the support, which is m over 2 times x dot squared, right? and plus m over 2 times, you know, l squared times c time dot squared, right? that's kinetic energy.
03:09
The dot means the time derivative.
03:11
And how about the potential energy of the system? well, the potential energy of the system is basically the potential nature of this mass, right? i'll take the zero potential energy reference point to be at the position of this support.
03:28
So that would actually be obviously given basically by the mass m times g, the acceleration gravity and times the height, which is basically this point.
03:41
And actually should take a minus sign because it's actually low.
03:44
So you can minus m g times that a.
03:46
And times obviously is because this is theta, so this is theta.
03:53
So this one is actually l -cosin -sita, right? so that is potential energy.
04:00
And then, of course, you can also find, you can find the like a range of the system, where l is simply given by e -canada energy minus e -p, right? so you'll find this to be mx .x squared over 2 plus m over l -squared.
04:18
The ceta dot squared and and of course plus m zl cosine cosine right um and then what you can do next is find equation...