0:00
Hi, everybody.
00:01
So for a, we need to find to express the initial in terms of m and l.
00:07
Okay, so we have the derivative, dx, lambda, x.
00:14
Okay, and basically we're going to have dm equals d.
00:21
Oops.
00:23
I mean, it's still going to be dx, but let me put this a little bit better, lambda x, dx.
00:28
There we go.
00:28
I wrote it weird.
00:29
So we have capital m, we're going to need to l and zero.
00:35
And pretty much it's just going to be lambda dx, and we're just going to take the intuple, okay? of the x, so we get x squared over square, over 2, l, zero.
00:48
So this we get one half, and i do not need that one there.
00:55
One over, lambda over two, l squared.
00:58
Okay? and now for b, b is asking us right here, what is, on the x is the right rod, how it results in b and c.
01:18
So basically we're kind of just doing the exact same thing as we did as a.
01:22
So we have the rod, and we need to find the inertia of the rod.
01:29
So we have m equals one -half and the l -square.
01:34
This is of the rod.
01:36
And we have lambda.
01:37
It's going to be 2m, l square.
01:40
Okay, we're not going to save that for later.
01:42
So now we have inertia.
01:44
It's going to be equal to l0 x squared d .m...