00:01
Locate the center of gravity of the given figure.
00:03
So here we have a rod of i'm assuming uniform density that is bent into shape.
00:11
We're given the dimensions for which we want to calculate the center of gravity.
00:19
So the x coordinate of our center of gravity, generally speaking, is given by the integral of xdl over the integral dl.
00:34
When integral d .l corresponds to the total length of our rod.
00:42
Similarly, to my component writes as y, dl over dl, the integral over dl.
00:49
And similarly, the z bar will be given by the integral of z, dl, over the integral over dl.
00:58
So if we were given something that was curved or a function to describe the pole, then the integral format is therefore much easier.
01:12
But because we're given rod that is bent straight lines, we can just use a little bit of geometry to obtain these in our mass because this will be right as length l1 times x1 bar plus length x2 times x2 bar so i'm going to dissect our bent here in text segments so oa is 1, ab is 2, hmm how to okay, so one, two, three, four.
02:12
And i tried to place the circles at, by i, seems to be the middle of the rod, which will correspond to the center of gravities of each segment.
02:48
And we can do this similarly for the rest, but i'm not gonna write just because it's long.
02:53
So let's start with the first segment of our rod...