00:02
So once again, welcome to a new problem.
00:05
This time we have a rod, or you can think of something that's horizontal.
00:16
But we're just going to call it about the extending a specific distance, and it still happens to have a length equivalent to capital l.
00:28
So this is the body itself, has the length of capital l.
00:32
And this extends from this side up until the left side up until the right side.
00:40
And it's always helpful to assume that there's an axis and x -axis that goes that way.
00:47
And then there's also y -axis that's perpendicular.
00:53
And then, you know, we have the density.
00:58
I'm going to talk about the density.
01:00
And this is the density connected to, and not just any density, but what we're saying is the mass density.
01:12
And the mass density is given by row x equals to row not plus row 1 minus row not x of a capital l.
01:28
And this is going to be squared.
01:32
The x that you're seeing, the x that you're seeing is this dimension, assuming, you know, maybe we want to start the whole rod at the x axis at the y axis, so right here, this is at the y axis.
01:50
So any x distance moves that way, and, you know, that's a reflection of what's going to happen with the mass density as we're progressing from the left up until the right.
02:03
And our goal in this particular problem is to figure out, well, the mass density, we already know it changes linearly as you progress from the left up until the right.
02:18
But we want to find the center of mass.
02:22
So our goal, and this problem is to figure out.
02:28
What the center of mass is going to be.
02:31
We do know that the row not and row 1 these are constants row not and row 1 these are constants and row that the center of mass is given by the reciprocal of all the masses, the particular masses combined while coming up the infinitesimal product of the masses and the radius from a specific position.
03:20
So assume that you're slicing these, you're slicing the radius, you're slicing the rod and then, you know, each one of this is infinitesimal and that's what we're calling mj remember there are a couple of them if you continue slicing you're going to have a bunch of them and that's what you're seeing in this particular problem of finding the center of mass this is obviously the central mass of the object that's the center of mass and then also but besides the center of mass we have the combined mass the combined mass that you're seeing and this is the mass of the jade particle.
04:20
So a single particle inside the road, if you have the road and you slice it, and assume this is the particle.
04:28
So that's the mass of that particle.
04:33
And obviously, n is the number of particles.
04:41
And is the number of particles if you see.
04:44
We also do know that the velocity of the center of mass is going to be the reciprocal of the combined masses turns to the product of the velocity and so this is the velocity of the center of mass.
05:08
That's the velocity of the center of mass that you're dealing with.
05:12
The rod itself has a density.
05:16
The rod itself has a density.
05:18
So row x is ro -not, ro -not plus row 1, row not, row not x over l, we want to square that, and assume on the road, the road is going that way the infinitesimal section of the road.
05:54
We're going to call that dx, that's the distance, and then this whole particle, dm so this is infinitesimal and infinitesimal you gotta get that right so this is um infinite tesson mass and this is an infinite tessimal distance so we have the intemesemem mass and we have the infinite decimal distance from the origin and forward and the distance in terms of x so remember if you think about volumes you know density is massive volume so obviously the mass has to do the density times per volume because you're multiplying both sides and so this is density volume is the area times the length think of it this way if you have some that's rectangular, this is the area and this is the land.
07:27
So at the intonetational level we can say that the n equals to row x, s dx, well this is the area, this is the length, and this is the density.
07:45
So it's similar to what you're seeing here, it's just that you're doing it with a bigger scale.
07:53
So s is the area of the board that you're dealing with.
07:58
The total mass which you're saying is the same as n is the integral of dm.
08:05
So what you're doing is you're summing up all these slices of dns.
08:11
We have a bunch of them.
08:14
But then we're using the integral sign because we're using summation.
08:17
Summation from i 1 up until n d n is the same as just getting the integral to get the total mass the vault itself runs a distance l so if you're doing the integral you're going to have to do it from zero up until l by dm but we've already seen that dm which is the infinitesimal mass is given by the density times the area times the intentesimal then in the x direction so we're going to change bm with those symbols that you see it's bx and then what happens is we can simplify this by the isolation process we already know what the density is going back you can see that the density is written in the specific using specific variable so we want to do replacement from that instead of having density in x i'll be write it in terms of constants.
09:25
So this is row 1 minus roe 0 x over l, squared up.
09:33
So this is x d x and this one becomes s is a constant so we can take it outside of the integral finer and it is still comfortable solving the problem.
09:49
The constants are helpful in simplifying what we're doing in the so this is 1 minus 1 l x over l squared d x and then this one becomes s we already took the s upside so now we'll get in the integral row not everything is in terms of d h and l for ro not l plus ro 1 x cubed over 3 l squared and that's going to go from 0 to l minus roanot x cubed of 3l squared again 0 to l.
10:47
So all we did is just to get the integral of all these items that we have and think of this here.
10:58
If you have roe -0 x not, the integral that is r00x and then if you plug in the l, it becomes roanaut l.
11:10
The same applies to row 1 minus row not.
11:16
If you do the integral of that it's going to be x squared.
11:24
It's going to be x.
11:27
So we have x squared and then we changed that to q3.
11:32
We already had an l and so in terms of just make sure these two are kind of isolated so we don't make any confusion so then we have row 1 and minus row not x cube over 3l and that's what you have at this point these two so we just fool them apart that's what we did and oh and we have to square this also because we have this squared so we need the distributive property and then we did the integral of that.
12:10
So at this point we've completed the integral and we can proceed with the simplification process of plugging in and so we have no not l plus row 1, l over 3 because we plugged in l into the equation and then minus 0 0 l over 3.
12:34
Proceed by simplifying.
12:39
So we have, we can pull the l out, this is an l, this is an l, this is an l, this is an l.
12:44
So we can pull the l out.
12:45
Pull the l out.
12:46
So inside we have r00 1 over 3 minus r0 1 over 3 and this becomes sl 2 r0 plus 1 over 3 so we're just combining everything together and since we wanted to get the center of mass we can go back to our original formula for finding the center of mass at the infinitesimal level, we're combining always values, and so by definition, if you're looking at the center of mass, it's the same as the, all the infinitesimal added together, divided by all the combined masses added together.
13:49
So all you're saying is that the denominator is the same as combining all the masses, and then the numerator is the same as combining the product of the distance of the intestinal level connected with the masses.
14:06
So now we're moving towards the integral after we get the summation.
14:10
And so this is going to become the same as integral from 0 to l, x, row x, sbx, and the reason for that is because dm, you know, it's like dm sending someone a dm on instagram or facebook.
14:31
That's the same as dm, so we're going to do the same thing with the denominator from 0 to l instead of dm, we do the replacement, which we have.
14:44
And so this whole thing, xpx, we have the numerator, so we're going to look at the numerator and the denominator...