A brass collar, of length 50 mm , is mounted on an aluminum rod of length 80 mm . Locate the cen

step 1 So once again, welcome to a new problem. This time we're still dealing with the center of mass,

A brass collar, of length 50 mm , is mounted on an aluminum...

Key Concepts

Center of Gravity

This is the point at which the entire weight of an object or composite body can be considered to be concentrated for the purpose of analyzing translational motion. In problems involving composite bodies, the center of gravity is typically found by using a weighted average of the component parts’ positions based on their masses.

Composite Body Analysis

When dealing with a body composed of different parts or materials, each with its own geometric and density characteristics, the composite body analysis involves calculating the overall center of gravity by considering the contributions (mass and position) of each part. This process allows complex bodies to be simplified into a single point of application for gravitational force.

Density and Mass Relationship

Density plays a key role in determining the mass of a component when its volume (or length in one-dimensional problems) is known. For composite bodies made of different materials, variations in density affect how much each part contributes to the total mass and consequently the location of the center of gravity.

Moment Calculation

The concept of moments is critical when determining the center of gravity. Here, a moment is the product of the mass of a component and its distance from a reference point. The overall center of gravity is found by summing these individual moments and then dividing by the total mass of the composite body.

Transcript

00:01 So once again, welcome to a new problem.

00:05 This time we're still dealing with the center of mass, and there's a rod that we're looking at.

00:15 It's extended, and it's, i mean, you can say it's rectangular in shape.

00:22 The distance from one side of the rod up until the other side happens to be the same as 50 centimeters.

00:34 That means obviously the middle part will be 25 centimeters, but the distance from one side to the other one is 50 centimeters, and it's made off of ion such that its density row is the same as the, you know, we can call the density, the density of ion, the same, the same as 8 grams per centimeter cubed.

01:10 That's the density of ion.

01:12 Or maybe we could use a different label saying the density is 8 grams per centimeter cubed.

01:21 That's lambda.

01:23 So we're using lambda for the density of iron, ferris.

01:28 And the rod itself is made up of 50 centimeters.

01:39 Of iron and of course the other portion that you're looking at is going to extend and that second portion is also 50 centimeters but this time the second portion is not iron but is aluminum this is aluminum which is also 50 centimeters and the density of aluminum is the same as 2 .7 grams per centimeter cubed.

02:16 That's the density of aluminum.

02:18 So our goal in this problem is to figure out what the center of mass is.

02:27 And being a rod, we're mostly interested in the x direction.

02:32 That's why we're saying, you know, what's the center of mass for this object? has two components.

02:40 One side is aluminum and the other side is ion.

02:45 It's one -dimensional.

02:48 If you think of it, it's one -dimensional.

02:51 So to get the center of mass, we obviously can recall, you know, m1, x1 plus m2, x2, all over m1 plus m2, but it's one dimensional.

03:06 So we have to use integrals because we're saying was summing up the product of infinitesimal masses of iron to infinitesimal distances of iron and infinitesimal masses of aluminum to infinitesimal masses of aluminum.

03:32 So here we're dealing with the radius.

03:34 We're dealing with the radius.

03:36 So if you have a measurement from a specific point, if this is, the origin for example, each piece has a gap of delta r.

03:48 I don't know why i'm doing delta d, but it's just delta r.

03:53 And then, you know, from one, from the origin up until the center of that portion, we have a radius r.

04:03 So that helps us capture the distance.

04:05 So we're going to have integral from 0 to 50 because our road extends from 0 to 50.

04:11 For each one, both the aluminum and the ion.

04:17 And recall that, you know, density, density is mass of volume.

04:31 So that means that mass is density times volume.

04:40 The density is lambda and the volume is, if it's a rod, if it's a rod, like this, then you can see that the volume is pi r squared times h, but then pi r squared, you know, you could take that to be delta r times r.

05:09 So on this one, if you compute the two, the mass times the distance being covered, is going to be lambda, f .e.

05:26 Times rdr.

05:28 And the f .e.

05:29 Terms rdr.

05:31 Radius times delta r.

05:34 Okay.

05:34 That's what we're saying in this problem.

05:37 And then we also have the second portion.

05:39 It's going from 0 to 50.

05:42 Well, or actually from 50 to 100.

05:46 Because, you know, if you can recall, the rod was extending from 0 to 50 and then from 50 to 100.

05:54 So then this becomes delta aluminum times radius dr.

06:02 Then we want to divide that by the masses of iron and also the masses of aluminum in that sense.

06:09 The next part of the problem is plugging in, the center of mass, from 0 to 50.

06:18 We plug in 8 grams per centimeter, 8 grams a centimeter.

06:28 And this is centimeter cubed and then times rdr plus going from 50 to 100, 2 .7 grams a centimeter cube, rdr.

06:51 We want to divide that by the mass.

06:53 Remember, mass is density.

06:57 Density is mass of a volume.

07:00 So mass has to be density times volume.

07:06 That's what we're going to do right here, 8 grams per centimeter cubed, times the volume which is measured in 0 .5 centimeter.

07:19 It's a singular road, so it doesn't have any thickness...

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