# 17.5 Worked Example – Center of Mass of a Uniform Rod

So now we want to

find the center of mass of a uniform rod. And we have the result for

a continuous body, which is that integral over the body

of dmr to that mass element dm divided by an interval. Now our goal is

to figure out how to apply this

result, specifically, to real physical objects. And the key, as always, is

choosing a coordinate system. So now I’ll draw

the object, again. And the first thing I’ll

do is choose an origin. I can pick my origin

anywhere I want. I can pick it in the middle. I can put it in the middle. I could put it at this end. I could put it that end. I could put it down here,

but I’ll choose it over here. Because the object

is linear, this is a very Cartesian system. I’m only doing a one

dimensional object. So I choose my

coordinate system plus x. That’s step one. Now my origin– now here

comes the crucial thing. In this argument, dm is the

infinitesimal mass element. And I want to pick that at an

arbitrary place in the object. I don’t want to pick

it at the origin. I don’t want to

pick it at the end. Note down here

this is x equals L. So I’ll arbitrarily pick

an infinitesimal element. I’ll shade it in dm. That represents–

this is what I’m going to make my summation

over when I do my integral. I’m going to add

up all these dm’s. And the point is that the

dm’s are different distances from the origin. So the vector– and

here’s the next step– is I draw a picture

of my vector rdm. So now I have these terms, at

least, explained in my diagram. The next step is to turn– is

to introduce an integration variable for both

of these quantities. So step one was the

coordinate system. Step two, was the

identification of dm. And step three and

I think this is absolutely the crucial one is

to introduce the integration variable. Now you’ll see that will

come in two different cases. So this is the quantity,

the distance from dm to the origin that’s changing. You can see for each of these

little elements, that changes. So what I’ll write [INAUDIBLE]

as a vector is x prime, which will be my integration

variable in the i hat direction. So the integration

variables x prime. That’s the first place that

I introduced the integration variable. And x prime, you

can see, will vary. And it varies from x prime

equals 0 to x prime equals L. And that will show up in terms

of the limits of my integral. Now the second place that the

integration variable comes in is dm. I want to express

in terms of x prime, which is a measure of

where this object is. And that’s how, if we

choose this length here to be dx prime, notice in terms

of the integration variable, then I have a relationship

between and dm and dx prime. dm is mass in this

little element. dx prime is the

length of the element. And if the whole object is a

uniform rod with a mass capital M and a length L, then its just

given by M over L dx prime. And this quantity M

over L is an example of a mass, linear

mass density, which we have a scale challenge about. So I have two places,

where my integration variable has been introduced. And now I can write up every

piece in this interval. So let’s now indicate

that we’re integrating from x prime equals

0 to x prime equals L. dm is M over L dx prime, and

our vector is x prime i hat. And downstairs, it’s

just M over L dx prime from x prime equals

0 to x prime equals L. And that’s how I

set up the integral for the center of mass. Both of these integrals are

now not difficult to do. Notice, it’s x prime dx prime. So this integral is

x squared over 2. And I get 1/2 M

over L, L squared. And downstairs, dx prime

from 0 to is just L. So the downstairs

integral is just M over L. And when you have

M over L’s cancel, we just are left

with a– this is– I’m sorry– this is just M not

M over L, dimension incorrect. So we get for the

position of the center of mass, the M’s cancel. One of the L’s cancel. And we have an i hat

in this expression so our answer is r equals

L over 2 I hat, which is exactly what we expected. We expected the center of mass

to be half way down the rod.

## 2 Replies to “17.5 Worked Example – Center of Mass of a Uniform Rod”

You're GOOD!! Very GOOD.. great presentation!!…

Thanks a lot!!