17.5 Worked Example – Center of Mass of a Uniform Rod

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.

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