Good morning. Previously we learned how to determine the center of mass of uniform, rigid
objects with shape and determined the x-position center of mass of a uniform right triangle. Today
we will use the same approach to determine the center of mass of a nonuniform, rigid object with
shape. But before we do that, let’s take a moment to discuss three different kinds of density.
Three different kinds of density? Yeah, I thought there was only one called…
Density ♪ Flipping Physics ♪
Right. Let’s start with the one you are most familiar with, which is volumetric
mass density and equals mass divided by volume and has a symbol of the lowercase Greek symbol
rho. This is actually not the only type of volumetric density which exists.
There is also volumetric weight density and volumetric charge density, for example.
However, today, we are only going to discuss mass densities. In addition to volumetric mass density,
there is also a mass density that uses the lowercase Greek letter sigma
which equals mass divided by area and is called surface mass density. And there is
also a mass density that uses the lowercase Greek letter lambda which equals mass divided
by length and is linear mass density. So, yeah. Lot’s of different types of mass densities.
Wow. Woah. Yeah. Today we are going
to do an example which uses mass per unit length, or linear mass density. Bo, please
read the problem, and Bobby, please translate. Determine the x-position center of mass of a
horizontally oriented rod with a length of 0.65 meters and lambda linear mass density
equal to 43 minus 21 x squared grams per meter. Well, our knowns are pretty obvious. Length
equals 0.65 meters, linear mass density equals 43 minus 21 x squared grams per meter, and we
are trying to find the x-position center of mass. Did you just make this problem up in your head?
What makes you think he did that? The numbers go 6, 5, 4, 3, 2, 1.
That seems made up to me. Yeah.
Why yes Bo, I did fabricate this problem. It is not real.
Then we won’t know if the physics works! But we will be able to tell if the physics
makes sense. How?
Well, if the rod were of uniform density, its center of mass would be at its geometric center,
right in its middle, or 0.325 meters from either end. Considering the rod has a linear mass density
of 43 minus 21 x squared grams per meter, Billy, where would you expect its center of mass to be
relative to the geometric center of the rod? Uh How would we know that?
When x equals zero, the linear mass density of the rod equals 43 grams per meter and as the
x-position increases, according to the equation, the linear mass density of the rod decreases.
Oh, that means the linear mass density of the rod decreases from left to right. So, we should
expect the rod to have more mass concentrated to the left of its center than to the right,
so the center of mass of the rod should be a bit to the left of the geometric center of the rod.
Yep. Exactly Billy. Bobby,
what equation are we going to use to determine the x-position center of mass of this rigid
object with shape of nonuniform density? Well. When we did the rigid object with shape
of uniform density, we used x-position center of mass equals the 1 over the total mass of
the object times the integral of x position with respect to mass. Do we use the same equation here?
Yes, Bobby. That is the equation we are going to use here. However, rather than starting
with that equation directly, we are first going to determine the total mass of the rod
because we will need to use that in the x-position center of mass integral equation.
In order to find the total mass of the rod, we are also going to use an integral. The
total mass of the rod equals the integral with respect to mass. Bo, how are we going to do this?
This is similar to what we did last time. We need to consider the rod to be made up of an
infinite number of infinitesimally small objects dm. Each piece dm is
going to be located a variable distance x from the left end of the rod. Each piece dm will have
an infinitesimally small width dx. Uh … Don’t forget to use lambda.
The linear mass density. Oh, right. Linear mass density equals
well for the whole rod it equals the total mass of the rod divided by the length of the whole rod L.
For the infinitesimally small piece dm, its linear mass density equals its mass, or dm, divided by
its width, which is the infinitesimally small width dx. Therefore, dm equals lambda times dx.
We can substitute that back into the equation for total mass of the rod.
Actually, now we have all the information we need. Lambda equals 43 minus 21 times x squared.
And this is a definite integral with limits for the x position of dm which go from the left end
of the rod to the right end of the rod or from zero to the length of the rod which is 0.65
meters The integral works out to be 43 x minus 21 x cubed over 3 with the same limits as before.
21 over 3 is seven. We can substitute in our limits and we get
24.9925 grams for the total mass of the nonuniform bar.
Wait a second. The linear mass density of the rod does not equal total mass of the rod divided by
length of the rod. It doesn’t?
No, the average linear mass density equals total mass divided by rod length.
Ohhh. Right. The
linear mass density changes as a function of position, so it cannot equal the total mass
over length because that would be just a single number and not change as a function of position.
Exactly. Oops.
Thank you everybody. Billy, please find the x-position center of mass of the rod.
At this point we use the equation x-position center of mass equals the inverse of total mass
of the rod times the integral of x position with respect to mass. This is very similar to what Bo
just did. We substitute in the limits of zero and 0.65 meters and lambda of 43 minus 21 x squared.
Multiply the x through to get 43 x minus 21 x cubed. Take the integral to get the inverse of
the total mass of the rod times the quantity 43 x squared over 2 minus 21 x to the fourth power over
4 with the same limits. Substitute in the limits and we get 0.325961 meters or 33 cm with 2 sig
figs and in centimeters. {waiting}
That’s not right. Yeah, it’s to the
right of the geometric center of the rod. We don’t even get to say the physics makes sense?
Nope. We made a mistake. Where?
We failed to use a calculator correctly. Bo, you squared 0.65 rather than cubed it. Our total mass
calculation is incorrect. The total mass of the rod, when calculated correctly, is 26.0276 grams.
Oh. Sorry. No Bo, this is not your fault. I have said
it many times. Y’all need to be doing all of these calculations yourselves. You need to be doing your
calculator calisthenics. If you were all doing these calculations, you would have caught the
mistake. When it comes to tests and exams, the last thing you want to discover is that you have
not been doing your calculator calisthenics and your calculator does not know what it is doing.
Way too many times I have had students who understand the physics, but fail to do basic
calculator calculations correctly, therefore, they get incorrect and illogical answers like this one,
and think they do not understand the physics, when in fact they do understand the physics,
they just have lazy calculators. So, please. Do not let your calculator be lazy,
do your calculator calisthenics. Calculator Calisthenics. Got it.
Sometimes my calculator is lazy. I’ll have to talk to them about that.
I don’t even have a calculator. You are a calculator.
Hold up. Then the physics makes sense.
The x-position center of mass of the rod is a bit to the left of the center of the rod,
just like we predicted. Yea! The physics makes sense, The physics makes sense, uh
That does not really work, does it. No.
Yeah. Bummer. I am glad the physics makes
sense though. Thank you very much for learning with me today, I enjoyed learning with you.