Mr. P: Good morning. Bo, please read the problem and Billy, please translate.
"Flipping Physics" Intro
Bo: Where is the center of mass of an "L" shaped, constant density constant thickness block
with the dimensions shown in the illustration.
Billy: Okay, well, um.
We know the equation for the center of mass of a system of particles.
So let's start there. The center of mass of a system of particles equals
mass 1 times the position of mass 1 plus mass 2 times the position of mass 2
plus as many of those mass times position expressions as we have objects,
all divided by the total mass of the system.
Other than that, I don't really know what to do because it's just one object.
Mr. P: Okay, that is reasonable. How about this?
Do you see any way we could separate this "L" shaped block into different pieces that are geometrically
symmetrical, so we will know where the centers of mass of those pieces would be?
Billy: Oh, I get it.
If we imagine the block to be made of two pieces, we can use the center of mass of the system of particles
equation to determine the center of mass of the block.
Bobby: Right. Two pieces. We could use would be
the lower rectangle that is 22 centimeters long and by 10 centimeters wide and the square in the upper right,
which is 6.8 by 6.8 centimeters.
Mr. P: Sure, that will work, but do realize that any two, or even more, shapes that we pick
will result in the same correct answer.
And you can see I've labeled the rectangle, piece 1, and the square, piece 2.
Bo, where will the centers of mass of both of these pieces be?
Bo: Well, the block has a constant density, so the center of mass of each piece will be in the geometric center.
For piece 1 that would be at the halfway point for each side. So, the center of mass for piece 1 is... uh.
Bobby: Actually, we have yet to identify where the zero reference point is.
Let's set the origin at the lower left-hand corner of the block. That should help.
Bo: Right, then the center of mass of piece 1 one is at x equals half of 22, or 11, centimeters
and y equals half of 10, or 5, centimeters.
The center of mass at piece 2 is at the center of piece 2 which is half of 6.8, or 3.4, centimeters
in the x and y directions from the lower left-hand corner of piece 2.
However, we need those distances from the origin.
The x-position of piece 2 is at 22 minus 3.4, or 18.6, centimeters.
And it's y-position is at 10 plus 3.4, or 13.4, centimeters.
Mr. P: Thank you Bo. Bobby, please solve the problem.
Bobby: Uh... We don't have the mass of the block.
Or, actually, we don't have the mass of anything.
So how do we find the center of mass without any masses?
Mr. P: So, really, you're going to let not knowing any masses get in the way of solving the problem?
Do you have a numbers dependency?
Bobby: Uh... Yeah, I think I probably do.
Billy: We know the density of each piece is the same, right?
Mr. P: That is correct, Billy. Please keep going.
Billy: So, density 1 equals density 2.
Density equals mass divided by volume.
So mass 1 divided by volume 1 equals mass 2 divided by volume 2.
Both volumes are area times thickness, but the thickness for both are the same so...
Billy & Bo: Everyone brought thickness to the party! Bobby: Everybody brought mass...
Mr. P: Absolutely, Billy. Keep going.
Billy: Well, we can solve for mass the 1 to get a relationship between mass 1 and mass 2,
however, I do not know how that helps us.
Bobby: I bet it would help if we substituted that equation for mass 1 back
into the center of mass of a system of particles equation.
Billy: Yeah, and when we do that...
Bobby: Everybody brought mass to the party!
Bo: Sort of.
Billy: It's kind of a mass party.
Mr. P: It's good enough for me.
And we end up showing that, when all the pieces of the object have the same density and thickness,
we can substitute in the area of each piece for the mass of each piece
in the center of mass of a system of particles equation.
This can be very helpful to remember. Bobby, please solve the problem.
Bobby: Well, at this point, all we need to do is substitute in numbers.
Oh, but I guess we also need to substitute in length times width for all the areas.
For the x-position center of mass, it works out to be 22 times 10 times 11 plus 6.8 squared times 18.6
all divided by the quantity 22 times 10 plus 6.8 squared.
That works out to be 12.31995, or 12, centimeters with two significant digits.
Mr. P: Correct. And we can do the same thing for the y position center of mass
only substituting in y values for the positions instead of x.
So, for the y-position for piece 1 we have 5 centimeters.
And for the y position for piece 2 we have 13.4 centimeters.
Which gives us 6.45889, or 6.5 centimeters with two significant digits.
So, the position of the center of mass of the "L"-shaped block is 12 centimeters to the right of and
6.5 centimeters above the lower left-hand corner of the block.
Bo, why is this a logical location for the center of mass of the "L"-shaped block?
Bo: Well, isn't the center of mass actually also half the thickness deep into the block?
Mr. P: Bo that's a good point.
So the block is not a two dimensional plane, so it does have a z dimension center of mass
which is halfway through the thickness of the block.
And Bo, what about the logic behind the center of mass of the block?
Bo: Um... The center of mass should be somewhere between the two centers of masses of the two pieces
and should be closer to the more massive piece 1 and that is where we determined the center of mass to be.
So that makes sense.
Mr. P: Great Bo, and if you recall we can determine the center of mass of any flat object
by hanging it from different points.
So, here it is hanging from one point. Add a vertical line on which the center of mass must be located.
And we can rehang the object from a different point and add another vertical line.
Those two lines intersect where the center of mass of the "L"-shaped block is located.
We can add a dot there to indicate it is the center of mass, and you'll notice it matches what we determined.
And then we can throw the block to put it into projectile motion.
And will you look at that? The "L"-shaped block not only rotates around its center of mass,
but its center of mass also describes a parabola just like all objects in projectile motion.
Billy, Bobby, Bo: Cool. Wow. Nice
Mr. P: I totally agree. Thank you very much for learning with me today, I enjoyed learning with you.