Mr.p: Good morning. Billy, please read the problem and Bobby, please translate.
"Flipping Physics" Intro
Billy: Where is the center of mass of a rectangular,
constant density, constant thickness block that has a hole in it?
Dimensions are shown in the illustration.
Bobby: Okay, this looks very similar to the problem we just did where we had a "L"-shaped block
we treated as two pieces, but...
Bo: Right, but this is a block with a hole in it. Where are the pieces we can identify?
Bobby: Maybe the hole is one piece?
Bobby: And the rectangular piece only without a hole in it is the other piece?
BBB: Maybe. Maybe. Maybe.
Mr.p: That is absolutely correct.
Let's identify piece 1 as the rectangular block,
which does not have a hole in it and is 15.7 by 6.5 centimeters.
And piece 2 as the hole, which has a diameter of 3.3 centimeters.
In order to make this work, realize piece 1 does not have a hole in it.
And, therefore we subtract piece 2, the hole, from piece 1.
Billy: That makes sense because that is what this block is. It is a rectangle with a hole removed from it.
So, we take piece 1, the rectangle, and remove piece 2, the hole, from the rectangle.
Mr.p: Bo, please identify the locations of the centers of mass of both pieces.
Bo: Well first, we need to identify the zero, zero location.
Let's say zero, zero is at the lower, left hand corner of the block.
Both pieces are symmetrical and have constant densities
so their centers of mass will be in their geometric centers.
Actually, the y-position center mass of both pieces is the same; it is at half of 6.5 or 3.25 centimeters,
so the y-position of the center of mass of the block must be the same, 3.25 centimeters.
The x center of mass of piece 1 is at half of 15.7 or 7.85 centimeters.
The x center of mass of piece 2 is at 15.7 minus 3.25 or 12.45 centimeters.
We do not have the equation for center of a system of particles written down yet, so let's write that down.
Center of mass equals mass 1 times position 1 plus mass 2 times position 2 all divided by the total mass.
There are only two objects, so that's the full equation, which equals question mark.
Actually Bo, because 6.5 only has two significant digits,
the answer for the y-position center of mass can only have two significant digits as well.
Because 3.25 ends in a perfect five, a five with no digits after it, the answer rounds to the even number.
So the y-position center of mass is 3.2 centimeters. Billy: That is the arcane rounding rule.
Mr.p: Billy is correct. Now, please solve the problem.
Well, just like last time, we do not have the masses of the pieces, however, we can determine their areas. And
we showed last time that we can replace each mass in the center of mass equation with its corresponding area;
do you need me to derive that again?
Mr.p: No, that's okay. If you want to review that derivation, you can watch the video. Please continue.
Billy: Okay, we substitute in equations for areas.
Area 1 equals length 1 times width 1 and area 2 equals pi times radius 2 squared.
We never did determine the radius of the hole, so radius 2 equals half of diameter 2, or 3.3 divided by 2,
which equals 1.65 centimeters. Now let's substitute in values.
We have 15.7 times 6.5 times 7.85 plus pi times 1.65 squared times 12.45 all divided by the quantity
16.5. I'm sorry, 15.7 times 6.5 plus pi times 1.65 squared. Which works out to be 8.2057 or
or 8.2 centimeters with two significant digits.
Actually, that cannot be correct. Relative to the center of piece 1,
the center of mass of the block should be farther away from the hole rather than closer to it.
Bo: We forgot to subtract the mass of the hole.
Bobby: Right, the hole is not there. So we have to subtract Mass 2, so we have to subtract area 2.
That's a pretty easy fix.
The correct answer is 7.4292 or 7.4 centimeters with two significant digits.
Mr.p: I should point out that,
because the block is three dimensional, it also has a center of mass in the z direction,
which is halfway through the thickness of the block.
However, in this example, we are treating it as a two dimensional object.
And we can of, course, test this answer by finding the center of mass, right?
Billy: Oh right! We hang the block by two different points and the center of mass of the block is at
the intersection of the vertical lines drawn directly below the hanging point.
Bobby: Then we can add a dot at the center of mass and it matches what we calculated.
Bo: And throw the block because it will rotate around its center of mass and
its center of mass will trace out a parabola because it is an object in projectile motion.
Will you look at that? The physics works.
Mr.p: The physics works, the physics works, uh huh, uh huh the physics works, the physics works...
Bobby: That's completely your fault, you know. He forgot last time.
Bo: I can take the blame.
Mr.p: The physics works,
the physics works,
uh huh, uh huh
the physics works, the physics works...
Thank you very much for learning with me today, I enjoyed learning with you.