Here's an example of how you could find the center of mass of a system,
and this one has to do with the sun, the moon and the earth.
Let's ask where the center of mass is of the earth-sun system.
In other words if I just consider the earth and the sun,
where is their center of mass between them, if we estimate the mass of the sun
is about 2 times 10 to the 30th kilograms and the mass of the earth
as 6 times 10 to the 24th kilograms and the distance between them
of 1.5 times 10 to the 8th kilometers
and then we'll do the same thing asking about the center of mass
between the earth and the moon.
And then again we'll have a mass for the moon which is much smaller than the earth
and then we'll have a distance of the moon from the earth.
So give this a shot using the equation that we just used
or down for center of mass in one dimension and see if you can find where the center of mass is of this object reside.
If you've try this problem hopefully it looks something like this.
We ask about the center of mass for two objects.
Now this object here in the first part of the problem would be the sun,
the bigger object and then the smaller object would be the earth.
So what we have to do is to use our center of mass equation
and say center of mass is equal to and then we multiply M1 times its position,
plus mass 2 times its position and then we'll divide it by the total mass.
In this case, we know that the mass 1 is the mass of the sun,
so in this case we have, actually, I'm just gonna write it like this.
What we can do in this problem, is pick any origin of our coordinates system.
So, it's important in this situation that you understand where your coordinates are.
In this picture, for example, we have x equals zero to the left of our first mass,
our first object the sun, but when they're asking us where is the center of mass,
we can give them an answer in terms of any position that we'd like.
So for example, if it's more convenient in the problem,
don't take a coordinate system that is inconvenient for you, pick your own coordinate system.
Here's how you would do that.
So we can say that the sun is at position x equal zero and then we can give our reference,
our center of mass in terms of what we're calling a zero in terms of the sun.
So we can say our center of mass is at such and such distance from the sun itself.
So let's do that and you can see why it will simplify our analysis greatly.
In this case, X1 the position of the sun is actually zero,
we're gonna call the sun at zero position and since X1 is zero,
we don't have to consider the mass term for the sun at all because its mass
is great but the position of it is zero by our definition.
Then all we have to do is write the mass 2, which is the mass of the earth
we said this was 6 times 10 to the 24th times its position
which we estimate as 1.5 times 10 to the 8th
and then we just need to divide it by the total mass.
This is the mass of the sun 2 times 10 to the 38th times, sorry plus,
be careful when you're doing this that you always multiply the numerators
but you're adding the denominators, 6 times 10 to the 24th.
Now there's one more simplification that we can make in this problem,
and that's by looking at the denominator and recognizing the vast difference
in the two numbers that we have in the denominator.
This is something that might be counterintuitive at first if we're just used to following the numbers and trying to find an answer,
but certainly in exam setting or any practical setting you really want to understand
which numbers overwhelm others in terms of significance.
So looking using scientific notation at these two numbers, we have 2 times 10 to the 38th
plus 6 times 10 to the 24th, notice that there are six orders of magnitude difference between these two.
This would be like my saying we're adding, we're taking the number one
and then adding to a .000001. The second number, the very small one
will not really change the one in any way and it's certainly won't affect our answer
towards the many, many decimal places.
So when you're given a problem like this, you should not ever spend extra time
trying to figure out this other term which has so many more zeros in front of it.
It will not change your answer in any way that you would ever see, in any option that you would be given.
So we're not gonna consider this very, very tiny number relative to the mass of the sun because it won't changed it.
And so we see, using a scientific notation, using 6 times 1.5 divided by 2, 6 times 1.5 is 9 divided by 2
and then we have our units 10 to the 24th times 10 to the 8th is 10 to the 32nd power
divided by our 10 to the 38th and so we see that we get 4.5 as 9 divided by 2
and then times 10 to the 2nd power, after we've cancel the 32 with the 30.
So our answer is that is 4.5 times 10 to the 2 and remember, what our units were
450 kilometers since these are the units we're using for distances away from the sun.
And we know the sun would be a huge object, so in the fact the sun is so big
that the center of mass still resides inside the sun, and doesn't even leaving.
So now we could use this exact same analysis to solve the earth-moon system
and it won't change dramatically at all but notice the similarities
and we have the center of mass for the earth-moon system
would be very similar. Where we could instead consider this to be the earth
and this one to be the moon and the center of mass would follow the exact same logic
And we have the mass of the earth, times its location so we can call that x
and then we have the mass of the moon times its location
and then we divide by the sum of the two objects masses
and we can do the exact same thing because again notice
if we have the position of the earth is defined as zero for us
we don't have to worry about that, then we'll just deal with the mass of the moon
which is 7 times 10 to the 22nd times its position which is 4 times 10 to the 5th
and then we divide it by the sum of the masses and we're gonna see the exact same sort of argument
6 times 10 to the 24th, plus 7 times 10 to the 22nd.
We have that this number here is still to or magnitude smaller.
In other words, 1% of the mass of the earth and so this won't change your answer
within 1% of the actual number.
And so again, it's much better not to spend a lot of extra time solving problems
when these numbers are changing your answer by such small amounts.
So we will again not consider this, because it will be such a small change
and we'll do the same sort of thing, we'll have 7 times 4 divided by 6 with our orders of magnitude upfront
so 10 to the 22nd times 10 to the 5, over 10 to the 24th.
Any numbers this is going to be approximately 4.5 again because we have 7 times 4 is 28
divided by 6 is close to 4.5 and then 22 plus 5 is 27 divided by 10 to the 24th,
we get a power of 3, so times 10 to the third kilometers.
In other words, about 4500 kilometers away from the earth,
since the radius of the earth is actually still greater than this.
We still haven't gotten our center of mass outside of the bigger object
this still a center of mass that is inside of the earth,
though not at the earth's surface.
What we'll going to do now is one last thing which is to consider
where is the center of mass for a two-dimensional system.
If we have objects that are not just stretch along one direction
but could also very along the y-direction, a vertical direction in this graph as well.
In this case, it's exactly what you might expected it,
which is that you could first find the center of mass in the x direction,
in which case you only take the masses times their x positions
and you keep doing this ones for each mass and then you divide it by
the sum of the masses to properly change your units again.
To get the y center of mass, you'll do the exact same thing and you can solve it separately.
So you have mass 1 times its position of the y-direction plus mass 2
times its position of the y-direction etc., and one more time you'll divide it by
the sum of the masses and these are now different.
So it's important to notice that in a problem like this, the y positions of many of these masses,
mass 1 has a y position of zero because it has no distance this way.
Y2 also has a position of zero and the x position of mass 3 is also zero.
So there are many simplifications that you can use in this problems
in the center of mass problems, in general, it's always a good idea
to try to take advantage of this sort of simplifications.
So in this lecture what we've done is we've gone through both circular motion,
a uniform circular motion and then discuss how to find the center of mass for a complicated system
or just a one-dimensional system, especially when one object is much, much heavier than the other.
Thanks for watching.