Good morning. Previously we learned that the x-position center of mass of a system of particles
can be described by the equation x-position center of mass equals the sum of mass of the ith particle
times the x-position of the ith particle all divided by the total mass
of the system. Where i goes from 1 to the total number of particles in the system.
But realize, we could also describe the location of the center of mass of a system of particles
in three-dimensions using this same equation. Oh yeah! All we have to do is have two more of
those equations and replace all the x’s with y’s for one of the equations and replace all the x’s
with z’s for the other equation. ♪ Flipping Physics ♪
What does “ith particle” mean? That’s just a fancy way of identifying a single
particle in an unknown total number of particles. Uh, what?
Like, if there were 8 total particles. We could refer to the 4th particle or 7th particle or the
8th particle, however, because the total number of particles is unknown, he is just referring to
a random particle as the ith particle because i ranges from 1 to the total number of particles in
the system. Hence, ith particle. Okay. Thanks.
You’re welcome. Exactly Billy and thank you Bo.
Now, we do not actually have to use 3 separate equations to identify the 3-dimensional position
of the center of mass of a system of particles. Instead, we can use the r position vector and
unit vectors. Recall that the position of a single particle can be defined as position
vector r equals x times unit vector i plus y times unit vector j plus z times unit vector k.
And, rather than identifying the location of a single particle, we can replace each of those, x,
y, and z position variables with their respective center of mass of a system of particles equations
instead and identify the 3-dimensional position of the center of mass of a system of particles
as the x-position center of mass times unit vector i plus the y-position center of mass
times unit vector j plus the z-position center of mass times unit vector k.
Bo, please substitute in our system of particles center of mass equations and tell me what you see.
Sure. Uh, for each of the position center of mass equations with the sum of the ith particle’s mass
times the ith particle’s position times a unit vector each divided by the total mass.
They all have total mass in the denominator, so we can combine all those terms and have the sum of
each of the sigma terms in the numerator and total mass in the denominator. And then I’m not sure. …
Well, we know the r-position of the ith particle … Nice.
… is x sub i times unit vector i plus y sub i times unit vector j
plus z sub i times unit vector k. And, in the numerator, there is
a common operator in each term, the sum of the mass of the ith particle.
Right, and that means we can substitute the r-position of the ith particle into the equation,
and we get the r-position of the center of mass of a system of particles equals the sum
of the mass of the ith particle times the r-position of the ith particle
all divided by the total mass of the system. But isn’t that just the equation we started with for
the x-position of the center of mass of a system of particles and we just replaced x with r? That
seems like a lot of work to just replace x with r. Ah, Bo. I can totally see that perspective,
however, please realize that we can now use one equation to identify the position of the center
of mass of a system of particles in 3-dimensional space. Again, when we have a system of particles,
we can describe the location of its center of mass in a coordinate plane using this single equation.
Clearly, I am only showing this in 2-dimensional space, however, you can see how this would
also be applicable to 3-dimensional space. To be clear, because the r position vector describes the
location of a single particle in three-dimensional space, the r center of mass position vector
describes the location of the center of mass of a system of particles in three-dimensional space.
Uh. I guess I get that, but, why does it matter? It matters Bo, because we can now do a
heck of a lot more by using this r-position of the center of mass of a system of particles equation.
For example, Bobby, please work through what we get when we take the derivative
of this equation with respect to time. Well, we know the derivative of position
with respect to time equals velocity. So, the derivative of the r-position of the center of mass
of a system of particles must equal the velocity of the center of mass of a system of particles.
But, what do we get when we take the derivative of that whole sigma term?
Uh. Mr.p, are we assuming the mass of the system does not change as a function of time?
(Yes Bobby, we are assuming the mass of each individual particle in the system does not change
as a function of time, so the mass of the system also does not change as a function of time.)
Okay, then we can take the total mass out from the derivative, and we can take the mass of the ith
particle out from the derivative and we are left with the derivative of the r-position of the ith
particle with respect to time. And that just equals the velocity of the ith particle.
Cool. The equation is actually just the same as the r-position of the center of mass of
the system of particles only we replace the r-position variable with a velocity variable.
Right. So now notice that we have a single equation
which describes the velocity of the center of mass of a system of particles. In other words,
we can have a system of particles where each particle is moving at a constant velocity,
however, we can describe the velocity of the entire system of particles using a single particle
instead and which is moving at the velocity of the center of mass of the system of particles.
Okay, that is kinda cool. We can actually go even further. If multiply
the velocity of the center of mass of the system of particles by the total mass of the system,
the total mass of the system cancels out, and we are left with the sum of the mass of the ith
particle times the velocity of the ith particle, which equals the linear momentum of the ith
particle. And when you sum the linear momentum of all of the particles, you get the total linear
momentum of the system of particles. In other words, the total linear momentum of a system
of particles is the same as the linear momentum of a single particle with a mass equal to the
total mass of the system which is moving at the velocity of the center of mass of the system.
Okay, that is cool. Yeah I bet we can
take the derivative of the velocity of the center of mass of the system of particles
with respect to time to get the acceleration of the center of mass of the system of particles.
Yep. And the total mass of the system is still constant, which means, just like with
the velocity, we can pull the total mass out from the derivative, and we get that the acceleration
of the center of mass of a system of particles equals the sum of the mass of the ith particle
times the acceleration of the ith particle all divided by the total mass of the system. Again,
just replacing velocity with acceleration in that equation. In other words, the acceleration
of the system of particles is the same as a single particle accelerating with the acceleration of the
center of mass of the system of particles. And we can multiply the equation for the
acceleration of the center of mass of the system of particles by the total mass of the system.
Total mass of the system cancels out again, and we are left with the sum
of the mass of the ith particle times the acceleration of the ith particle.
(And class, what does mass times acceleration equal?)
Net force! That’s Newton’s second law!
Exactly. Newtons second law applied to a system of particles instead of a single particle or
object. Now, the forces acting on any particle in the system will include both internal and
external forces. In other words, internal forces will occur when particles within the
system collide with one another, however, those forces will form a Newton’s third law force pair
and will therefore cancel one another out. That means the sum of the internal
forces acting on the system will always be zero and summing the internal and external
forces acting on the system will equal only the external forces acting on the system.
Therefore, when the net external force acting on a system equals zero,
as it does in this example, the acceleration of the center of mass of the system equals zero,
therefore, the velocity of the center of mass of the system will remain constant, and the linear
momentum of the system will also remain constant. So, when you look at this example, you can see
there are collisions occurring between particles in the system, however, there is no net external
force acting on the system, therefore the system continues moving with a constant linear momentum.
And that linear momentum is the same as the linear momentum of a single particle with a mass equal
to the total mass of the system moving at the velocity of the center of mass of the system.
That is correct Billy. And I will say that, again, the linear momentum of the system of particles
remains constant only when the net external force acting on the system of particles equals
zero. Thank you very much for learning with me today, I enjoy learning with you.
Does anyone else find it strange that he used 8 small spheres to have roughly the same mass
as 1 large sphere? Why strange?
I feel like there should be fewer than 8 small spheres to have the same mass as 1 large sphere,
assuming the same density for all of the spheres. Actually, because the large sphere has roughly
twice the radius of the small spheres, it does take 8 spheres to have the same mass.
Nah. Okay. Well,
we know they have the same volumetric mass density and volumetric mass density equals mass divided
by volume. That means mass equals density times volume. So, to have the same mass, because they
have the same density, the small spheres need to total the same volume as the volume of the single,
large sphere. Yeah.
And the radius of the large sphere is twice the radius of the small spheres.
The ratio of the volume of the large sphere to the volume of a single small sphere equals. Well,
we can substitute the volume of a sphere, or four thirds pi times radius cubed, in for each of those
volumes. four thirds pi cancels out. We can substitute 2 times the radius of a small sphere
in for the radius of the large sphere. 2 cubed is eight and the radius of the small sphere
cubed cancels out. In other words, the volume of a sphere with twice the radius and equal density
is 8 times larger than the volume of the smaller sphere. So, it takes 8 small spheres to have the
same mass as 1 large sphere. Oh.
Yep. Yeah.