### Video Transcript

Stokes’ law describes sedimentation
of particles in liquids and can be used to measure viscosity. Particles in liquids achieve
terminal velocity quickly. One can measure the time it takes
for a particle to fall a certain distance and then use Stokes’ law to calculate the
viscosity of the liquid. Suppose a steel ball bearing —
density 7.8 times 10 to the third kilograms per meter cubed, diameter 3.0
millimeters — is dropped in a container of motor oil. It takes 12 seconds to fall a
distance of 0.60 meters. Calculate the viscosity of the
oil.

As we solve this problem, we’re
gonna make a few assumptions. One, based on the fact that we know
particles in liquids achieve terminal velocity quickly, we’re going to assume that,
for the entire length of this steel ball bearing’s journey, it is moving at terminal
velocity. Second, we’re going to assume that
𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared.

Let’s take a moment now to
highlight some of the critical information given in this problem statement. First, we’re told the density of
the steel ball bearing, 7.8 times 10 to the third kilograms per meter cubed, and
we’re also told the ball bearing’s diameter, 3.0 millimeters.

Further, in 12 seconds of time, the
ball bearing drops a distance of 0.60 meters. We’re on the lookout to solve for
the viscosity of the oil. Now let’s begin by drawing a
diagram of the situation.

We’re told that we have a container
filled up with motor oil and that into that motor oil we drop a steel ball bearing
and that, over time, this ball bearing slowly drops through the oil a certain
distance. Now we’re told that the time the
ball bearing takes to drop is 12 seconds — we’ll call that 𝑡 sub drop — and that,
in that time, the ball bearing drops a distance of 0.60 meters — we’ll call that
𝑑.

We’re given some more information
about the steel ball bearing. We’re told its density, which we’ll
call 𝜌, is 7.8 times 10 to the third kilograms per meter cubed. And we’re also told that the
diameter of the ball bearing is 3.0 millimeters. We’ll symbolically represent that
term as dia to distinguish it from the distance of the drop, so dia is equal to 3.0
millimeters. And if we convert this value to
meters, that’s equal to 3.0 times 10 to the negative third meters.

So this is our given
information. And what we’re after is the
viscosity of the motor oil. We represent that viscosity using
the Greek letter called 𝜂, and it looks like this.

Now we’re told in this problem that
we can use Stokes’ law to help us calculate this viscosity of the oil, so let’s
recall what that law says. This law lets us solve for the drag
force represented by 𝐹 on a sphere as it travels through a fluid like, in our case,
motor oil.

The terms involved in this equation
are 𝜋, a constant; 𝑟, the radius of the sphere that’s moving through the fluid;
𝜂, the viscosity of the fluid; and 𝑣, the velocity of the sphere or ball as it
moves through the fluid.

Since we now we want to solve for
𝜂, let’s apply Stokes’ law to our scenario by rearranging for that term. To do that, let’s divide both sides
of our equation by six times 𝜋 times 𝑟 times 𝑣.

When we do that, all four of those
terms cancel out on the right side, leaving us with 𝜂, the fluid viscosity, by
itself. We can thus rewrite this equation
so that it reads 𝜂 equals force divided by six 𝜋𝑟𝑣.

Let’s begin to solve for the terms
on the right side of this equation, beginning with the drag force 𝐹. We’ll rely on one of the
assumptions we made at the outset of this problem. We said that we’re going to assume
that the steel ball bearing reaches terminal velocity instantly when it reaches the
motor oil.

So if we draw a free-body diagram
of the ball bearing, we know there is the force of gravity acting down on the ball
bearing — we’ll call that 𝐹 sub 𝑔 — and there’s also the drag force pushing up on
the ball bearing, and we’ve called that 𝐹.

Now since the ball bearing under
our assumption is moving at its maximum velocity and indeed that velocity is staying
constant, that means that the ball bearing under these force conditions is not
accelerating, or that its acceleration is zero.

As we recall Newton’s second law
and how it connects force, mass, and acceleration, we realize that, in our case, if
we say that motion up is positive, then Newton’s second law, as it relates to the
motion of the ball bearing, could be written 𝐹 minus 𝐹 sub 𝑔 equals zero.

And again, we say that these forces
combined that way equal zero because of our assumption of terminal velocity. So looking at this equation, if we
add 𝐹 sub 𝑔 to both sides, we’ll realize that 𝐹, the drag force, is equal to 𝐹
sub 𝑔.

That’s good news because we can
solve for 𝐹 sub 𝑔, the force that gravity exerts on the ball bearing. That force is equal to the mass of
the ball bearing times 𝑔, the acceleration due to gravity.

So what is the mass of the ball
bearing? Let’s recall the relationship
between density, mass, and volume.

We know that the density of an
object is defined as its mass divided by its volume. So the mass of the ball bearing 𝑚
is equal to the density of the ball bearing times its volume.

Now we’ve been given that density
value 𝜌 and we also know the diameter of the ball bearing, which would let us solve
for the volume of the ball bearing. We recall that the ball bearing is
a sphere and therefore its volume 𝑣 is equal to four-thirds times 𝜋 times the
radius of the sphere cube.

So the volume of our ball bearing
equals four-thirds 𝜋 times half the diameter cubed.

When we plug in our diameter value
of 3.0 times 10 to the negative third meters, then we have an expression for the
volume of our sphere. Now we take this volume and
multiply it by 𝜌, the density of our ball bearing, to get the mass of the ball
bearing.

Before we enter these numbers into
our calculator, take a look at the units and notice that the meters cubed term
cancels out of our final answer, leaving us with units of kilograms.

That’s a good sign because we’re
solving for mass, whose base unit is kilograms. When we type all these numbers in,
we end up with a mass of our ball bearing of 1.10 times 10 to the negative fourth
kilograms.

We can now take this mass value;
insert it into our equation for 𝐹, the drag force; and solve for that force. So 𝐹, the drag force acting on the
ball bearing, is equal to the mass of the ball bearing times the acceleration due to
gravity 𝑔.

We plug in our mass value and then
insert 9.8 meters per second squared for 𝑔. And we find that the drag force
acting on our steel ball bearing is equal to 1.08 times 10 to the negative third
newtons.

Now that we have this value, we’re
able to revisit our equation for 𝜂, the viscosity of the fluid that we’re trying to
solve for. Now we have all the components on
the right side of this equation and can move towards solving for that viscosity.

So let’s begin inserting the
numerical values represented by the symbols on the right-hand side of this equation
for 𝜂. We know the drag force which we
just solved for, 1.08 times 10 to the negative third newtons, and that force is
divided by six times 𝜋 times the radius of the ball bearing, which is half the
diameter multiplied by 𝑣, the speed that the ball bearing moves through the
oil.

We’re told the diameter of the ball
bearing, 3.0 times 10 to the negative third meters, so let’s plug that in to our
radius term. And now all we have left to solve
for is 𝑣.

When it comes to that speed, we are
told 𝑣, but we are told the time it takes for the ball bearing to drop — we’ve
called that 𝑡 sub drop — and we’re also told the distance that it moves over that
time.

Remembering that speed, which we
can represent by 𝑣, is equal to distance divided by time, we can replace our symbol
𝑣 with the distance 0.60 meters divided by the time to drop of 12 seconds. That distance divided by that time
is equal to the speed of the ball bearing, which we assume is constant throughout
its descent.

We now have all the ingredients
we’ve needed to calculate 𝜂, the viscosity of the fluid. When we plug all these numbers into
our calculator, we find a result of 0.76 kilograms per meter times second.

That is the viscosity this fluid
must have for this type of ball bearing to behave the way it’s been described.