### Video Transcript

In this video, we’re going to learn
about rotational variables. We’ll see what these variables are,
how to calculate them, and how they relate to linear or translational variables.

To get started, imagine that you
work for a car tire manufacturing company. Your role is to track the wear and
tear on tires as they’re used. One day, you hear about a driving
competition, where drivers outfitted in cars with tires made by your company will
drive once all the way around the world. From your work study in tire wear,
you have a rough idea of how much a tire wears down each time it rotates. To use this information to figure
out how much the tires on the cars that drived once around the world will wear down,
we want to know something about rotational variables.

To get an idea for rotational
motion, imagine that we have a large wheel that rotates about an axis through its
centre and say further that we put a mark on the outer edge of the wheel. Over time, as the wheel rotated,
this mark would travel a linear distance—an arc length—along the outer edge of the
wheel we can call 𝑠. 𝑠 is a linear distance measured in
linear units such as metres. What’s interesting is we also have
a rotational displacement as this mark on the edge of the wheel moves around. We can say that rotationally we’ve
travelled a distance 𝜃, where because 𝜃 is an angle, it’s measured in units of
radians.

We know that in a complete circle,
there are two 𝜋 radians. But that’s not an upper limit on
the rotational distance this mark might travel no matter how many times the mark
moved around in a circle. With a starting angle, we can call
it 𝜃 sub 𝑖, and an ending angle, we can call it 𝜃 sub 𝑓, we’re able to calculate
the rotational distance travelled no matter how many times the wheel turns. So the arc length 𝑠 measured in
metres and the angular distance 𝜃 measured in radians are analogues one of
another. In the linear world, we could call
it, we talk about 𝑠 and in the rotational world, we talk about 𝜃.

Now, imagine that instead of just
wanting to know the distance travelled by the mark, we want to know its
velocity. Say at the point in time where
we’ve drawn it last. We know that velocity will point
tangent to the edge of the wheel and that in SI units is measured in metres per
second. We also know though that in
whenever time it takes for the mark to move from its initial to its final position,
not only has the dot moved a linear distance, but it’s moved through an angular
distance. And we can call that angular
distance in some amount of time—that is a value in radians per second—the angular
velocity of the wheel. We typically represent this
variable with the Greek letter lowercase 𝜔. Just like 𝜃 and 𝑠, 𝜔 and 𝑣 are
analogues.

We can imagine further that not
only do we wanna know the velocity of the mark, but we also want to know its linear
acceleration. We know that acceleration 𝑎 is the
time rate of change of velocity 𝑣. When we want to solve for 𝑎, we
take the time derivative of 𝑣 and the units of these two values bear this out. If we do the same thing with the
units of angular velocity, we get a result in units of radians per second
squared. That may sound like an angular
acceleration and that’s exactly what it is—often represented by the Greek letter
𝛼. For every linear motion term then,
there is an analogous rotational term. Let’s consider further how these
pairs of terms correspond.

We know that linear velocity is
equal to change in position over change in time. It turns out we can write the
angular velocity 𝜔 in a similar way. It’s equal to the change in the
angular position Δ𝜃 over a change in time. And just as linear acceleration 𝑎
is equal to Δ𝑣 over Δ𝑡, so the angular acceleration 𝛼 is equal to Δ𝜔 over
Δ𝑡. We can go even further with
this. Remember the kinematic equations of
motion. These are equations which help
describe object motion when acceleration 𝑎 is constant. They let us solve for object motion
variables such as displacement or initial velocity. It turns out there are an analogous
set of rotational kinematic equations.

These four equations are written
exactly like their linear counterparts with the rotational variables substituted in
for the linear ones. And just like the linear kinematic
equations, the rotational ones require constant acceleration—in this case, angular
acceleration. Now that we have an idea of where
rotational variables come from and how they’re related to their linear
correspondents, let’s get some practice working with them through a couple of
examples.

A piece of dust falls onto a
spinning compact disc and sticks in place. The spin rate of the disk is 5.00
times 10 to the two rpm and the piece of dust lands 4.30 centimetres from the disk’s
centre. What is the total distance
travelled by the piece of dust in 1.80 times 10 to the two seconds of rotation? Assume that the disk is negligibly
slowed by the dust falling on it and that the time taken for the dust to accelerate
to the same speed as the part of the disk that it lands on is negligible.

We’ll call this total distance
travelled by the piece of dust lowercase 𝑠. In this scenario, we’re told that a
piece of dust falls at distance—we’ve called 𝑟—of 4.30 centimetres from the centre
of a rotating disc. The disc is spinning at an angular
speed—we’ve called capital Ω—5.00 times 10 to the two revolutions every minute. After the dust has been on the
rotating disk for a time of 1.80 times 10 to the two seconds, we want to know the
linear distance 𝑠 it has travelled. The linear distance 𝑠 will equal
the number of times the disc rotates in this amount of time multiplied by two times
𝜋 times 𝑟.

We’re given the distance 𝑟 from
the centre of rotation at which the dust settles. But we don’t yet know how many
times the disc rotates in our amount of time 𝑡. To figure that out, we can use our
angular rotation rate capital Ω. To make capital Ω as useful as
possible to us, we’ll convert it from its given units of revolutions per minute to
revolutions per second. To do that, we can multiply capital
Ω by one minute per 60 seconds, which is effectively multiplying it by one. But what it does do is it cancels
out the units of minutes and leaves us with time units of seconds. The total number of rotations our
disk will make in the time 𝑡 is equal to that time multiplied by the rotational
rate.

When we plug in our given value for
𝑡, we see that the units of seconds cancel out and the units of our number of
rotations will indeed be in revolutions. When we plug these numbers in for
𝑁 rotation in our equation for 𝑠 and then insert the value of our radius 𝑟
converting it from units of centimetres to units of metres, we see looking at this
overall expression that the units of revolutions cancel out. And we’ll be left with an answer in
units of metres. When we calculate 𝑠, we find that
to three significant figures, it’s 405 metres. That’s the distance this piece of
dust would travel in the amount of time 𝑡.

Now, let’s try a second example
involving rotational variables.

A phonograph turntable rotating at
33 and a third rpm slows down and stops in 1.00 minutes. What is the turntable’s angular
acceleration assuming it is constant? How many complete revolutions does
the turntable make while stopping?

We’ll call the turntable’s angular
acceleration 𝛼 and the number of complete revolutions it makes while stopping
𝑁. In this scenario, we have a
rotating turntable which initially has an angular speed we’ve called capital Ω, 33
to third revolutions per minute. Over a time span of 1.00 minutes,
it slows to a stop. Since we’re told that its angular
acceleration as it does so is constant, that means that the kinematic equations of
motion apply to the motion of the turntable. In particular, we’re helped by the
equation which says that 𝜔 sub 𝑓, the final angular speed, is equal to 𝜔 sub
zero, the initial angular speed, plus 𝛼, the angular acceleration, times time.

In this expression, our angular
speeds are written in units of radians per second. When we apply this relationship to
our scenario, we know that because the turntable ends up at rest, 𝜔 sub 𝑓 is
zero. And we want now to solve for 𝜔 sub
zero, the initial angular speed of our disc, in units of radians per second. To do that, let’s start by writing
down capital Ω. If we multiply capital Ω by the
conversion factor two 𝜋 radians are in every revolution, then we’ve effectively
changed the units of this expression now to radians per minute. To get the units to radians per
second, we can multiply by the conversion factor one minute is in every 60 seconds,
cancelling out the units of minutes. We now have our expression for 𝜔
sub zero. It’s 33 and a third times two 𝜋
over 60 radians per second.

With that value known, we can now
rearrange our kinematic equation to solve for the angular acceleration 𝛼. 𝛼 is equal to negative 𝜔 sub zero
over 𝑡. Plugging in these values, we see 𝛼
is negative 33.33 repeating 𝜋 over 30 times one over 60 in units of radians per
second squared. This is equal to negative 0.058
radians per second squared. That’s the angular acceleration of
the disk as it slows down.

Next, we wanna solve for the number
of complete rotations the phonograph goes through as it slows to a stop. Considering what we know: we know
the initial angular speed of the phonograph. We know its angular
acceleration. We know its final angular speed
zero. And we essentially wanna calculate
an angular distance that is travelled. We can begin solving for 𝑁 by
working from a kinematic equation that lets us solve for the angular displacement
𝜃. The total angular displacement of
the phonograph as it comes to a stop is 𝜃. And if we divide 𝜃 by two 𝜋
radians, then that will equal a total number of revolutions 𝑁 we complete.

Since we’ve solved for 𝜔 sub zero
and 𝛼 and we’re given 𝑡 in the problem statement, we’re ready to plug in and solve
for 𝑁. When we have our values for 𝜔 sub
zero and 𝛼 and 𝑡 in units of seconds plugged into this equation, notice as we go
from left to right that the units of seconds cancel out in our first term. The units of seconds squared cancel
out in our second term. And because we’re dividing all of
this by two 𝜋 radians, the units of radians also cancel out, leaving us with a
unitless result. When we enter this expression on
our calculator, we get a result of approximately 16.7. But since 𝑁 is the number of
complete revolutions the phonograph makes before it comes to a stop, that means that
𝑁 is 16. And so, we’ve used the rotational
kinematic equations to solve for the number of times the phonograph spins before
coming to a stop.

Let’s summarize what we’ve learned
about rotational variables. We’ve seen that rotational
variables exist for each linear motion variable. For a linear displacement 𝑠,
there’s angular displacement 𝜃. For linear velocity 𝑣, there’s
angular velocity 𝜔. And for linear acceleration 𝑎,
there’s angular acceleration 𝛼. We’ve also seen that the
mathematical relationships between linear variables match those between rotational
variables. For example, just as linear
velocity equals change in displacement over change in time, so angular velocity
equals change in angle Δ𝜃 over Δ𝑡. And finally, we saw that the
kinematic equations of motion apply for both linear and rotational variables, so
long as acceleration whether linear or rotational is constant.