### Video Transcript

In this lesson, we’re going to
learn how to relate the angular momentum of an object to the linear momentum of the
same object. We’ll start with the special case
of objects moving in circles and then expand those results to cover objects moving
in any direction.

First, though, let’s review linear
momentum. Say we have some object and it’s
moving in a particular direction. The rate at which the object is
moving is its speed, and we’ll use the letter 𝑣 to denote speed. If the object has a mass of 𝑚,
then the magnitude of the linear momentum of this object is the object’s mass times
its speed. In symbols, we write 𝑃 is equal to
𝑚 times 𝑣, where 𝑚 and 𝑣 are the mass and speed like we previously defined and
𝑃 is the symbol that we use for linear momentum. This only gives us the magnitude of
the linear momentum but not its direction. The direction of the linear
momentum is the same as the direction of motion. And we’ll need to consider this
direction when we calculate angular momentum.

Let’s consider an object moving
along a circular path represented by the dashed line. We’ll call the mass of the object
𝑚. And because we’ll need the radius
of the circle later, we’ll call it 𝑟. Since the object is moving in a
circle, whether it moves clockwise or counterclockwise, at any given moment, its
direction of motion is tangent to the circle. To finish up our description, we’ll
call the rate of motion 𝑣. And, considering the direction of
motion, we’ll call this the tangential speed. Furthermore, as the object moves
around the circle, even though its direction of motion changes, the size of the
tangential speed remains constant. We’re now almost ready to define
the angular momentum of this object relative to the center of the circle.

The last thing we need to observe
is that because the direction of motion is always tangent to the circle, it is also
always perpendicular to one of the radii of the circle. Keeping all this in mind, we define
the magnitude of the angular momentum of an object moving in a circle as the radius
of the circle times the linear momentum of the object or, equivalently, the radius
of the circle times the mass of the object times its tangential speed. To arrive at the second formula,
all we did was replace linear momentum with its definition as mass times linear
speed.

As things stand right now, these
formulas come with a few caveats. Firstly, these formulas only give
the size of the angular momentum. The angular momentum does have a
direction associated with it, but not one that can be found from our formulas as
we’ve written them. Secondly, in writing the formulas
this way, we’ve assumed that the direction of motion is perpendicular to 𝑟. This is fine for a circle because,
as we mentioned, 𝑟 is a radius of the circle. And since the direction of motion
is always tangent to the circle, it is indeed always perpendicular to one of the
radii. Finally, we’ll illustrate one last
thing we have to be careful about with angular momentum by way of a general
statement about circular motion.

An object moving in a circle with a
constant tangential speed has a fixed angular momentum about that circle center. Looking back at our formulas, if
tangential speed is constant, then linear momentum is constant. So angular momentum can only be
fixed if 𝑟 is also constant. But remember that for our
particular case, we find 𝑟 by measuring the distance between the center of the
circle and the object. But this is just the radius of the
circle, which is in fact always constant. Since we measure 𝑟 from the center
of the circle to the object, we see that this angular momentum is about the circle
center. And unless otherwise explicitly
stated, the angular momentum of an object moving in a circle is assumed to be about
the center of the circle.

However, as you’ll see shortly, we
can actually define the angular momentum of any object about any point. So when using our formulas, we know
that 𝑟 is measured to the object. But we have to be careful about
where 𝑟 is measured from. In technical language, we say that
𝑟 is measured from a reference point to the object. And in the case of circular motion,
the assumed reference point is the center of the circle. Let’s look at two illustrative
examples to get a better understanding of the dependence of angular momentum on 𝑟
and the reference point.

In this picture, we have two
identical objects. And rather than specify their mass
and tangential speed, we’ve simply specified that they have the same linear momentum
𝑃. Each of these objects is moving
along a circular path, one with radius 𝑟 one and one with radius 𝑟 two. And the two paths have the same
center point. Using our formula for the size of
the angular momentum reference to the center of a circle, we can write the angular
momentum of these two objects. 𝐿 one is the angular momentum of
the object at radius 𝑟 one, and its size is 𝑟 one times 𝑃.

Similarly, 𝐿 two, the size of the
angular momentum of the object at radius 𝑟 two, is 𝑟 two times 𝑃. If we call the center of the
circles 𝑥, we can specify that these are the angular momenta of the object about
𝑥. These results tell us the two
objects with the same linear momentum can have different angular momenta about a
point if their distances to that reference point are different. To get more insight into this
relationship, let’s imagine that we slid the circle of radius 𝑟 two over to the
left so that the two identical objects exactly coincide it.

In this new diagram, we have a
single object at a single position instead of two identical objects at different
positions. As a result, our two circular
trajectories now have different center points, whereas in our previous diagram, they
were concentric and shared a center point. To mark this difference, we’ll call
the center of the circle of radius 𝑟 one 𝑥 one and the center of the circle of
radius 𝑟 two 𝑥 two. Since these two diagrams are
physically similar, when we calculate angular momentum in our new diagram, we ought
to get similar answers to what we got in our previous diagram. And indeed we do. If the object is moving along the
smaller circular trajectory, then it’s an object of linear momentum 𝑃 at a distance
𝑟 one from the center of the circle. Our formula then tells us that the
size of the angular momentum is 𝑟 one times 𝑃 about 𝑥 one.

On the other hand, if the object is
following the larger circular trajectory, then it is an object of linear momentum 𝑃
at a distance 𝑟 two from the center of the circle 𝑥 two. Here then, the size of the angular
momentum would be 𝑟 two times 𝑃 about 𝑥 two. Look at the similarity between
these two pairs of results. In fact, the only difference is
that the first set of results applies to a single reference point and an object at
two different positions, while the second set of results applies to a single object
at a single position in space but with two different reference points. In coming up with this second set
of results, we assumed that the object was on one of two circular paths. This assumption is not actually
necessary, and we can easily see why by slightly redrawing the diagram.

We’ve redrawn the radii 𝑟 one and
𝑟 two so that we can explicitly see that 𝑃 is perpendicular to 𝑟 for both of our
reference points. Remember, our angular momentum
formula works under the assumption that 𝑃 and 𝑟 are perpendicular. Drawn this way, we can clearly see
that we have all of the necessary information, namely, linear momentum of the
object, distance to the reference point, and the fact that the direction of motion
of the object is perpendicular to the line connecting the reference point to the
object that we need to calculate the angular momentum. Furthermore, we still have all this
information even if we erased the two possibilities for the circular
trajectories.

Without the circles, we can also
see from this diagram that a single object with linear momentum 𝑃 has a different
angular momentum depending on the choice of reference point. This is, of course, why we keep
specifying that the angular momentum is about some particular point. Paying attention to reference
points is particularly important when we want to compare angular momenta. Furthermore, now that we’ve seen
that we can have a well-defined angular momentum for an object that doesn’t have an
explicitly circular trajectory, we can define angular momentum for any object with
any trajectory.

As we just saw, there are really
only two things we need to define the angular momentum of any object. We need to know the location of the
object relative to some reference point. And we need to know the size and
direction of the object’s linear momentum, in other words, its direction of motion
and its speed. There are then three possible
orientations for the direction of motion relative to the line connecting the
reference point and the object. The two could be perpendicular,
which will be almost exactly like the situation of circular motion that we’ve
already discussed. The two could be parallel, which is
a new situation that we’ll need to address. Or the two could be neither
perpendicular nor parallel.

This last situation requires a bit
more advanced mathematics than what we’re going to deal with here. Suffice to say that any neither
perpendicular nor parallel motion can be treated as a combination of two motions,
one of which is perpendicular and one of which is parallel. Once we’ve successfully split up
the motion into its parallel and perpendicular components, we treat the parallel
part with the parallel rules and the perpendicular part with the perpendicular rules
and then combine those two results. So since every motion is equivalent
to a combination of perpendicular and parallel motions, we’ll still get a full
picture by only considering these two cases.

A direction of motion perpendicular
to the line connecting the reference point and the object is exactly the
relationship we had for circular motion. To see this more clearly, we’ve
drawn part of a circular path. As we can see, even though we
haven’t specified a trajectory for the object, at this moment, its linear momentum
and position are the same as an object moving in a circle. Therefore, at this moment, it is
still effectively moving around 𝑥, and its angular momentum would be given by 𝐿
equals 𝑟 times 𝑃 just like for a circle. On the other hand, when the
direction of motion is parallel to the line connecting the reference point to the
object, the object only ever moves closer or farther from the reference point, but
never around it.

Drawing in a hypothetical circular
trajectory, we can see that the object’s motion is never along this path. So, because there is no motion
about the reference point, the angular momentum is simply zero. There is one important note that we
need about the validity of these two formulas. Except for the special case of
circular motion with constant speed, we can only be sure that these formulas will
give us the correct angular momentum for the moment in time at which we are
calculating them. For example, if an object is
speeding up or slowing down, the size of its linear momentum is changing. But we can still use the linear
momentum any particular instant to calculate the angular momentum at that
instant.

Lastly, it’s worth noting that
since the angular momentum is zero for parallel motion, it’s very easy to calculate
the angular momentum for a general direction of motion. This is because when we break up a
general direction of motion into its parallel and perpendicular components, the
contribution to the angular momentum from the parallel component will be zero. So the total angular momentum will
just be 𝑟 times the perpendicular component of the linear momentum. Where the somewhat advanced
mathematics comes in is in actually breaking up a general direction of motion into
its parallel and perpendicular components. All right, let’s now apply what
we’ve learned to some examples.

A 2.0-kilogram cast iron weight is
swung around in a circle at 1.5 meters per second at the end of a rope. The rope has a length of 1.2
meters. What is the magnitude of the
angular momentum of the iron weight?

Since this is a mechanics problem
about a weight moving in a circle, let’s draw a picture to organize our
information. Here is our label diagram. We have the cast iron weight
labeled with its mass of 2.0 kilograms. We’ve also labeled the speed of the
cast iron weight as 1.5 meters per second. We’ve arbitrarily chosen to draw
the direction of motion as clockwise because we aren’t told the direction in the
question. And anyways, we only want the
magnitude of the angular momentum. Finally, we also have the rope
labeled with its length of 1.2 meters. Notice that the rope forms the
radius of the circular path traced out by the weight. We’ve also written a label for the
unknown size of the angular momentum since that’s what we’re looking for.

Let’s now recall the formula for
the angular momentum of an object moving in a circle. We have for the angular momentum of
an object moving in a circle that the size of the angular momentum is equal to the
radius of the circle times the mass of the object times the tangential speed of the
object. Recognizing that mass times
tangential speed is the same as the linear momentum of the object, we can see that
this formula also agrees with the definition of angular momentum as the radius of
the circle times the linear momentum of the object. Remember, this formula gives the
angular momentum about the center of a circle. Since our question didn’t
explicitly state otherwise, we can assume that the angular momentum that it’s
looking for is referenced to the center of the circle.

Since we already have the mass of
the object, its tangential speed, and also the radius of the circle, all we need to
do is plug in values. We have that the angular momentum
that we’re looking for is 1.2 meters times 2.0 kilograms times 1.5 meters per
second. Collecting the numbers first, 1.2
times 2.0 times 1.5 is exactly 3.6. And now collecting units, meters
times kilograms times meters per second is kilograms meters squared per second,
which is the units of angular momentum. So the magnitude of the angular
momentum of the cast iron weight is 3.6 kilograms meters squared per second. This is the magnitude of the
angular momentum about the center of the circle since that’s the assumed reference
point for our problem.

Okay, let’s see another
example.

A car is being driven around a
circular road. The car has a mass of 2300
kilograms, and the road has a radius of 40 meters. If the car has an angular momentum
of 7.36 times 10 to the fifth kilogram meters squared per second, what is its
tangential speed?

This question centers around a car
on a circular road. We’re given several pieces of
information: the radius of the road, the mass of the car, as well as the car’s
angular momentum. We’re also asked to find the car’s
tangential speed. Let’s draw a diagram to help
organize all of this information. Here’s our diagram labeled with
values from the question. We have the circular road with its
radius of 40 meters. We have the car with its mass of
2300 kilograms. And we’ve also marked the angular
momentum of the car 7.36 times 10 to the fifth kilograms meters squared per
second. Finally, since the question doesn’t
specify, we’ve chosen to draw the direction of motion of the car as clockwise. And we’ve marked the unknown
tangential speed, which is what the question asks us to find.

Since we’re given the angular
momentum of the object and the presumed reference point is the center of the circle
since the question doesn’t specify otherwise, let’s write down the formula for the
angular momentum of an object moving in a circle. We have that the size of the
angular momentum about the center of the circle is equal to the radius of the circle
times the mass of the object times its tangential speed. To solve this expression for
tangential speed, we need to divide both sides by radius times mass. When we do this, on the left-hand
side, we just get 𝐿 divided by 𝑟 times 𝑚. On the right-hand side, 𝑟𝑚
divided by 𝑟𝑚 is just one, and we’re left with the tangential speed.

Before we plug in values to this
expression, note that 7.36 times 10 to the fifth is just 736000. So, taking the values that we’re
given, we have that the tangential speed is equal to 736000 kilograms meters squared
per second divided by 40 meters times 2300 kilograms. Simplifying units, the factor of
kilograms in the numerator cancels the kilograms in the denominator. And the factor of meters in the
denominator cancels one factor of meters in the numerator. We’re left with final units of
meters per second, which tells us we’re on the right track because meters per second
is a unit of speed. Okay, all that’s left to do is plug
736000 divided by 40 times 2300 into a calculator. This comes out to exactly
eight. And carrying over the units, we
have that the tangential or linear speed of the car is eight meters per second.

Okay, now that we’ve seen how to
calculate angular momentum from other quantities and how to calculate other
quantities from angular momentum, let’s review what we’ve learned in this
lesson. We started by recalling that the
size of the linear momentum of an object is the object’s mass times its linear speed
and the direction of the linear momentum is the same as the direction of the
object’s motion. We then considered the special case
of circular motion with an object of mass 𝑚 and momentum 𝑃 moving in a circle of
radius 𝑟. In this case, we called the linear
velocity of the object the tangential velocity because the direction of motion is
always tangent to the circle. With this in mind, we defined the
angular momentum of the object as the radius of the circle times the linear momentum
of the object.

Implicit in this definition is that
the reference point for the angular momentum is the center of the circle. This ensures that the line
connecting the reference point and the object is always perpendicular to the
direction of motion and also that the angular momentum of the object is
constant. We then tried to generalize angular
momentum to objects moving along any trajectory. Although we couldn’t guarantee that
the angular momentum would be constant in such a situation, we still saw that we
could calculate the angular momentum at any particular moment.

We considered two special cases:
when the direction of motion is perpendicular and when the direction of motion is
parallel to line connecting the reference point and the object. Although these are two very
specific cases, they’re sufficient for describing any direction of motion. For the perpendicular case, we saw
that at any given moment, the object looks like it’s moving around a circle of
radius 𝑟. So the angular momentum about the
reference point will be the same as for circular motion, 𝑟 times 𝑃, with the
caveat that this value is not necessarily constant in time. For the parallel case, the object
only moves towards or away from the reference point, but not around it. So in this case, the angular
momentum is simply zero.

Finally, but crucially, the angular
momentum of an object is defined with respect to a reference point. So if we have two reference points,
we can measure two different angular momenta of an object at a single instant in
time, one for each reference point. We can see this visually in the
diagrams we’ve already drawn. We have two similar objects with
similar directions of motion. However, one choice of reference
results in an angular momentum of 𝑟 times 𝑃, while a different choice of reference
result in an angular momentum of zero.