# Lesson Video: Angular Momentum in Terms of Linear Momentum Physics

In this video, we will learn how to calculate the angular momentum of an object that is moving along a circular path from its instantaneous linear momentum.

16:44

### 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.