Video: Angular Momentum

In this video we learn what angular momentum is and how to express it in linear and rotational variables, as well as the rotational version of the Impulse-Momentum Theorem.

15:01

Video Transcript

In this video, we’re going to learn about angular momentum. We’ll learn what this term means, how it relates to linear momentum, and how to work with it practically.

To get started, imagine that you’re the owner of a wind farm which uses wind power generators to make electricity. Today is the day of the company’s picnic. And since all the employees will be at the picnic with no one able to monitor and attend to the wind farm, per safety regulations you’ll need to shut the farm down while it’s unattended. With the wind turbines currently rotating as they usually do in the wind, you’d like to know just how hard it will be to bring them to rest. To figure this out, it will be helpful to know something about angular momentum.

To get started talking about angular momentum, let’s imagine a planetary orbit scenario where a smaller planet orbits a larger one in a circular path. If we were to write down some of what we know about this scenario, we know that this smaller planet has some mass we can call 𝑚. And we also know that at the instant we’ve shown it, it has a tangential velocity we can call 𝑣. Considering just these two terms, we see that if we combine them, we can form the linear momentum of this rotating planet.

We’ve seen before that linear momentum 𝑝 is equal to an object’s mass times its velocity. Since the orbiting planet is also moving in a circular path, we can draw a vector that goes from the point around which it moves to the planet itself. And we’ll call that vector the distance vector 𝑟.

By adding in this distance vector, which, we can see, will change as the position of the planet in its circular orbit changes, we’ve introduced an angular, or rotational, aspect to the motion of this orbiting planet. After all, if we draw in the position of the planet at some later time, we can see that’s it’s moved through an angular distance we can call 𝜃 during that time.

When we combine these terms we’ve used to get linear momentum with the distance of vector 𝑟, that’s when we arrive at an angular momentum. Angular momentum is symbolized by a capital letter 𝐿, and it’s a vector quantity with both magnitude and direction. Angular momentum is equal to a cross product of our distance vector 𝑟 — remember, this vector points from the point of rotation to the object that is doing the rotating — and the linear momentum of that rotating object 𝑝.

Since linear momentum 𝑝 is equal to 𝑚 times 𝑣, we can also write our angular momentum as 𝑟 cross 𝑚 times 𝑣, or since mass is a scalar quantity 𝑚 times the quality 𝑟 cross linear velocity 𝑣. With this new quantity we’re calling angular momentum, we’d like to know what’s useful about it or why we’d be interested in an object’s angular momentum.

The great thing about angular momentum is, as we’ll see in a later segment, it’s a conserved quantity just like linear momentum 𝑝. That means that over any physical interaction or process, we know that the angular momentum at the beginning is the same as the angular momentum at the end. We’ve said that angular momentum has to do with rotation. And in fact, we can write the expression for angular momentum in terms of rotational variables.

Moving from speaking of vectors to speaking of scalars, we know that linear speed 𝑣 is equal to the radial distance an object moves from its axis of rotation multiplied by its angular speed 𝜔. If we assume that the vectors 𝑟 and 𝑣 are perpendicular to one another, then we can write that the magnitude of 𝐿 is equal to 𝑚 times 𝑟 times 𝑣 using simple multiplication. Then, we can substitute 𝑟𝜔 in for 𝑣 so that now 𝐿 is equal to 𝑚𝑟 squared 𝜔.

And if we look at 𝑚 times 𝑟 squared, we may recall that this is the expression for the moment of inertia of a point mass. Under the condition of working with a point mass then, where its velocity and radial distance are perpendicular to one another, we can say that the magnitude of angular momentum is equal to moment of inertia times 𝜔.

Let’s take a moment at this point and consider 𝜔 as a vector quantity, that is as angular velocity. Over here on our diagram of the orbiting planet, if we were to draw in 𝜔 the angular speed of the planet, we might do it with a curving arrow like this to show that the planet is moving in a counterclockwise direction in a circle. Here is something that’s interesting though. If we consider this 𝜔 as a vector, that is as an angular velocity, then the actual direction of 𝜔 under these circumstances is out of the page at us.

That might seem strange like it contradicts the direction of the arrow that we drew showing the planet’s motion direction. But going back to our convention we’ve sometimes used before of a right-hand rule, if we rotate our fingers in the direction of the planet’s motion, that is, in some sense, the direction of its angular position, then our thumb points in the direction of 𝜔, the angular velocity. That’s why this counterclockwise rotating planet has an angular velocity that points out of the page at us.

The reason we say all this is because we can actually go back to our equation for angular momentum in terms of rotational variables, and we can write in 𝜔 as a vector and 𝐿 as a vector, and this equation is correct as written. That’s to say, the direction of an object’s angular momentum is the same as the direction of its angular velocity, just like we saw for linear momentum 𝑝 matching up with the direction of linear velocity 𝑣.

Let’s consider one more thing about angular momentum. What if we were interested not in the angular momentum at a point in time but in a change in angular momentum Δ𝐿. To figure out what this quantity might be, let’s go back to the linear analog of angular momentum, that is linear momentum 𝑝.

We can recall that there’s a theorem with the long name, the impulse-momentum theorem. This theorem says that the change in an object’s momentum Δ𝑝 is equal to the net force acting on that object times the time over which this average force acts. Now knowing that linear variables such as these typically have corresponding angular or rotational variables, how would we rewrite the impulse-momentum theorem for angular momentum rather than linear momentum?

We would replace Δ𝑝 with Δ𝐿, and that would equal not force but angular force, which is torque, multiplied by Δ𝑡. Time is the same regardless of whether we’re talking about rotational or linear motion. This relationship, which we could call the rotational analogue of the impulse-momentum theorem, is an accurate statement of changes in angular momentum Δ𝐿. Realize that the torque in this expression 𝜏 is a net average torque. We assume that it’s constant over the time Δ𝑡 that it acts.

Now that we have a sense for what angular momentum is in terms of linear and rotational variables and we also have a feel for how it changes and what it depends on, let’s get some practice with these ideas through an example.

Three particles move independently of each other, and each particle is subject to a force perpendicular to the direction of its instantaneous velocity. What is the total angular momentum about the origin of the particles? What is the rate of change of the total angular momentum about the origin of the particles?

We can call the total angular momentum of the particles capital 𝐿 and the rate of change of that total Δ𝐿 over Δ𝑡. To start off solving for 𝐿, the total angular momentum of our three particles 𝑚 one, 𝑚 two, and 𝑚 three about the origin, we can recall the mathematical definition for angular momentum. 𝐿 as a vector is equal to its position vector crossed with its momentum. And knowing that 𝑝 momentum is equal to mass times velocity, we can also write this as 𝑚 times the quantity 𝑟 cross 𝑣.

In our situation, we have three masses at different locations, and, therefore, each one has its own different position vector. We can write out what those position vectors are by observing our mass positions relative to the origin. We see that 𝑚 one is at a position of negative two meters on the 𝑥-axis and positive one meter on the 𝑦. So, we write its position vector as negative two 𝑖 plus 𝑗 meters.

The position of mass 𝑚 two relative to the origin is plus four meters in the 𝑥-direction and plus one meter in the 𝑦. So, it’s position vector is four 𝑖 plus 𝑗 meters. And 𝑚 three is at a position of positive two meters in the 𝑥-direction and negative two meters in the 𝑦-direction relative to the origin. So, we can write its position vector, 𝑟 three, as two 𝑖 minus two 𝑗 meters.

Since we’re given the velocities and masses of each one of our three masses, we now have all the information we need to calculate the angular momentum of each one. Since angular momentum 𝐿 is a vector and we’ll take a cross product to compute that vector, let’s remind ourselves of the rule for crossing two different vectors, we’ll call them generically 𝐴 and 𝐵, and assume they’re three-dimensional.

Given two three-dimensional vectors 𝐴 and 𝐵, their cross product is equal to the determinant of the three-by-three matrix where the first row of the matrix is the unit vectors 𝑖, 𝑗, and 𝑘 of our three-dimensional space. The last two rows are the 𝑖, 𝑗 and 𝑘 components of these two vectors 𝐴 and 𝐵 respectively. We’ll use this rule to calculate the angular momentum of mass one, we’ll call it 𝐿 one, and also the same for mass two, we’ll call it 𝐿 two, and the same thing for mass three. Then, we’ll add together those three angular momenta to solve for 𝐿, the total.

Starting with 𝐿 one, that’s equal to 𝑚 one times 𝑟 one cross 𝑣 one. We know 𝑚 one, the mass, and 𝑣 one, the velocity, from our diagram. And we’ve solved for 𝑟 one, the position vector of that mass relative to the origin. Plugging in for our value 𝑚 one and following our cross-product rule for setting up the relationship between 𝑟 one and 𝑣 one, we know that because 𝑟 one and 𝑣 one only have 𝑥- and 𝑦-components, that means the direction of their cross product will only have a 𝑧-component. Therefore, 𝐿 one will only have a 𝑘-hat direction.

This means that when we evaluate this matrix, we only need to evaluate the 𝑘th component because the 𝑖- and 𝑗-components will be zero. When we make that evaluation, we take negative two meters and multiply it by 4.0 meters per second and subtract from that one meters times zero, which is zero. Overall then, 𝐿 one is 2.0 kilograms times negative 8.0 meters squared per second in the 𝑘-direction, or negative 16 kilograms meter squared per second in the 𝑘-direction.

Next, we move to calculating 𝐿 two, the same value but for 𝑚 two, 𝑟 two, and 𝑣 two. Again calculating only the 𝑘-component because that’s the only nonzero component to 𝐿 two, we find it’s equal to 4.0 kilograms times negative 5.0 meters squared per second in the 𝑘-direction, or negative 20 kilograms meter squared per second 𝑘. Then, we do the same thing for 𝐿 three with 𝑚 three, 𝑟 three, and 𝑣 three. This vector comes out to 6.0 kilograms meter squared per second again in the 𝑘-direction.

Now that we’ve solved for 𝐿 three, 𝐿 two, and 𝐿 one, we’re able to add all these three together and solve for the overall angular momentum 𝐿. This is equal to negative 16 minus 20 plus 6.0 kilograms meter squared per second in the 𝑘-direction, or negative 30 kilograms meter squared per second in the 𝑘-direction. That’s the total angular momentum of this system of masses about the origin.

Next, we wanna solve for the time rate of change of that total angular momentum. We recall from the rotational version of the impulse-momentum theorem that the change in angular momentum 𝐿 is equal to the total average torque 𝜏 multiplied by Δ𝑡. This tells us that Δ𝐿 over Δ𝑡 is equal to the total torque 𝜏. Each one of the three masses experiences a torque. So, to solve for the overall torque 𝜏, we’ll need to solve for the individual torques and sum them.

Since torque is equal to 𝑟 cross 𝐹, and we’ve already solved for the position vectors of our three masses and are given their forces, we can use the same cross product rule to solve for the overall torque. Just like with our angular momentum, since all of our position vectors and all of the forces acting on our particles have only 𝑥- and 𝑦-components, that means all of our torques 𝜏 one, 𝜏 two, and 𝜏 three will only have 𝑘-components. And that means we only need to evaluate the 𝑘-component of our three-by-three matrix.

In the case of 𝜏 one, plugging in for 𝑟 one and 𝐹 one by component, this is equal to 6.0 newton meters in the 𝑘-direction. Moving onto 𝜏 two, the torque on particle two, this is equal to the cross product of 𝑟 two and 𝐹 two, or 40 newton meters in the 𝑘-direction. And lastly, we calculate 𝜏 three, plugging in for 𝑟 three and 𝐹 three in our matrix. This gives us a result of negative 16 newton meters in the 𝑘-direction.

With values now for 𝜏 three, 𝜏 two, and 𝜏 one, we can add these together to solve for the overall torque 𝜏. This is equal to 6.0 plus 40 minus 16 newton meters in the 𝑘-hat direction, or 30 newton meters in the 𝑘-direction. And since this is the overall torque acting on our system of masses, by the impulse-momentum theorem, it’s also equal to the time rate of change of angular momentum.

Let’s summarize what we’ve learned about angular momentum. We’ve seen that angular momentum is the vector product of an object’s position vector with its linear momentum. As an equation, 𝐿 is equal to 𝑟 cross 𝑝. We’ve also seen that angular momentum can be written in terms of rotational variables, that it’s equal to an object’s moment of inertia times its angular velocity. And finally, we’ve seen there is a rotational version of the impulse-momentum theorem. It says the change in an object angular momentum equals the torque acting on that object times the time over which it acts.

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