Lesson Video: Angular Impulse Physics • 9th Grade

In this video, we will learn how to calculate the angular impulse given to an object when a torque is applied to it for a period of time.

15:32

Video Transcript

In this video, our topic is angular impulse. We’re going to learn what this term means as well as how to calculate angular impulse using an equation.

We can begin our discussion by looking at Newton’s second law of motion. And specifically, this is what we could call the linear form of this law. 𝐹 equals 𝑚 times 𝑎. 𝑎, the acceleration, is defined as a change in velocity over a change in time, which means we can take this fraction and substitute it in for the 𝑎 in Newton’s second law. And then, notice this: we have the object’s mass that we’re considering multiplied by its change in velocity. That product, 𝑚 times Δ𝑣, is equal to the object’s change in linear momentum. And this variable on the left, Δ𝑝, has a particular name. We call it the impulse.

We can remind ourselves of what impulse means physically by taking Δ𝑝 and substituting it in for 𝑚Δ𝑣 in our Newton’s second law equation. We now see that 𝐹 is equal to Δ𝑝 divided by Δ𝑡. And this implies that the change in an object’s momentum, Δ𝑝, also called its impulse, is equal to the force it experiences times the time over which it experiences that force. As an example of impulse, imagine that we have a rocket ship moving up through Earth’s atmosphere. If we burned fuel at a faster rate using the engines, then we would provide a greater thrust force. In other words, 𝐹 would go up in our impulse equation.

Applying this thrust force over some amount of time, Δ𝑡, would give the spaceship an impulse, which we can see increases as Δ𝑡 goes up, in other words, as we burn fuel at the same rate over a longer period. And it would also go up as the thrust force, 𝐹, increases. And then, of course, we could diminish Δ𝑡 and 𝐹 as well. We could burn fuel for a shorter period of time, thereby diminishing Δ𝑡, or we could burn it at a lower rate diminishing the thrust force, 𝐹. Either one of these changes would decrease the impulse, Δ𝑝, delivered to the space craft.

Now, so far, all the variables we’ve been talking about are linear variables. That is, they apply to motion in a straight line. But we know that there could be a connection between motion in a straight line and rotational motion. For example, say that we have a mass 𝑚 which is moving at a linear speed 𝑣, but it’s also moving in a circular arc around this circle here. This means that not only does this mass have a linear speed 𝑣 at any point along its path, but it also has an angular speed we call it 𝜔. And we can recall that the two speeds are related like this: 𝑣 is equal to 𝑟, the circle’s radius, times 𝜔.

The point, though, is that linear speed, 𝑣, has what we could call an angular or a rotational analog in angular speed. And speed isn’t the only variable where there is both a linear and a rotational version of that variable. For example, this symbol 𝑝 represents linear momentum, but we know there’s also such a thing as angular momentum, capital 𝐿. And likewise, the symbol capital 𝐹 refers to a force, specifically a linear force. But then, if we apply a force some distance away from an object’s rotation axis, we know that creates a torque. And if we multiply a torque by the time interval over which that torque acts, then that product is equal to the change in angular momentum an object experiences.

What we find then is that this whole equation for the impulse experienced by an object has both a linear and a rotational version. And just as we saw earlier that Δ𝑝 is referred to as impulse, so Δ𝐿 is sometimes called angular impulse. Now, we say “sometimes” because the typical symbol for angular impulse is actually this, an uppercase 𝐻. Oftentimes then, it’s this equation in the box that we will see, but it’s helpful to recall that 𝐻 is simply Δ𝐿, the angular impulse applied to some object.

Now, we’ve seen what a linear impulse might involve, but what about an angular one? If we think back to the situation on our opening screen, there we had a door that was slightly ajar. And here, we’re seeing that door from an aerial view. The plan was to throw a ball against the door like this so that as it bounced back, it would apply a force on the door. But then, notice that that force is being applied some distance away from the axis of rotation of the door, that is, the door hinge, which means that the ball would apply a torque to the door. That’s the symbol 𝜏 in our equation for angular impulse. And it would do this over some amount of time Δ𝑡. This torque, applied while the ball was in contact with the door, would give it an angular impulse. And if this impulse was great enough, it would cause the door to shut.

Now, not only can we calculate angular impulse using an equation, we can also do it working from a graph. Say, for example, that we had a graph like this, with torque in newton meters on the vertical axis and time in seconds on the horizontal. And say further that our graph was marked out like this and that we had some physical process taking place where the torque applied to an object over time looked like this. So, we see that at the origin, this torque starts out at zero but then climbs steadily up to four newton meters over the first two seconds and then over the next two seconds stays constant at that value and then drops down to zero.

Now, here’s our question. Based on this graph, what is the overall angular impulse, 𝐻, delivered to whatever object we’re considering? If we look back to our equation for angular impulse, we see that it’s equal to torque multiplied by a change in time. This means that on our graph, to solve for angular impulse, we’ll want to multiply each torque value by the corresponding time interval over which that torque is exerted. And then, we’ll take each one of those products, torque times time, and add them all together to get the total angular impulse delivered, 𝐻. So we’re multiplying torque by time, and what that does on our graph is it calculates this area here. We sometimes call it the area under the curve, that is, the area between our curve and the horizontal axis.

Now, one way to make calculating this total area easier is to break it up into two pieces, one piece is this triangle here and the other is this rectangle. If we call the total angular impulse delivered in our first area 𝐻 one and the total angular impulse delivered in the second area 𝐻 two, then we can write that the total angular impulse delivered, we’ll just call it 𝐻, is equal to 𝐻 one plus 𝐻 two. If we consider first how to calculate the area 𝐻 one, since this is a triangle, we can recall that in general the area of a triangle is one-half the triangle’s base times its height. So then, 𝐻 one is equal to one-half the base of this triangle, which we can see is two seconds, multiplied by the height of the triangle, which we see is equal to four newton meters. This comes out to four newton meter seconds. And if we recall that a newton is equal to a kilogram times a meter divided by a second squared, then when we make that substitution, our units become overall kilograms meter squared per second.

So, that’s 𝐻 one. And then, we move on to calculating the second area, 𝐻 two. To do this, we can recall that the area of a rectangle is equal to its base times its height. And the base of this particular rectangle is its length along the time axis, that’s four seconds minus two seconds or simply two seconds, and its height is the same as the height of our triangle, four newton meters. So, the area 𝐻 two is eight newton meter seconds or eight kilograms meter squared per second. And therefore, the total angular impulse delivered to this object is 12 kilograms meter squared per second. And note that we figured this out just by calculating the area under our torque-versus-time curve.

Knowing all this about angular impulse, let’s get some practice now through an example exercise.

A torque of 0.2 kilograms meter squared per second squared is applied to an object for 20 seconds. Then, a torque of 0.5 kilograms meter squared per second squared is applied to the object for five seconds. What is the total angular impulse given to the object? Give your answer to two significant figures.

Okay, in this example, we have torques acting on an object. And we may as well say that this here is our object. We can imagine, say, that it’s a disc capable of spinning around its center. What happens first is we apply a torque, we can call it 𝜏 one, to this object. And that’s the torque given as 0.2 kilograms meter square per second. And this torque is applied for an amount of time we can call 𝑡 one, which is 20 seconds. Then, after all this, a second torque, we can call it 𝜏 two, is applied to our object. This is the torque of 0.5 kilograms meter squared per second squared. This second torque is applied for a period of time we can call 𝑡 two, which is given as five seconds. Knowing all this, we want to calculate the total angular impulse given to the object.

Now, at this point, we can recall that angular impulse, typically symbolized with a capital 𝐻, is equal to a change in angular momentum. And using a rotational form of Newton’s second law of motion, we can say that Δ𝐿, the change in angular momentum and therefore the angular impulse, is equal to the torque applied to an object multiplied by the time period over which that torque is applied. All this means that to calculate the total angular impulse delivered to our object, we’ll call that 𝐻, what we’ll want to do is add together these two torques multiplied by the times over which the individual torques act.

Torque one, we saw, acts for a time 𝑡 one, and torque two acts for a different time 𝑡 two. Now, each one of these products by itself is an amount of angular impulse. But to calculate the total, we’ll need to add them. So, what we do then is substitute in the values for 𝜏 one, 𝑡 one, 𝜏 two, and 𝑡 two. Note that in both cases, we have units of kilograms meter squared per second squared multiplied by units of seconds. And so, in both terms, one factor of seconds cancels out. This means the two separate angular impulses applied to our object are 0.2 times 20 kilograms meter squared per second and 0.5 times five kilograms meter squared per second.

0.2 times 20 is four. And 0.5 times five is 2.5. So, to two significant figures, our answer is 6.5 kilograms meter squared per second. This is the total angular impulse given to the object.

Let’s look now at a second example exercise.

A car accelerates for 20 seconds, during which time a torque of 32 newton meters is applied to each of the wheels. What angular impulse was given to each wheel over the time it accelerates for?

Okay, so, in this exercise, we’re considering a car that’s accelerating. But our focus is really on the wheels of the car, which each have a torque of 32 newton meters applied to them. We can just represent this torque as 𝜏 sub w. And we know it’s applied for a time of 20 seconds. And we’ll call that time 𝑡. We want to know, then, what is the angular impulse given to each wheel over this time of 20 seconds.

To start figuring this out, we can recall that the angular impulse, 𝐻, delivered to an object is equal to that object’s change in angular momentum. And that, by what we could call a rotational form of Newton’s second law of motion, is equal to the torque experienced by the object multiplied by the time over which that torque acts. In the case of our car wheels, then, the angular impulse that each one experiences will equal the torque, 𝜏 sub w, multiplied by the time, 𝑡. So then, that’s 32 newton meters multiplied by 20 seconds.

Before we calculate this product though, let’s recall that a newton is equal to a kilogram meter per second squared. If we substitute that in for newtons and then consider just the units in this expression for a moment, we see that one factor of seconds cancels from numerator and denominator. And if we collect all our units off to the right, we get kilograms meter squared per second. So, 32 times 20 kilograms meter squared per second is the angular impulse delivered to each wheel. And this is 640 kilograms meter squared per second. This is the angular impulse given to each wheel over the time of the car’s acceleration.

Let’s look now at one last example exercise.

A washing machine applies a torque of 8.0 kilograms meter squared per second squared to the drum of the machine over a period of time, giving the drum an angular impulse of 960 kilograms meter squared per second. For how long was the torque applied to the drum?

Alright, let’s say that this is our washing machine drum. We’re looking down on it from above. And so, here’s our drum full of clothes as they’re about to be washed. And we’re told that this drum experiences a torque, we can call it 𝜏, of 8.0 kilograms meter squared per second squared. This constant torque is applied over a period of time we’ll call Δ𝑡. And as a result, an angular impulse is delivered to the drum. We’ll call this angular impulse 𝐻. And we’re told it’s equal to 960 kilograms meter squared per second. The question is, what is Δ𝑡? That is, for how long do we apply this torque to our drum so that we got this resulting angular impulse?

To figure this out, we can recall a mathematical relationship that connects these three variables, time, torque, and angular impulse. The angular impulse delivered to some object, also equal to its change in angular momentum, is equal to the torque applied to that object times the time over which the torque is applied. For our purposes, we can shorten this expression to read 𝐻 is equal to 𝜏 times Δ𝑡. But then, it’s not 𝐻 we want to solve for but in our case Δ𝑡. If we divide both sides of this equation by the torque 𝜏, we find that Δ𝑡 is equal to 𝐻 divided by 𝜏.

And now, we see how to go about solving for Δ𝑡 in our particular scenario. It’s equal to the angular impulse experienced by the washing machine drum divided by the torque applied to create that impulse. Plugging in for our values of 𝐻 and 𝜏, we find that significant unit cancellation goes on. The units of kilograms and meter squared cancel from top and bottom, as does one factor of one over seconds. This means that in the end, our units simplify to seconds, which is good because we’re calculating an amount of time. 960 divided by 8.0 is 120. And so, here we have our answer. Torque was applied to this washing machine drum for 120 seconds.

Let’s summarize now what we’ve learned about angular impulse. In this lesson, we recall that linear impulse, Δ𝑝, is equal to the force applied to an object multiplied by the time over which that force acts. Expanding on this, we then saw that angular impulse, represented by capital 𝐻, which is equal to an object’s change in angular momentum, is equal to the torque applied to the object times the time over which that torque acts. In this way, we saw that angular momentum, 𝐿, is the rotational analog of linear momentum 𝑝.

And likewise, torque is the rotational or angular analog of force. And so, angular impulse, a torque applied over an amount of time, is the rotational or angular analog of a linear impulse, a force applied over an amount of time. And lastly, we saw that angular impulse can be calculated from looking at a graph of torque versus time. On such a graph, the total angular impulse delivered is equal to the area under the curve. This is a summary of angular impulse.

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