Lesson Video: Torque of a Force Physics • 9th Grade

In this video, we will learn how to use the formula 𝑇 = 𝐹𝑑 to calculate the torque of a force about a rotation axis, and also to calculate the net torque.

15:31

Video Transcript

In this video, we will be looking at the concept of torque and how it’s directly related to the rotational motion of objects. And regarding this opening screen, Billy is actually very right to learn about torque by spinning his pencil. Try it yourself. Try spinning your pencil on different surfaces or by exerting a force at different points along the length of the pencil. And see what you can learn about torque.

But before you do that, let’s actually understand what torque means. We’re actually going to use Billy’s tactic and learn about torque using a pencil. So let’s start by thinking about this pencil just sitting on a table. At the moment, the pencil is at rest. That is, it’s not moving at all. Now, before we can do anything to this pencil, let’s start by labelling its centre of mass. Remember, the centre of mass is a special point at which the gravitational force acting on the object, in other words, the weight of the object, can be said to act. Now, since we’re imagining looking at this pencil from above. In other words, it’s sitting on a table. So let’s say that this is our table.

What this basically means is that the weight of the pencil will be acting into the screen. And specifically, the point at which the weight of the pencil will be acting will be at the centre of mass. So now, let’s take our pencil and apply a force on it in this direction. And this force is applied some distance away from the centre of mass. Let’s say that the force that we apply has a magnitude 𝐹. And the distance between the centre of mass and the point at which the force is applied we will call 𝑑. Now as we can imagine, applying this force 𝐹, a distance 𝑑 away from the centre of mass, will firstly cause the centre of mass of the pencil to move in this direction.

In other words, the entire pencil will move along the table in this direction. But as well as this linear motion, the pencil will actually rotate counterclockwise. And that rotation of the pencil will be about the centre of mass. Now in this situation, because we haven’t pinned down the pencil in any way, we have to deal with the motion of the centre of mass of the pencil. In other words, we have to deal with the linear motion of the pencil, as well as the rotational motion of the pencil, whereas what we really want to do is to just deal with the rotation of the pencil.

So one thing that we can do is to take a nail and hammer it into the pencil at any point along its length. It doesn’t have to be at the centre of mass of the pencil. We can do it anywhere along its length. When we do this, the pencil will no longer rotate about its centre of mass. But instead it will rotate about the point at which we’ve nailed it to the table because that is the point at which the pencil cannot move. And so the rest of the pencil will move around that point.

Therefore, the relevant distance 𝑑 that we now need to think about is actually the distance between the point at which we’ve nailed the pencil and the point at which the force is applied. Now, the benefit of nailing this pencil to the table is that this force, 𝐹, will not cause any linear motion of the pencil. We will only be dealing with rotational motion now. In other words, we’ll just see a counterclockwise rotation of the pencil. And at this point, we can define something known as the torque applied to the pencil. We can say that the torque applied to the pencil, 𝑇, is equal to the magnitude of the force applied, 𝐹, multiplied by the distance that we’ve labelled, 𝑑.

And it’s worth noting by the way that we may often see the torque applied to an object denoted by the Greek letter 𝜏 rather than the English letter 𝑇. But in this case, we will just stick with using letter 𝑇. Now, the distance 𝑑 is actually the distance between the point at which the object rotates. And this special point is known as either the fulcrum or the pivot point and the point at which the force is actually applied. And an important point to remember is that if we think about the direction in which we measure this distance 𝑑, then that distance has to be perpendicular to the direction of the force 𝐹.

So to simplify, the force 𝐹 that’s causing a rotation in this pencil has to be perpendicular to or at 90 degrees to the direction in which we measured the distance 𝑑. The reason for this is that only the perpendicular component of any force applied to an object will cause a rotation in that object. Let’s say for example, we took our pencil and applied a force in this direction. And once again, we had nailed it to the table here. Well in that case, the distance between the pivot point and the point at which the force is applied is measured in this direction. And so we’d have to break up the force, which we’ve called 𝐹 new, into a parallel component and the perpendicular component.

Let’s call this perpendicular component 𝐹 subscript perpendicular. And when trying to work out the torque applied on our pencil, the force in this equation would actually have to be the perpendicular component of the total force applied to this pencil. The parallel component would actually only work towards causing a linear motion. But as we’ve seen, this is not possible because the pencil is nailed to the table. But anyway, so the way that we write this important condition is to say that the force 𝐹 must be perpendicular to the distance 𝑑, where in this case the force 𝐹 that we’re talking about is the force in this torque equation. Now as we’ve already seen, exerting a torque on an object causes it to rotate. And actually the larger the torque, the larger the angular or rotational acceleration of the object.

Now, before we exerted this force 𝐹 onto the pencil, we said that the pencil was sitting stationary on the table. Then we exerted the force 𝐹 at this particular point on the pencil. That caused the pencil to rotate about the fulcrum. We can actually measure how fast this pencil begins to rotate in terms of an angular acceleration. Before we exerted the force, the pencil was stationary. And so it was moving at zero degrees every second. The pencil wasn’t moving in an angular direction. However, after we exerted this force and therefore a torque on the pencil, the pencil started moving counterclockwise at a rate of a certain number of degrees every second say.

So now, it has a certain angular velocity, where angular velocity means the number of degrees moved every second. And in order to go from a zero angular velocity, where the pencil is not rotating, to having an angular velocity of a few degrees per second, that pencil will have undergone an angular acceleration. In other words, this was its initial angular velocity, zero degrees per second. And let’s say its final velocity was something like 10 degrees per second. And it went from here to here in a period of one second. Then the angular acceleration was the final angular velocity, 10 degrees per second, minus the initial angular velocity zero degrees per second divided by the time it took to get there which was one second.

And so it’s angular acceleration was 10 degrees per second squared. And this angular acceleration was caused by this torque. In other words, if we recall newton’s second law of motion, we can see that a force applied to an object will create an acceleration on that object. And that force is equal to the mass of the object multiplied by the acceleration. So our force causes a linear acceleration. And in the same way, a torque causes an angular acceleration where this time we don’t multiply by the mass but something known as the moment of inertia of an object. And so torque behaves in a similar way for angular acceleration to forces in terms of how they generate linear accelerations.

Now, we’re not going to worry too much about what the moment of inertia of an object is. But we just need to know the parallels between linear motion and angular motion. Our torque causes a change in the angular velocity of an object. Now in this example, we’ve said that the object is initially stationary. But it could for example have already been rotating at, let’s say, five degrees a second. But that is a constant angular velocity, until we apply this force 𝐹 at which point it’s now going to start rotating at, let’s say, 10 degrees per second. And that’s caused an angular acceleration.

So this is how torque works. Now in practice, there are a couple of different ways to change the torque exerted on an object. Let’s say we’re going to increase this torque. Well then, we could firstly increase the force exerted on the object, which in this case we’ve called 𝐹. In other words, we could just push this pencil harder in this direction. Or we could keep this force the same and actually increase the distance 𝑑, in other words, exert the same force here rather than here.

So this is where it gets interesting. We can actually generate a larger angular acceleration with the same force, if we just increase the distance between the fulcrum and the point at which the force is exerted. This is why for example if we tried to turn a nut with our fingers, it often gets to a stage where we can no longer turn the nut. But if we instead use a spanner, than we’re probably exerting the same force with our hands, where the distance between whether forces are exerted and the pivot point is actually much larger. And so we can turn the nut a little bit further. So this is how exerting the same force can result in a larger torque, if we just increase the distance between the pivot and the point at which the force is exerted.

Now at this point, it’s worth noting that when we think about forces acting in opposite directions, then we arbitrarily choose one direction to be positive and the other to be negative. And we do the same thing for torques. Any torques causing a counterclockwise rotation of an object are said to have the opposite sign to torques causing a clockwise rotation. For example, out of convenience, we may choose that clockwise torques are positive. And so counterclockwise torques must be negative or vice versa. So let’s now think about the following scenario.

Let’s say that two people are playing a game where one of them applies a force to the pencil that tries to turn the pencil clockwise. So let’s say they exert a force 𝐹 one. And the other person, who’s a little bit weaker and so exerts a smaller force 𝐹 two, trying to turn the pencil counterclockwise. Now, let’s say that the first person exerts the force 𝐹 one a distance of 𝑑 one away from the fulcrum. And the second person exerts 𝐹 two, a distance of 𝑑 two away from the fulcrum. Now, let’s think about putting in some numbers.

Let’s say that the stronger person exerts a force 𝐹 one, which is five newtons, whereas the weaker person exerts a force 𝐹 two, which is two newtons. But let’s say that the distances 𝑑 one and 𝑑 two are 0.02 metres and 0.05 metres, respectively. When in this situation, we can say that the total clockwise torque which we’ll call 𝑇 subscript clockwise, and this torque is exerted by the stronger person, is equal to the force applied which is 𝐹 one multiplied by 𝑑 one. And we can also say that the counterclockwise torque is equal to the force trying to turn the pencil counterclockwise which is 𝐹 two multiplied by the distance 𝑑 two.

Then we can plug in the numbers. And then when we evaluate the right-hand sides of both equations, we find that both torques have a magnitude of 0.1 newton metres. So there are a couple of things to note here. Firstly, we can see that torque has units of newton metres. That’s units of force multiplied by units of distance. And secondly, as we said earlier, clockwise torques and counterclockwise torques must have the opposite signs. So although we worked out the magnitude here, we need to account for directions. And, therefore, we notice that torque is a vector quantity.

So the actual clockwise torque, if we say, is positive 0.1 newton metres. Because we’ve arbitrary chosen that clockwise is positive, then the counterclockwise torque must be negative 0.1 newton metres. And so the net torque on the object, which we’ll call 𝑇 subscript net, is equal to the sum of the clockwise torque and the counterclockwise torque. So in other words, it’s positive 0.1 newton metres minus 0.1 newton metres. And hence, the net torque is zero newton metres on the pencil. Therefore, the pencil will not rotate in any direction because the torque trying to turn it clockwise is cancelled by the torque trying to turn it counterclockwise. Just like how if there is a net force of zero on an object, then there is no acceleration. If there is a net torque of zero on an object, then there is no rotational acceleration.

So even though the stronger person exerted a larger force because they were closer to the pivot, the force exerted by the smaller person actually manages to cancel out that effect. So these two torques have exactly managed to cancel each other out. And hence, the angular velocity of the pencil stays the same. In this case, because the pencil wasn’t moving in the first place, the angular velocity remains at zero degrees per second. It’s also worth noting interestingly that we don’t need to nail something down to a table or otherwise restrict its motion for the object to only have angular velocity but no linear motion. For example, if we consider a ruler where all of its mass is perfectly distributed over its entire length, then its centre of mass will actually be at the geometrical centre of the ruler.

Now, if we exert two identical forces, let’s say at the ends of the ruler, but acting in opposite directions, so let’s say they both have magnitude 𝐹 but one is acting upwards and the other is acting downwards. Then we can say that each of these forces is acting at distance 𝐿 divided by two away from the centre of mass, where 𝐿 is the total length of the ruler. In this case, the upward force is trying to turn the ruler clockwise. And the downward force is also trying to turn the ruler clockwise.

So if we decide that clockwise rotation is positive, then we can say that the net torque on the ruler, which we’ll call 𝑇 subscript net, is equal to the force 𝐹 multiplied by the distance 𝐿 divided by two plus the force 𝐹 multiplied by the distance 𝐿 divided by two. And this simplifies to 𝐹 multiplied by 𝐿. Therefore, there is a net rotational clockwise torque on the ruler. And so the ruler will rotate clockwise. But because we’ve got two identical forces acting on the ruler in opposite directions, the net linear force is actually zero on the ruler because this upward Force 𝐹 is balanced by this downward force 𝐹.

And so the net force on the object 𝐹 subscript net is equal to zero. Hence, the centre of mass of the ruler will not experience a linear acceleration. It will stay in the same position. And all that happens is that the ruler rotates around the centre of mass. So now that we’ve understood a little bit about torque, let’s take a look at an example question.

How much torque is produced by a 30-newton force acting on an object at a distance of 0.15 metres from the point about which the object can rotate?

Now in this question, we haven’t actually been told what the object in question is. So let’s make life easy for ourselves and think about it as a simple object, let’s say a ruler. Now for this ruler, if the mass is evenly distributed along its length, then the centre of mass will be the geometrical centre of the ruler. And actually this is the point about which the ruler can rotate. Now, we’ve been told that a 30-newton force is exerted. So let’s say, this is the 30-newton force. And this force is exerted at distance of 0.15 metres from the point about which the ruler can rotate.

So here is the point about which the ruler can rotate, the centre of mass. And here is the point at which we’re exerting the force. If we assume that the direction in which we measure this distance, the 0.15-metre distance, is actually perpendicular to the direction in which the force is applied, then we can recall that the torque applied on an object is equal to the force exerted on that object multiplied by the distance between the point at which the force is applied and the point at which the object can rotate, where the important thing to remember is that the force 𝐹 in this equation is perpendicular to or at right angles to the distance 𝑑.

And so we can say that the torque applied 𝑇 is equal to the torque applied which is 30 newtons multiplied by the distance which we know is 0.15 metres. And once again this works because we’ve assumed that the force is perpendicular to the distance 𝑑. Then when we evaluate the right-hand side of the equation, we find that the magnitude of this torque is 4.5 newton metres.

Okay, so now that we’ve looked at an example question, let’s summarise what we’ve talked about in this lesson. Firstly in this lesson, we’ve seen that torque is defined by 𝑇 is equal to 𝐹 𝑑, where 𝑇 is the torque exerted on an object. 𝐹 is the force applied to that object. and 𝑑 is the distance between the point at which we are exerting the force and the point at which the object can rotate. Now, the important thing to remember is that the 𝐹 in this equation is a force that’s perpendicular to the distance being measured.

So if we’ve got a force that’s not perpendicular to this distance, then we need to break it up into the perpendicular and parallel components. So let’s say here is our object. And it rotates about this point here. And we exert a force in this direction, let’s say. Then we need to break that force down into the parallel component and the perpendicular component. Then we see that the distance being measured is the distance between the point at which the object rotates and the point at which the force is exerted. And we only consider the perpendicular component of the force. So the torque exerted is equal to this perpendicular component multiplied by this distance 𝑑.

And it’s also worth noting that often torque is symbolised using the Greek letter 𝜏 rather than the English letter 𝑇. Secondly, we saw that torque has units of newton metres. Thirdly, we saw that a net torque causes an angular acceleration. And in many ways, torque can actually be thought about as an angular analogue to forces. Just like how a net force on an object causes a linear acceleration, think Newton’s second law of motion, a net torque causes an angular acceleration. So this is an overview on torque.

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