Lesson Video: Centripetal Force Physics • First Year of Secondary School

Video Transcript

In this video, we’re talking about centripetal force. This is a force which causes an object to move in a circular arc. We can get started discussing this force by talking about the word centripetal. It means center seeking. So if we have an object, like this one, that’s moving in a circular path, then this centripetal force it experiences to keep it on that path is always pointed toward the center of the circle. In that sense, this force, usually abbreviated 𝐹 sub c, is center seeking. Something very important to see about centripetal force is that it’s not a new force we could say that we’re introducing. This is simply the name we give to any force that acts on an object towards the center of a circular arc the object is traveling in.

For example, let’s say that this object here that’s moving in a circular path is a car driving around a horizontal circular road. When we think about what force is it that keeps the car on the road as it turns to this path, we know that that force is friction between the tires and the road surface. So in this example, it’s the frictional force which is a centripetal or center-seeking force. Or picture this scenario instead. Say that we have a satellite that’s in circular orbit around the Earth. The force that draws the satellite in to the center of its circular path is the force of gravity. So in this instance, gravity is the centripetal force acting on the satellite. We’ll see more examples of these various center-seeking forces as we go along.

The point is, in each case, the centripetal force is not a new or separate force that we’re adding in to the picture. Rather, it’s already there, present in the particular physical circumstances. Now, whenever we talk about forces acting on objects, we can be reminded of Newton’s second law of motion. Recall that this law tells us that the net force acting on an object is equal to that object’s mass multiplied by its acceleration.

So when an object moving in a circle experiences a force, the centripetal force, then if that’s the only force acting on the object, this force must be equal to the object’s mass multiplied by its acceleration. But here’s something interesting about this Newton’s second law equation. This is actually a vector equation. That’s because force, as well as acceleration, is a vector. So if we write this out to recognize that, we would say that the vector of force is equal to an object’s mass times that object’s acceleration vector.

Since force and acceleration are the only vectors in this equation, that must mean they point in the same direction. And therefore, this acceleration here, caused by the centripetal force acting on our mass, must also point towards the center of the circle that this mass is moving around. Therefore, we can call this acceleration a centripetal acceleration. And we’ll call it 𝑎 sub c. Okay, so these objects that we’ve identified that are moving in a circle do so because they’re experiencing a center-seeking force. And therefore, they accelerate towards the center of these circles too.

But here’s something interesting. The velocity vectors of these objects do not point in the same direction as their acceleration. That is, they don’t point towards the center of these circular paths. Rather, they point tangent to the part of the circular path where the object currently is. So the velocity of our car, 𝑣 sub c, points like this. And the velocity of our satellite, 𝑣 sub s, points like this.

Notice that this means that the velocity of each one of these objects is perpendicular to each one’s acceleration. Now, since both of these objects are moving in circles, there’s one other thing we can say about each scenario. In each scenario, the circular our object is moving in has a particular radius. We’ve called the radius of our car circle 𝑟 sub c and the radius of our satellite circle 𝑟 sub s. Now, we bring this up because there’s another way of expressing centripetal acceleration, the acceleration of an object moving in a circle. It’s equal to the tangential velocity of the object, at any point along the circular path, squared and then divided by the radius of the circle the object moves in.

If we look at this equation and recall that velocity is a vector, we might think that our object’s centripetal acceleration will depend on which velocity value we choose to use in this equation. For example, considering our car driving in a circle, what if, instead of using the car’s velocity when it’s here, we use its velocity when it’s here in the circle or here or here or anywhere else? It’s certainly true that the directions of these velocity vectors are not the same. But their magnitudes are. So regardless of which one we use, since we’re squaring that term in our equation for centripetal acceleration, we’ll get the same result no matter what. And this points to the fact that this equation is showing us the magnitude of centripetal acceleration.

Now, if we take the right-hand side of this equation and substitute it in for 𝑎 sub c here, then we can start to see some interesting connections between the speed of our object and the radius of the circle it travels in and the centripetal force it experiences. We can see, for example, that if we were to increase our object’s speed without changing the radius of the circle it moves in, then the center-seeking force our object experiences would go up by a greater factor than the factor by which our speed increased. That’s because we take our speed and square it to help us calculate the centripetal force.

On the other hand, say we were to keep our speed the same as before. But now we decrease the radius of the circle our object moves in. Once again, this would lead to an increase in centripetal force. For a smaller radius and with the same speed, it takes more force to keep an object moving in a circle. Now, so far, we’ve been talking only about linear variables, linear speed and linear acceleration. But we know that there are angular versions of these variables too. For example, if we have an object moving in a circle of radius 𝑟 with some tangential, that is, linear speed 𝑣, then we can say that this object has an angular speed symbolized using the Greek letter 𝜔. And the relationship between this object’s linear speed 𝑣 and angular speed 𝜔 is that 𝑣 is equal to 𝑟 times 𝜔.

Now, if we consider the units involved for these factors on the right-hand side, SI base units for linear distance are meters. And then angular speeds are often given in radians per second. But then, and this is an important note, the unit of radians is dimensionless. So when we multiply the radius 𝑟 by the angular speed 𝜔, the units that result are simply meters per second, with no unit of radians involved. These are the units of the speed 𝑣. So understanding that 𝑣 is equal to 𝑟 times 𝜔, we can take the right-hand side of this expression and substitute it in for 𝑣 in our equation for centripetal force. When we do this, we can see that squaring 𝑟 times 𝜔 will give us two factors of 𝑟, one of which cancels with the 𝑟 in the denominator. And our equation simplifies to 𝐹 sub c is equal to 𝑚 times 𝑟 times 𝜔 squared.

And now let’s recall something about this equation from earlier. Remember that in line with Newton’s second law, this equation used to read 𝐹 sub c is equal to 𝑚 times 𝑎 sub c, the centripetal acceleration. The form of the equation now tells us that 𝑟 times 𝜔 squared is equal to that centripetal acceleration. And we can indicate that fact down here. Now, before we get some practice working with centripetal forces, let’s consider this specific scenario. Say that we’re holding one end of a rope, right here, and tied to the other end of the rope is a small stone. And say that we then start to swing this rope around so that it moves in a vertical circle.

Now, one characteristic of an object moving in a circle is that its speed is constant. So the speed of our stone here is the same as it is here and here and here and everywhere else in this arc. So recalling this form for the centripetal force acting on our stone. If the stone’s speed is constant all throughout, and it is, and the radius of the circle the stone moves in is constant as well, and it is. Then the center-seeking force experienced by the stone must also be constant in magnitude. As we mentioned earlier though, every centripetal force has some physical cause for it, whether it’s the friction between tires and a road or gravity between a satellite and Earth. There’s always some physical mechanism creating a center-seeking force.

In this case, we can see that one of those mechanisms is the tension in our rope. This tension force is always pulling the stone towards our hand to the center of the circle. Therefore, it’s a centripetal force. But, and here’s where things get interesting, because this is a vertical circle that we’re swinging the stone in, the force of gravity is also involved here. The important difference here between our tension force and gravity is that the tension force always points towards the center of this stone’s circular path, whereas gravity always points straight downward. This means that when the stone is at the bottom of its circular arc, gravity is tending to move it out of a circular path. That is, it’s pushing it away from the center of its circle, whereas, by contrast, when the stone is at the top of its arc, gravity and the tension force in the rope are acting in the same direction, downward.

In cases like these where objects are moving in vertical circles, it’s very important to realize that so long as they still continue to move in a circle, the center-seeking force acting on them is constant in magnitude. If the object’s speed and the radius of the object’s circular path don’t change, then neither does the centripetal force. But what does change is the magnitude of the forces involved in creating the center-seeking force. To show what we mean by that, let’s consider the stone at these two points, at the bottom of its arc and at the top. We know that in each of these cases, the magnitude of the centripetal force acting on it, and we can draw this force vector in in orange, is the same. That doesn’t change.

But then, looking at the force of gravity and the force of tension, when our stone is at the bottom point of its arc, we can see that these forces act in opposite directions on the stone. The tension force is acting up, while gravity is acting down. Let’s label this tension forest 𝑇 sub b, the tension in the rope when the stone is at the bottom of its path. And we’ll call the gravitational force 𝐹 sub g. Looking at our three vectors 𝐹 sub c, 𝑇 sub b, and 𝐹 sub g, we can say that 𝐹 sub c is the resultant of 𝑇 sub b and 𝐹 sub g.

In other words, the centripetal force is equal to the difference between the tension force and the force of gravity acting on the stone. If we decide that forces acting upward are in the positive direction, then we could write that the tension force, which acts up and therefore is positive, minus the gravitational force, the one that points down on the stone, is equal to the centripetal force on the stone when it’s at the bottom of its arc. So that’s our equation showing the forces on the stone at this point in its path. Now let’s consider when the stone is at the top of its arc. Here, all three of the forces acting on it point downward and by our convention, therefore, are negative.

Now we call the tension force in our rope 𝑇 sub b when our stone was at the bottom of its path. So let’s call our rope tension 𝑇 sub t when the stone is at the top of its arc. Now, as we mentioned, based on our direction choice, all three of these forces, 𝐹 sub c, the force of gravity represented here by this blue vector, and 𝑇 sub t, are negative. So then, the equation showing the forces acting on our stone when it’s at the top of its arc would look like this. Looking at this equation, what if we multiplied both sides by negative one? That would switch all the negative signs to positive signs.

And now, look at this. We have two different expressions for the centripetal force, which we’ve said must be the same in magnitude on our stone at any point. And so we can equate these two expressions. If we do that, we get a result that looks like this. And as a last step, let’s say that we add 𝐹 sub g to each side of the equation. Now we’re assuming that the magnitude of the gravitational acceleration acting on the stone is constant all throughout its path. And since the stone’s mass is also constant all throughout, that would mean that on the left-hand side of our equation, we’re adding and then subtracting the same amount of gravitational force. That means these terms cancel out. While on the right-hand side, 𝐹 sub g plus 𝐹 sub g can be written as two times 𝐹 sub g.

And here’s where we see the result of all our labor. The force of tension in the rope is different when the stone is at the bottom of its arc versus when it’s at the top. Indeed, we can say that the tension force is less when the stone is at the top of its arc because we need to add two times the force of gravity on the stone to that tension in order to equal the tension in the rope when the stone is at the bottom. So when the object on the end of a string or a rope or a cord is moving in a vertical circle, while the force of gravity is essentially the same on the object all throughout. And the centripetal force certainly is all throughout its circular movement. It’s the tension force in that rope or string or cable that changes as the object moves around. Okay, having said all this, let’s get to an example exercise.

What is the magnitude of the centripetal force that must act on an object of mass 1.0 kilograms to make it move along a circular path of diameter 1.0 meters, completing a circle every 1.0 seconds?

Well, okay, so in this scenario, we have a circular path. And we’re told that the diameter of this path is 1.0 meters. Along with this, we have an object whose mass, that we can call 𝑚, is given as 1.0 kilograms. We want this object to move around the circular path so that it completes a circle every 1.0 seconds. One time around the circle is a complete revolution. So that means we can refer to this time of 1.0 seconds as our period of revolution. The question is, in all this, what’s the magnitude of the centripetal force that must act on this mass? That is, for the mass to be moving in a circle, there must be some force making it tend towards the center of that circle. That’s the centripetal force, and we want to figure out how strong it is.

We can start doing this by recalling that the centripetal force on an object is equal to the object’s mass times its centripetal acceleration. Now, an object’s centripetal acceleration is given by its speed squared divided by the radius of the circular path it moves in. But then we can also recall that for an object moving in a circle, its linear speed, its tangential speed around that circle, is equal to the circle’s radius times the object’s angular speed 𝜔. If we substitute 𝑟 times 𝜔 in for 𝑣 in our equation for centripetal acceleration, we find it’s equal to the quantity 𝑟 times 𝜔 squared divided by 𝑟 or, more simply, 𝑟 times 𝜔 squared. We can then substitute 𝑟 times 𝜔 squared in for 𝑎 sub c in our equation for centripetal force. And now we have an expression for this force that we can work with in terms of our given parameters.

Starting with our object’s mass, we’re given that it’s 1.0 kilograms. And then the radius of the circle our object is moving in is equal to the diameter divided by two. And then lastly, there’s this factor here, the angular speed 𝜔 squared. We can recall that the units of 𝜔 are radians per second. And since we’re told that our object completes a circle every 1.0 seconds, recalling that a full circle is equal to two 𝜋 radians, we can say that the angular speed of our object is equal to two 𝜋 radians divided by 1.0 seconds. This is its angular speed. So then, substituting in to our equation for centripetal force, here’s what we get. Our mass as we saw is 1.0 kilograms. Our circle’s radius is its diameter, 1.0 meters, divided by two. That’s 0.5 meters. And then our object’s angular speed is two 𝜋 radians in 1.0 seconds. And this factor of 𝜔 is squared.

When we calculate all this out, we find a result of 19.73 and so on and so forth newtons. But notice how all of the values we’re given in our problem statement have two significant figures. That’s how many we’ll keep in our answer. And to two significant figures, our result is equal to 20 newtons. That’s the magnitude of the centripetal force acting on our object.

Let’s recall some key points now about centripetal force. We learned in this lesson that centripetal force, true to its name, always points toward the center of a circular arc. We also saw that for an object of mass 𝑚 moving with a speed 𝑣 in a circular arc with radius 𝑟, the center-seeking or centripetal force acting on that mass is equal to 𝑚 times 𝑣 squared over 𝑟.

Furthermore, we identified this factor, 𝑣 squared over 𝑟, as a centripetal acceleration. And we also recalled the connection between linear and angular speeds, that 𝑣 is equal to 𝑟 times 𝜔, and saw that this implies that centripetal acceleration can be written as 𝑟 times 𝜔 squared. And lastly, we saw that when an object moves in a vertical circular path, the effect of gravity must be considered. This is a summary of centripetal force.