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Lesson Video: Centripetal Force

In this video we learn that objects moving in a circle experience a centripetal, or center-seeking, acceleration, and that by Newton’s 2nd Law a corresponding centripetal force exists.


Video Transcript

In this video, we’re going to learn about centripetal force. We’ll learn what this force is, where it comes from, and how it got its name.

To start on this topic, imagine that you are at an amusement park and you’ve just stepped into a new ride, called the spinning chamber. This ride consists of a circular room, where riders will lean up against the outside wall. When everyone’s ready, the circular room begins to rotate about its center in faster and faster circular motion.

As the room spins faster, you begin to feel the wall behind you press harder and harder into your back. As the ride continues to spin faster, eventually, to your amazement, you see the floor beneath your feet drop down. But you stay in place. You no longer need the floor to support your weight. To better understand this spinning chamber ride, it will be helpful to know something about the centripetal force.

Imagine that you have a large rotating disc. And on the outer edge of that disc, you put a mark. And what if you wanted to know the acceleration of that point on the desk, what direction it pointed, and what its magnitude is. We know that the velocity of that point points tangent to the edge of the desk. But it turns out its acceleration points in the direction we might not expect. It points inward right towards the center of the circle.

And if we call the radius of this rotating disc 𝑟, we can say that the acceleration experienced by our mark on the edge of the desk is equal to its linear speed 𝑣 squared divided by the radius of the disc 𝑟. This acceleration is called centripetal acceleration. And it’s often symbolized 𝑎 sub 𝑐. Why, we may wonder, is this acceleration given this name centripetal?

Well, the word centripetal simply means center-seeking. So, the fact that this acceleration points towards the center of the circle it moves around gives it its name. If we give our rotating disc an angular velocity, we can call it 𝜔, then there’s another way we can write centripetal acceleration. Recalling that linear velocity 𝑣 is equal to 𝑟 times angular velocity 𝜔, centripetal acceleration is also equal to 𝑟𝜔 squared.

Here’s something that’s interesting about centripetal acceleration. Because acceleration is a vector defined as the change in velocity over change in time, that means that anytime an object is on a circular path, its direction must be changing, and, therefore, it has a centripetal acceleration. Even if the linear speed is constant since direction is constantly changing, the object is accelerating.

Now, let’s take this expression for centripetal acceleration and connect it with another expression we know. Newton’s second law of motion tells us that the net force acting on an object is equal to the object’s mass times its acceleration. If the acceleration of our object is a centripetal acceleration 𝑎 sub 𝑐, then we can call the net force acting on our object 𝐹 sub 𝑐 a centripetal force. This force depends on an object’s mass, its linear speed, and its rotational radius 𝑟. We can also write this in its rotational analogue 𝑚𝑟 𝜔 squared.

Since the net force acting on an object acts in the same direction as its net acceleration, that means that the centripetal force on the mark on our rotating disc acts in the same direction as the centripetal acceleration, towards the center of the circle. So, we see why the force is called center-seeking itself. Let’s get some practice with these ideas of centripetal acceleration and centripetal force through a couple of examples.

What is the magnitude of the acceleration of Venus toward the Sun, assuming a circular orbit with a radius of 1.082 times 10 to eleventh meters, and an orbital period of 0.6152 years. Use a value of exactly 365 for the number of days in a year.

We can call this acceleration magnitude of Venus toward the sun 𝑎. And if we make a sketch of the planet Venus orbiting circularly around the sun, we know that the distance between the center of the sun and the center of Venus is given as 𝑟 1.082 times 10 to eleventh meters. And that the planet Venus makes it once all the way around this orbit in a period we’ve called capital 𝑇 of 0.6152 years.

Since the planet Venus is moving in a circular orbit, that means it will accelerate centripetally towards the center of the circle. We recall that an object’s centripetal acceleration is equal to its linear speed squared divided by the radius of the circle it moves in. Recalling further that an object’s speed is equal to the distance it travels divided by the time it takes to travel that distance, we can say that the linear speed of Venus is equal to the distance it travels the circumference of the circle two times 𝜋 times its radius divided by the period 𝑇.

This means that the centripetal acceleration of Venus is equal to two 𝜋𝑟 over 𝑇 quantity squared all divided by the radius 𝑟. This simplifies to four 𝜋 squared 𝑟 over 𝑇 squared. Since we’re given both the radius 𝑟 and the period 𝑇, we have all the information we need. But before we plug in and solve for 𝑎, we like to convert the period 𝑇 from units of years to units of seconds.

To make that conversion, we’ll take 𝑇, which is given in years, multiply it by the number of days in a year multiply that by the number of hours in a day and multiply that by the number of seconds in an hour. This then will give us a time value in units of seconds.

With the radial distance 𝑟 plugged into our expression, we’re already to calculate 𝑎. When we do, we find a value of 1.135 times 10 to the negative two meters per second squared. That’s the magnitude of the center-seeking acceleration of the planet Venus as it moves in its circular orbit.

Now, let’s look at an example involving centripetal force.

A car that has a mass of 900.0 kilograms drives along a circular unbanked curve at a constant speed of 25.000 meters per second. The radius of the curve is 500.0 meters. What magnitude force acts on the car to maintain its speed and direction as it follows the curve? What is the minimum possible value for the coefficient of static friction between the car’s tires and the road surface?

We can label the force we want to solve for capital 𝐹 and the minimum possible coefficient of static friction we’ll name 𝜇 sub 𝑠. We can start on our solution by drawing a diagram of the situation. Our car in this example, with a mass of 900.0 kilograms, moves on a circular section of road with a radius of curvature 500.0 meters. While it drives on this section, it maintains a steady speed we’ve called 𝑣 25.000 meters per second.

Knowing all this, we wanna solve first for the magnitude of the force acting on the car that keeps it moving in a circular path on this unbanked road. A force that keeps an object moving in a circular path is a centripetal force. And it’s equal to the mass of the object times its speed squared over the radius of the circle it moves in. In our case, we’ve been given the radius 𝑟, the speed 𝑣, and the mass of the car 𝑚. So, we’re ready to plug in and solve for 𝐹.

When we enter this expression on our calculator, we find that 𝐹 is 1125 newtons. That’s the center-seeking force needed to keep this car on a circular path. If we drew a picture of our car from behind while it was turning on this circular path, the direction of the force 𝐹 would be to the left, pointing it towards what from this perspective is the center of the circle. The physical mechanism that creates this force 𝐹 is the friction force on the tires of the car from the road.

Wanting to solve for the coefficient of static friction between these two materials, we can recall that the force of friction on an object is equal to the coefficient of friction it experiences multiplied by the normal force acting on it. In our case, we can write that the force of friction on the car is equal to the coefficient of static friction because the wheels don’t slide on the road’s surface multiplied by the mass of the car times the acceleration due to gravity 𝑔. Where 𝑔 we’ll treat as exactly 9.8 meters per second squared.

This frictional force of the road’s surface on the tires is the physical mechanism by which the centripetal force 𝐹 sub 𝑐 is exerted on the car. So, we can say that the coefficient of static friction times 𝑚𝑔 is equal to 𝑚𝑣 squared over 𝑟. And we see that the mass of the car cancels out from this expression. If we then divide both sides of this equation by 𝑔, we see that 𝜇 sub 𝑠 is equal to 𝑣 squared over 𝑟 times 𝑔.

When we plug in for the given values of 𝑣, 𝑟, and 𝑔, we find that, to four significant figures, 𝜇 sub 𝑠 is 0.1276. That’s the minimum value that the coefficient of static friction can be between the road’s surface and the tires in order for this magnitude centripetal force to be exerted on the car.

Let’s summarize what we’ve learned so far about centripetal force. We’ve seen that objects in circular motion experience acceleration toward the center of the circle they’re moving in. We’ve seen that this acceleration called centripetal, or center-seeking, acceleration is equal to the linear speed of the objects squared divided by the radius of the circle they’re moving through. We also saw that this is equal to 𝑟 times the angular velocity 𝜔 squared.

We’ve also seen that Newton’s second law of motion says that this net centripetal acceleration corresponds to a force. We call it the centripetal force. By the second law, this net force is equal to 𝑚 times 𝑎 sub 𝑐, or 𝑚 times 𝑣 squared over 𝑟. And finally, we’ve learned that centripetal force always has a physical mechanism behind it, for example, the tension in a string or the friction force caused by road surface on car tires.

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