In this video, we’re going to learn about sources of centripetal force. In particular, we’ll see what are the physical mechanisms that cause the center-seeking force.
In all kinds of different physical situations, whether it be a planet orbiting a star or a ball being swung in a circle held by a string or a car navigating a turn on the road, the force that causes these objects to move towards the center of their circular path is mediated by a physical mechanism. In the case of the orbiting planet, that force is the force of gravity. In the case of the ball being swung around on a string, it’s the tension force that creates the center-seeking force. And for our car making a turn in the road, the friction force pushes that car towards the center of its circular path.
The centripetal or center-seeking force is always due to some physical cause. This means that whenever we draw a free body diagram of an object that’s moving in a circular path, we should be able to identify the force pushing it towards the center of the circle as something other than 𝐹 sub 𝑐, the centripetal force. The centripetal force in a given situation can be caused by gravity, tension force, spring force, friction, and the list goes on. The name “centripetal force” is the name we give to one of these physically caused forces, which happens to make an object move towards the center of a circular path.
For an object moving in a circle, the centripetal or center-seeking force acting on it is equal to the mass of the object times its linear speed squared divided by the radius of the circular arc it’s moving in. This relationship is true regardless of what force is physically causing the center-seeking force. This is helpful to us because it’s a bit like having a completed Newton’s second law expression before we’ve even included the effects of the physical force causing the circular motion. Let’s get some practice now, connecting up this expression for centripetal force with the physical forces that create the circular motion.
Railroad tracks follow a circular curve of radius 500.0 meters and are banked at an angle of 5.00 degrees. For trains of what speed are these tracks designed?
We can call the speed we want to solve for 𝑣. And we’ll start off with a diagram of this train track. If we draw a cross section of the train track with a train car on it, we’re told that the distance of the train car from the center of its curvature—we’ve called 𝑟—is 500.0 meters and that the track is banked at an angle—we’ve called 𝜃—of 5.00 degrees. Based on the condition of the track, we wanna solve for the speed of the train 𝑣 at which it could navigate this turn without any external forces being applied, like the force of friction, to keep the wheels from sliding on the track. Under this constraint, there are only two forces that are acting on the train car. There’s the force of gravity acting straight down on it. And then there’s the normal force pushing up on the car perpendicular to the track surface.
If we decide that motion vertically upward from this cross section is positive and that motion down the bank is positive in the 𝑥-direction, then we can break the force of gravity into its 𝑥- and 𝑦-components. And since the angle at the top corner of this right triangle is the angle 𝜃, then we can write that the 𝑥-component of the gravitational force is 𝑚 times 𝑔 times the sin of 𝜃, where 𝑔, the acceleration due to gravity, we treat as exactly 9.8 meters per second squared. Since this component of the gravitational force is the only force acting in the 𝑥-direction in our scenario, we can take advantage of the fact that Newton’s second law says that this force is equal to the mass of the car times its acceleration in the 𝑥-direction. When we write this expression, we’ll label that acceleration 𝑎 sub 𝑥.
There’s another clue we can use to solve for the speed of the train car 𝑣. And that’s the fact that the train car is moving in a circular arc. That means that the force pushing it to stay in this arc, in this case a component of the gravitational force, is a centripetal or center-seeking force equal to the mass of the train car times its speed squared over the radius of the circular path in which it moves. So in our expression, we can replace 𝑚 𝑎 sub 𝑥 with 𝑚𝑣 squared over 𝑟. And we see that the mass of the car cancels out from both sides of the equation. We then rearrange to solve for the speed 𝑣. We find it’s equal to the square root of the radius 𝑟 times 𝑔 times the sin of the angle 𝜃.
Since we know all three of these values, we’re ready to plug in and solve for 𝑣. When we enter this expression on our calculator, we find that 𝑣 is 20.7 meters per second. That’s the speed the train should maintain through this curve to most easily navigate this circular arc. Now let’s look at an example involving centripetal force in a vertical arc.
A ball of mass 1.3 kilograms at the end of a 1.5-meter string swings in a vertical circle. At its lowest point, the ball is moving with a speed of 9.5 meters per second. What is the speed of the ball at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the top of its circular path? What is the magnitude of the tension in the string when the ball is at the bottom of its circular path?
In this three-part problem, we can call the speed of the ball at the top of its path 𝑣 sub 𝑡, the tension at that point 𝑇 sub 𝑡, and the tension in the string at the bottom of the ball’s path 𝑇 sub 𝑏. Let’s start by sketching out this vertical circular path. This ball, with a mass of 1.3 kilograms we’ve called 𝑚, moves in a circular vertical arc on a string of length 𝑟, which is 1.5 meters. We know that when the ball is at the bottom of the circular arc, its speed is given as 9.5 meters per second. Given all this information, we wanna solve for the speed of the ball when it’s at the top of its arc.
Considering this system as a whole, since no energy is added to it or taken from it, we can use the conservation of energy to solve for the velocity of the ball at the top of its arc. We can say that the initial kinetic plus potential energy of our system is equal to the final kinetic plus potential energy, where our initial point is at the bottom of the circular arc and the final point is at the top. Recalling that kinetic energy is given as one-half an object’s mass times its speed squared and that gravitational potential energy is 𝑚 times 𝑔 times ℎ, where the acceleration due to gravity, 𝑔, we’ll treat as exactly 9.8 meters per second squared.
We can write that one-half the ball’s mass times 𝑣 sub 𝑏 squared is equal to one-half its mass times 𝑣 sub 𝑡 squared plus 𝑚 times 𝑔 times two 𝑟, the radius of the circular arc. We see that the mass 𝑚 cancels from this expression. And when we rearrange to solve for 𝑣 sub 𝑡, we find it’s equal to the square root of 𝑣 sub 𝑏 squared minus four times 𝑔 times 𝑟. Since we’re given 𝑣 sub 𝑏 and 𝑟 and know 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑡. Entering this expression on our calculator, we find that 𝑣 sub 𝑡 is 5.6 meters per second. That’s the speed of the ball at the top of its circular arc.
Next, we wanna solve for the tension of the string when the ball is at point 𝑓, at the top of its arc. If we draw a free body diagram showing the forces acting on the ball at this point, the two forces acting on the ball at this point are the gravitational force, 𝐹 sub 𝑔, and the tension force, 𝑇 sub 𝑡. Recalling Newton’s second law of motion that the net force on an object is equal to its mass times its acceleration and letting motion in the downward direction be positive in our example. We can write that 𝑚𝑔, the gravitational force, plus 𝑇 sub 𝑡, the tension force from the balls at the top of its arc, is equal to the mass of the ball times its acceleration, 𝑎.
Because the ball is moving in a circle, we know that it accelerates centripetally, that is, towards the center of the circular path, and that the magnitude of this acceleration is equal to 𝑣 squared over 𝑟. So we can write that 𝑚𝑔 plus 𝑇 sub 𝑡 is equal to 𝑚 times 𝑣 sub 𝑡 squared over 𝑟. This expression on the right-hand side is the centripetal force acting on this ball. We want to rearrange and solve for 𝑇 sub 𝑡. When we do, we see it’s equal to 𝑚 times the quantity 𝑣 sub 𝑡 squared over 𝑟 all minus 𝑔. When we plug in for 𝑚, 𝑣 sub 𝑡, 𝑟, and 𝑔 and enter this expression on our calculator, we find that, to two significant figures, 𝑇 sub 𝑡 is 15 newtons. That’s the tension in the string when the ball is at the top of its path.
Lastly, we wanna solve for the tension in the string when the ball is at the bottom of its arc. To do this, we can once again draw a free body diagram of the forces on the ball at this point. There are again two forces on the ball: the tension force, 𝑇 sub 𝑏, and the weight force. But now they point in opposite directions. We can write that the gravitational force minus 𝑇 sub 𝑏 is equal to negative 𝑚𝑣𝑏 squared over 𝑟, where the centripetal acceleration is negative because it points up. When we rearrange this expression to solve for 𝑇 sub 𝑏, we find it’s equal to 𝑚 times the quantity 𝑣 sub 𝑏 squared over 𝑟 plus 𝑔. When we plug in for these values and enter this expression on our calculator, we find that 𝑇 sub 𝑏 is 91 newtons. That’s the tension in the string when the ball is at the bottom of its arc. Let’s summarize what we’ve learned about sources of centripetal force.
We’ve seen that centripetal force always has a physical cause, such as the tension in a string or the pull of gravity. For an object moving in a circle, that object experiences a centripetal force equal to the mass of the object times 𝑣 squared over 𝑟, its linear speed over the distance from the object to the center of the circle. We’ve seen that this quantity, 𝑣 squared over 𝑟, is equal to the net acceleration acting on the object. And finally, we’ve seen that free body diagrams and Newton’s second law of motion are two tools that help us solve for centripetal force.