Centripetal force - Wikipedia
The goal of this three-part activity is to develop understanding of the relationship between centripetal force and its factors: circular speed, mass, and radius. If r is the radius of the path, and we define the period, T, as the time it In circular motion many people use the term centripetal force, and As always, the place to start is with a free-body diagram, which just has two forces. This force is called the centripetal force which means "center seeking" force. The force has the Force Calculation. Centripetal force = mass x velocity2 / radius.
And over here, the velocity vectors are tangential to the path, which is a circle. They're perpendicular to a radius. And you learned that in geometry-- that a line that is tangent to a circle is perpendicular to a radius. And it's also going to be the same thing right over here. And just going back to what we learned when we learned about the intuition of centripetal acceleration, if you look at a1 right over here, and you translate this vector, it'll be going just like that.
It is going towards the center. So all of these are actually center-seeking vectors. And you see that right over here. These are actually centripetal acceleration vectors right over here. Here we're talking about just the magnitude of it. And we're going to assume that all of these have the same magnitude. So we're going to assume that they all have a magnitude of what we'll call a sub c. So that's the magnitude. And it's equal to the magnitude of a1.
That vector, it's equal to the magnitude of a2. And it's equal to the magnitude of a3.Uniform Circular Motion: Crash Course Physics #7
Now what I want to think about is how long is it going to take for this thing to get from this point on the circle to that point on that circle right over there? So the way to think about it is, what's the length of the arc that it traveled?
The length of this arc that it traveled right over there. The circumference is 2 pi r. So that is the length of the arc.
And then how long will it take it to go that? Well, you would divide the length of your path divided by the actual speed, the actual thing that's nudging it along that path. So you want to divide that by the magnitude of your velocity, or your speed. This is the magnitude of velocity, not velocity.
This is not a vector right over here, this is a scalar. So this is going to be the time to travel along that path. Now the time to travel along this path is going to be the exact same amount of time it takes to travel along this path for the velocity vector.
What is centripetal acceleration? (article) | Khan Academy
So this is for the position vector to travel like that. This is for the velocity vector to travel like that. So it's going to be the exact same T. And what is the length of this path? And now think of it in the purely geometrical sense.
We're looking at a circle here. The radius of the circle is v. The circumference of this circle is 2 pi times the radius of the circle, which is v. Now what is nudging it along this circle? What is nudging it along this path? What is the analogy for speed right over here?
Speed is what's nudging it along the path over here.
Visual understanding of centripetal acceleration formula (video) | Khan Academy
It is the magnitude of the velocity vector. So what's nudging it along this arc right over here is the magnitude of the acceleration vector. So it is going to be a sub c. And these times are going to be the exact same thing. The amount of time it takes for this vector to go like that, for the position vector, is the same amount of time it takes the velocity vector to go like that.
What is centripetal acceleration?
So we can set these 2 things equal each other. And now we can simplify it a little bit. Get rid of that. We can divide both sides by 2 pi, get rid of that. Let me rewrite it. And now you can cross multiply.
And so you get v times v. So I'm just cross multiplying right over here. And cross multiplying, remember, is really just the same thing as multiplying both sides by both denominators, by multiplying both sides times v and ac.
So it's not some magical thing. If you multiply both sides times v and ac, these v's cancel out. These ac's cancel out.
You get v times v is v squared, is equal to a sub c times r. And now to solve for the magnitude of our centripetal acceleration, you just divide both sides by r.
And you are left with-- and I guess we've earned a drum roll now-- the magnitude of our centripetal acceleration is equal to the magnitude, our constant magnitude of our velocity. Centripetal force is a misleading term because, unlike the other forces we've dealt with like tension, the gravitational force, the normal force, and the force of friction, the centripetal force should not appear on a free-body diagram.
The centripetal force is not something that mysteriously appears whenever an object is traveling in a circle; it is simply the special form of the net force.
Newton's second law for uniform circular motion Whenever an object experiences uniform circular motion there will always be a net force acting on the object pointing towards the center of the circular path. This net force has the special formand because it points in to the center of the circle, at right angles to the velocity, the force will change the direction of the velocity but not the magnitude. It's useful to look at some examples to see how we deal with situations involving uniform circular motion.
Example 1 - Twirling an object tied to a rope in a horizontal circle. Note that the object travels in a horizontal circle, but the rope itself is not horizontal.
If the tension in the rope is N, the object's mass is 3. As always, the place to start is with a free-body diagram, which just has two forces, the tension and the weight.
It's simplest to choose a coordinate system that is horizontal and vertical, because the centripetal acceleration will be horizontal, and there is no vertical acceleration. The tension, T, gets split into horizontal and vertical components. We don't know the angle, but that's OK because we can solve for it. Adding forces in the y direction gives: This can be solved to get the angle: In the x direction there's just the one force, the horizontal component of the tension, which we'll set equal to the mass times the centripetal acceleration: We know mass and tension and the angle, but we have to be careful with r, because it is not simply the length of the rope.
It is the horizontal component of the 1. Rearranging this to solve for the speed gives: