This video is for those of you who are studying for tests on uniform circular motion. So what I'm going to do in this video is I'm going to give you a quick review of the formulas that you need to know for this chapter. So uniform circular motion.
So we're dealing with objects that are moving in a circle. Now, keyword uniform. These are objects that are moving at constant speed. So let's say if the object is going in the counterclockwise direction, the velocity vector is going to be upward at this point. And there's going to be an acceleration vector known as centripetal acceleration that always points towards the center of the circle.
So when the object is here, the velocity vector will be directed west, but the centripetal acceleration will always be directed towards the center of the circle. Now the centripetal acceleration is equal to the square of the velocity divided by the radius of the circle. So as you double the velocity, 2 squared is 4, the centripetal acceleration increases by a factor of 4. Now, keep this in mind.
The object is moving in a circle at constant speed. But because it's changing direction, the velocity changes. And whenever the velocity changes, there's going to be an acceleration.
acceleration is the change in velocity divided by the change in time. Remember velocity is speed with direction. The magnitude of velocity is constant for uniform circular motion but the direction changes changes.
And so because the direction is changing, the velocity is changing, which means that there is an acceleration. So anytime you have any object that moves in a circle, you're always going to have some kind of centripetal acceleration. Now, according to Newton's second law, the net force acting on an object is equal to the mass times the acceleration. So anytime you have an object moving in a circle, there's going to be something called a centripetal force. And it's equal to the mass times the centripetal acceleration.
So replace an AC with V squared over R. We get the formula for the centripetal force, which is m V squared over R. So make sure you add these two formulas to your list. And you can also put this one as well.
Okay, let's just redraw this picture real quick. Now, going back to this equation. Let's talk about how we can calculate the velocity for an object moving in uniform circular motion. Now we know that velocity is displacement over time and the distance that the object travels in a circle is going to be the circumference. of that circle and the circumference is 2 pi times the radius.
The time it takes for the object to make one complete revolution around the circle that is known as the period which is capital T. Both lowercase t and capital T that measured in units of time like seconds minutes or hours. The frequency is 1 over the period.
Now, if we were to take this equation and plug it in for v, we will get that the centripetal acceleration is 4 pi squared times r over t squared. So that's how you can calculate, you could use this formula to calculate the central book's elevation if you know the radius of the circle and the period of the object in circular motion. If you know the period, you can also calculate the velocity using this formula. Now the period is equal to the time divided by the number of cycles. So for instance, let's say if it takes 30 seconds for this object to make 5 revolutions.
The period would be 30 seconds divided by those five cycles, so that's six seconds per cycle. So the period will be six seconds. The frequency is the reciprocal of that. It's the number of cycles divided by the time. The frequency is measured in hertz, or one over seconds.
Now sometimes you may need to calculate the tension of an object that's moving in circular motion. So let's say you have a ball with negligible mass and it's attached to a rope and it's moving in a circle. And we want to calculate the tension force at these different points.
Let's say at point A, B, C, and D. At points A and C, the tension force is approximately equal to mv squared over r. In those cases, the centripetal force is provided by the tension force in a rope. Now, at point D, at the bottom of the vertical circle, the tension force is going to be the sum of the centripetal force and the weight force.
Because not only does the rope have to support the weight of the object at point D, but it also needs to apply a centripetal object to turn it from point D to point C. Now, at point B, the tension force is going to be the weakest. And it's going to be the difference between mv squared over r, the centripetal force, and the weight force. So make sure you understand this.
At point D, the tension force is at a maximum. At point B, the tension force is at a minimum. Now let's say if you have a similar situation, but instead of the ball moving in a vertical circle, let's say if it's moving in a horizontal circle.
If it's moving fast, the tension force is going to be approximately equal to the centripetal force. But let's say if it's not moving very fast, such that it's not very horizontal, there's an angle to it. So let's say we have this situation. But it's sort of moving in a horizontal circle. So this is going to be the tension force.
It's going to be an x component, which is usually going to be significantly larger than the y component. And we have an angle here. In this case, the tension force in a rope is going to be the square root of Tx squared plus Ty squared. Now notice that Tx is horizontal and it points towards the center of the circle. So Tx provides the centripetal force, which is mv squared over r.
If this object is moving fast enough, this angle is going to be close to zero. And T is going to be approximately equal to Tx, which is the centripetal force. That's why I have this approximate symbol.
But if it's not moving fast enough and you could see a significant angle, then you need to calculate T this way. Now, T y supports the weight of the object. So T y is going to equal mg. And tangent theta is going to equal the ratio of T y over T x.
So using these formulas, you can calculate the tension in the rope. now for those of you who want to see how to use these formulas with example problems feel free to check out the links in the description section below I'm going to be posting some other videos on this chapter where you can practice using these equations so you know what to do on an actual exam Now there's other formulas, but there's one more formula I want to give you. And this has to do with, let's say if you have an object that is moving on a hill. And let's say in a valley. At the bottom of the valley, let's say it moves with speed v. We want to calculate the normal force exerted by the ground on the object and also when it's at the top of the hill.
So at the bottom of the hill, the normal force is going to be the sum of the centripetal force plus the weight force. So keep in mind, this object, it's turning. And anytime it changes direction like that, there's a centripetal force that is directed towards the center of the circle. Because there's a centripetal acceleration that points towards the center of the circle.
In this case, the center of the circle is over here. So this will be like the radius of this circle. And here is the centripetal force pointing towards the center of the circle. So at the bottom of the hill, or at the valley, the normal force is going to be at a maximum. because it has to support the weight force and it has to provide the centripetal force to cause the object to turn.
At the top, the normal force is going to be the difference between the weight force and the centripetal force. If the object is moving too fast, such that this quantity is greater than this quantity, where you get a negative result, what that means is that the object loses contact with the ground, and it's basically flying off. If you get a positive result for the normal force, that means the object remains in contact with the ground. But as V increases, if it gets too high, the normal force that you'll get will be negative, which means that there is no normal force.
The object is off the ground at that point. So at the top, the normal force will be at a minimum for this kind of problem. So I'm going to stop the video here.
And if you want to see an example problem on this particular situation, feel free to check out the links in the description section below. I have a video that's entitled Normal Force on a Hill. You can look it up on YouTube, but you'll also find it in the description section. So feel free to take a look at that, and thanks for watching.