Hey guys. So now that we've covered the basics of uniform circular motion, it's time to start talking about forces. When you have forces in circular motion. Those are called centripetal forces. So the whole idea with this video is that the way that we have solved forces problems in the past we have forces along straight lines. These are called linear force problems is gonna be the same way that we solve circular or centripetal forces problems. But there's a couple of differences I want to go through, so and then and then we'll do a quick example. So let's check this out. So just to recap, whenever we had linear forces problems, we had forces along fixed X and y axis. We draw a coordinate system like this and let's say we had a couple of forces in the X axis for simplicity. I've got these two forces like this, right. If we had multiple forces, then basically the net force is going to produce an acceleration, right? So you have an acceleration and force those point along the same direction. And the way that we saw this is just by using F equals M A and the X and the y axis. Now let's talk about circular motion, right. We know that in circular motion you have some tangential velocity and your acceleration always points towards the center of the circle. But the same principle applies. If you're accelerating towards the center, there has to be some net force that's pushing you towards the center. The difference is that depending on where you are in the path, those directions are going to change, Right. So if you are here, the acceleration and therefore the net force is going to change. So it actually doesn't really make a lot of sense for us to use a fixed X and Y coordinate system like this because the directions are always changing in circular motion. So what we do is we just have forces along the centripetal direction, which really just means that wherever you are on the path, those forces are gonna be pointing toward or away from the center. But ultimately we still have the same variables F nets and A C. So really, we're just going to use f equals M A to solve these kinds of problems. The only difference is that we're going to use the sum of all forces in the centripetal access equals mass times acceleration, centripetal. That's really all there is to it. It's still f equals m A. But now everything is just in circular motion. So the way we solve these problems is actually be very similar to how we solve problems before we're gonna right, We're gonna draw a free body diagram and the right of equals M a right. So let's go ahead and take a look at this problem. So we have this 3 kg block that's tied to a string and it slides around the circle. So I've got this little diagram like this basically, this little this little block is gonna be sliding around in a circle unless horizontal tabletop. Now we want to calculate the tension on the string so that variable is going to be T. But the problem is, is that we already have a variable for tea. And that's the period, right? So we're gonna have multiple variables involving T. So just to avoid any confusion, we're actually gonna be solving for the force of tension in this problem. All right, so we're trying to solve for a force we know we're going to have to draw a free body diagram. So let's go ahead and get to it. Right? So if we were to take a look at this box here, the perspective is kind of weird because we're looking at it like, sort of diagonally on the tabletop. So we know we want to start off with the wait for us, but where does that force act? So what I would like I would like to do is like you like to draw a little side view. Imagine that you were looking at this table top basically from the side like this. And so what you would see is the tabletop like this, the box, And so therefore you would have a wait for us. That's straight down. That's your mg. And then you would have the force of tension that's basically caused by the string. And then you'd also have a normal force like this. Now, if you were to look at this thing from the top, so this is like a top view. If you were looking at this table top from the tap top down, basically, you just see the path, the circle, and at any point where the object is, you would just see the tension force It points towards the center. This is going to have tension. Force are normal in our weight force are basically along our direction of sites as we look down, so it's hard to draw them. So now that's our free body diagram. So now we just want to write f equals m A. So we have f equals m A. But now in this centripetal axis. So now we have to expand all of our forces and we have three of them to consider. We have the normal, the weights and the tension. But the thing is that the normal and the tensions are sort of the normal and the weight force, or only acting along the vertical plane like this when you look at it from the top, the only thing that's actually keeping this thing going in a circle is really just the force of tension. And I'm sorry, actually, I sort of, uh, I should have drawn. I should have written. This is f T. So this is really just r f t right here. So here's our f t and that's the force of tension. So when we expand our F equals m a, the only force, it's actually causing this thing to accelerate. Centripetal e is our force of tension. So this is equal to mass times acceleration. And that actually brings me to the second thing. So oftentimes what happens is that we're going to have to write this a c this acceleration from trip. It'll as v squared over r, remember, that's a really, really important equation that you should remember. So what happens is we're gonna write this ft here and this is equal to M, and we just replace this with the squared over R. So if we want to figure out the tension force, we're gonna have to figure out all the other variables. So now let's get to it. So we have the mass and we also have the radius. So I just have to go ahead and find the velocity here. So what's the velocity? And I actually have a couple of equations for velocity. Remember that velocity is really just the circumference divided by the period, and so therefore we can use to power of the tea. Or we can use two pi r f either one of them will get those get us the right answer. So if we take a look at our question, we have that the block completes a rotation every four seconds so we can use either one of these two pi r of A t or two pi r f Now, really, what happens is if this completes one rotation every four seconds, then that actually kind of gives us information about the period. Remember that to solve your period, it's just the seconds over the cycles. So really, what it's telling us is that it takes four seconds to complete one cycle, and so a period is actually just four seconds. You wanted to solve for the frequency, you would have gotten 1/4, right? You basically the inverse of that? Okay, so that means that our velocity here is really just two pi times the radius, which is too divided by the period, which is four. And you get 3.14 Remember, the reason we went to solve for this is because we have to plug this back into M. V squared over r. And now we can do that. So basically, we have Our tension force is our mass, which is three. And we have V which is 3.14 We're gonna square this divided by the radius, which is to if you go ahead and plug this in, you're going to get 14.8 Newtons. Alright, So basically the same exact steps free body diagram f equals M a and then just solve. But now you just have a couple of more equations involving the centripetal acceleration they might have to go solve for All right, So that's it for this one, guys.