Guys, 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. The whole idea with this video is that the way we have solved forces problems in the past, we had forces along straight lines, these are called linear force problems, is going to be the same way that we solve circular or centripetal forces problems. But there are a couple of differences I want to go through. So, let's do a quick example and check this out.

To recap, whenever we had linear forces problems, we had forces along fixed x and y axes. We drew our coordinate system like this, and let's say we had a couple of forces in the x axis for simplicity. If we had multiple forces, then basically the net force is going to produce an acceleration. You have an acceleration and force those point along the same directions. And the way that we solve this is just by using F=ma in the x and y axes. Now let's talk about circular motion. We know that in circular motion, you have some tangential velocity and your acceleration always points towards the center of the circle. 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. It actually doesn't really make sense for us to use a fixed x and y coordinate system 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 going to be pointing toward or away from the center. But ultimately, we still have the same variables, F_{net} and A_{c}. So, we're just going to use F=ma to solve these kinds of problems. The only difference is that we're going to use the sum of all forces in the centripetal axis equals mass times acceleration centripetal. That's really all there is to it. It's still F=ma, but now everything is just in circular motion. The way that we solve these problems is actually very similar to how we solve problems before. We're going to draw our free body diagram and then write F=ma.

Now, let's consider a problem where we have this 3-kilogram block that's tied to a string, and it slides around the circle. Essentially, this block is going to be sliding around in a circle on this horizontal tabletop. We want to calculate the tension on the string. To avoid confusion with multiple variables involving 'T' (as it also represents the period), we're going to solve for the force of tension. We're going to start off with the weight force. Suppose we draw a side view as if looking at the tabletop from the side. You would see the tabletop, the box, and the weight force that's straight down (your mg), and then you would have the force of tension caused by the string, and then you'd also have a normal force.

If you were looking at the tabletop from the top, you would just see the path of the circle. At any point, the tension force points towards the center. Our forces to consider are the normal, the weights, and the tension. But the normal and the weight force are only acting along the vertical plane. When you look at it from the top, the only thing that's actually keeping this object moving in a circle is really just the force of tension. So here's our ft, and that's the force of tension. When we expand our F=ma, the only force that's actually causing this to accelerate centripetally is our force of tension. Then we express this acceleration centripetal, A_{c}, as v2/r which is crucial.

In this example, the block completes a rotation every 4 seconds, allowing us to calculate that v is 2πrt. If we calculate using the given radius and period, we get a velocity of 3.14 m/s. Plugging this back into mv2/r, we get a tension force of 14.8 newtons. Just like that, using the same steps of free body diagram and F=ma, but now with additional equations involving the centripetal acceleration that we might have to solve for.