 Hello everybody. My name is Brad Langdell. I'm going to talk to you about vertical circular motion and this video is brought to you by uniform circular motion, which is the unit that this topic comes up in. And so we're talking about things that are in motion. Okay, that's good in physics. We like things that move. We're talking about things that have a circular motion, which means they move in a circle, and then it's uniform as well. The uniform part is we're talking about the speed. The speed of these objects is the same throughout their entire motion. So it's uniform circular motion. And here's the problem that I'd like to give you today. A shoelace can hold a force of 135 Newtons before it breaks. If a 2 kilogram brick is tied to the end of the shoelace, and the total length we've got to deal with here is 1.10 meters, how fast can the brick be spun vertically before the string breaks? So here's kind of the situation that we've got here. Here's the person who has got this brick on a string, and there's their arm, and there is the string, and the string is 1.1 meters long. And on the end of the string is a brick. And the brick is being spun in a vertical circle kind of like this. Good thing it missed that guy's growing. And you can see the length of the string is the same as the radius of the circle. Okay, that's our radius. Now, we're actually interested at the point where the string breaks. So we have to start thinking about where is the string going to break? The string is not going to break here where I drew the brick, is not going to break at the top or anywhere else. The string is going to break more than likely down here at the bottom. It's going to break at the bottom because here, something very interesting is happening to the forces acting on the brick. Well, let's talk about those forces. We've got a force of gravity acting downwards on the brick. And then we also have a force of tension acting on this brick from the rope. And these aren't necessarily drawn to scale, okay? So here we've got kind of a tug of war between our forces. And the fact that gravity is acting on the tension at this point, that they're acting in opposite directions, is what leads to the force of tension being overcome and the string eventually breaking. Okay, so we want to find how fast this brick is going when that string breaks. So if we're talking about that, and we're talking about a unit about uniform circular motion, we're probably going to be talking about the centripetal force, the force which dictates how objects move in a circle. And the centripetal force is always equal to some other force. I tell my kids in class it's equal to some other force, or in this case, a pair of forces. In this case, the centripetal force is equal to the force of gravity. So I'll write that in there, force of gravity. But it's also equal to the force of tension. The centripetal force is kind of like the net force acting on this brick. Okay, now it's substitution time. Let's actually get to putting some numbers in. So in place of centripetal force, I'm going to put in the equation for centripetal force, or one of them anyways, mv squared over r. Where m is the mass of the brick, and v is the velocity I'm spinning it at, and r is the radius. For force of gravity, I'm going to put in mg, mass times the acceleration due to gravity. And I actually know what the tension is, it's 135 newtons, so I'll just stick that in. Now I can start to do a few other substitutions here really quickly, so I can put in my mass, that was two kilograms. I'm looking for my velocity, so I'll leave that as a v. Got my radius, that was given in the question. Got my mass, and the acceleration. Now the one thing I got to be careful of here, and that a lot of students make mistakes with, in my experience, is vectors up and down, plus and minus, the force of gravity is going down. So I got to make sure that I make this force of gravity vector negative. When I'm done, the force of gravity, the mg, the force of gravity will be a negative number, which is what my diagram tells me it should be. So now I can go and I can type it all through my calculator. Out comes the fancy graphing calculator, let's type this through. So I'm going to start by going 2 times negative 9.81 plus 135. So there's my whole right hand side of the equation. Now I'm going to times that by 1.1 to move the denominator over to the other side. I'm going to divide it by 2 to get v squared by itself, and then I'm going to square root. And I don't want that negative sign in there now that I think about it, so I guess I'll take that out. So this object, this brick is going to move at 7.97 meters per second. Yes, and we do want three siktics. So there is a great example of a vertical circular motion type problem. Kids, you're going to see these in the future, so practice, practice, practice. If you'd like more information about vertical circular motion, you can check out my package of notes at LDindustries.ca.