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concept

## Intro to Centripetal Forces

6m

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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.

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Problem

A small 4kg block is tied to the end of 3m string and slides around in a circle on a frictionless table. Suppose the string will break if the tension exceeds 50N. Find the maximum speed the block can have without breaking the string.

A

37.5 m/s

B

6.1 m/s

C

8.7 m/s

D

2.8 m/s

3

example

## Conical Pendulum

5m

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Hey, guys. Got a really fun problem here for you. So the idea here is that we have this ball that's on a cord and were spinning the ball around in a horizontal circle only. But we're doing this basically so that the cord always makes a 30 degree angle with the verticals. This is 30 we want to do is we want to calculate the ball's speed. It's tangential speed as it's going around in a circle like this. So this is really our VT. That's what we're trying to find in this problem. So how do we do this? Well, we're gonna go ahead and stick to the steps. We want to draw a free body diagram, and so let's go ahead and do that. So we've got the ball like this. We're gonna have the weight force that's acting straight down. We draw that a little bit bigger, so we have the weight force like this. And then we also have. There's no applied forces. There's no normal or friction, but there is a tension because of the cord. So we have this tension force here like this. Now, this tension force is two dimensional. So we have to break this up into its X and Y components. So we have a T y and then we have a T X. So what's going on here is that we know that this T X components for the X component of the cord as the ball is going around in a circle right is always gonna point towards the inside of the circle. So r TX here actually points like this and this is basically responsible for the balls centripetal acceleration. This is a C equals V squared over R. So if you want to find out the V tangential, we're gonna have to use a C because we know that's related to V squared over r, right. So we're gonna have to go into our f equals m a. Let's go ahead and do that. So our f equals m a in the centripetal direction. Now we only have one force to consider. Remember, R T X is the only force that's causing the ball to accelerate some trip Italy. So we've got our t X like this, and this is equal to M and we have v squared over R. So this is our V squared over r we're trying to find here. Uh, we're actually trying to find v squared, so let's go ahead and write out an expression for T X. Right, So this T X here is going to be related to, uh, sign and co signs. We're gonna have to break up this t here, Um, using sine and cosine equations. So how do we do that? Well, the angle that we were given here is this 30 degrees. So if we take a look, we kind of just draw a little right triangle like this. This is actually not the angle that we want, because this is with respect to the vertical. We always want our angles with respect to the horizontal. So we want this angle right here. And the idea here is that if we know that this angle is, uh, 30 And that means that if this is a right triangle and this has to be a 60 degree angle, and that's the one that we're gonna use for t x and T y. So this is really gonna be t co sign of 60 and t y is gonna be t times the sine of 60. All right. So when we write this T X, we've got t times What's not sign? Uh, this is gonna be t times the co sign of 60 and this is equal to M V squared over R. So if you take a look here, right, we're trying to find the velocity that's our final variable. But if we actually have a lot of other unknowns in this equation, we don't know what the tension is. We also don't know what the radius of that circle is. The radius would actually basically be this distance right here, which is our Okay, so we're gonna have to go find both of those before we find the velocity. The easiest thing to do is going to be solved the tension force, because we know if we want the tension, we can always just go to the other access. We can look at all the UAE forces, some of all forces in the Y axis equals m A Y. Now remember, this ball is only going around at a horizontal plane, meaning the only acceleration of this ball has is in the centripetal horizontal direction. It doesn't accelerate in the vertical, right? So what happens here is that these forces actually have to cancel. So we know that the acceleration equals zero and your t y minus mg have to cancel out. So that means that your tee times the sine of 60 right? That's the expression that we wrote here has to equal mg. So we can do here is we can rewrite this equation in Sulfur thi this is really just gonna be 0.5 times 9.8 divided by the sine of 60 and you're good attention of 5.7. All right, so now we actually do know what this tension is now we have to do is go and solve for the radius, right? That's the last unknown that we can go ahead and plug everything in. So how do we do that? Well, if you take a look here, this is Radius is really just gonna be one of the sides of the triangle that we've built. We know this is 4 m, right? That's the length of the chord. We also know that this angle here is 30 degrees. So if we have the hypotenuse and we have the angle, you can always find another side by using sine and cosine. Now we're trying to find the opposite side to the angle, so we're gonna use sign. So really, this is just gonna be four times the sine of 30. And so you you work this out. This is gonna be too. So now we're good to go here. So we basically have 5.7 times the co sign of divided by. Actually, this is gonna be 5.60. So we're gonna move the radius up like this, so we're gonna multiply this by two, and then we're gonna divide by the mass, and so we're gonna divide by 0.5. So this is basically what V squared is equal to. So then when you take the square roots, this really just becomes the squared of 11.4 and you get 3. m per second. So that's how fast this ball is traveling around in its horizontal path as you're basically dangling it like this, spinning in a circle. All right, so that's it for this one. Guys, let's move on

Additional resources for Centripetal Forces

PRACTICE PROBLEMS AND ACTIVITIES (10)

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