Skip to main content

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

Related Videos

Related Practice

© 1996–2023 Pearson All rights reserved.