8. Centripetal Forces & Gravitation
Centripetal Forces
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Hey, guys. I've got a really fun problem here for you. So the idea here is that we have this ball that's on a cord and we're 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. What we wanna do is we wanna calculate the ball's speed, its 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 wanna 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. Let me draw that a little bit bigger. So we have the weight force like this. And then we also have well, 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 2 dimensional. So we're gonna have to break this up into its x and y components. So we have a t y and then we have a tx. So what's going on here is that we know that this tx component or 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 our t x here actually points like this, and this is basically responsible for the ball's centripetal acceleration. This is ac equals v squared over r. So So if we wanna find out the v tangential, we're gonna have to use AC 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 centripetal direction. Now we only have one force to consider. Remember, our tx is the only force that's causing the ball to accelerate centripetally. So we've got our tx like this and this is equal to m and we have v squared over r. So this is our v squared over r that we're trying to find here. We're actually trying to find so v squared. So let's go ahead and write out an expression for tx. Right? So this TX here is gonna be related to, sine and cosine. So we're gonna have to break up this t here 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 kinda just draw out 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 is, 30, then that means that if this is a right triangle, then this has to be a 60 degree angle. And that's the one that we're gonna use for tx and ty. So this is really gonna be t cosine of 60 and ty is gonna be t times the sine of 60. Alright? So when we write this tx, we've got t times oops. Not sine. This is gonna be t times the cosine of 60, and this is equal to m v squared over r. So if we 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 r. Okay. So we're gonna have to go find both of those before we find the velocity. The easiest one to do is gonna be to solve the tension force because we know if we want the tension, we could always just go to the other axis. We can look at all the y 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 that 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 0 and your ty minus mg have to cancel out. So that means that your t times the sin of 60, d, right, that's the expression that we wrote here, has to equal mg. So we can do here is we can rewrite this equation and solve for t. This is really just gonna be 0.5 times 9.8 divided by the sine of 60, and you get a tension of 5.7. Alright? So now we actually do know what this tension is. Now what we have to do is go ahead 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 we take a look here, this radius is really just gonna be one of the sides of the triangle that we've built. We know that this is 4 meters. Right? That's the length of the cord. We also know that this angle here is 30 degrees. So if we have the hypotenuse and we have the angle, we 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 sine. So really, this is just gonna be 4 times the sine of 30. And so you go you work this out, this is gonna be 2. So now we're good to go here. So we basically have 5.7 times the cosine of 60 divided by oops. Actually, this is gonna be 5.60. So we're gonna move the radius up like this. So we're gonna multiply this by 2 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 square root of 11.4 and you get 3.38 meters 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. Alright? So that's it for this one guys. Let's move on.
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