A 20 kg child is on a swing that hangs from 3.0-m-long chains. Wh...

Question

A 20 kg child is on a swing that hangs from 3.0-m-long chains. What is her maximum speed if she swings out to a 45 degree angle?

Accepted Answer

Hey, everyone in this problem, we have an iron ball with a mass of 50 kg tied to the ceiling using a two m long chain. And we are asked to determine the top speed of the iron ball. If it's released at an angle of 30 degrees with the vertical, we're given four answer choices all in meters per second. We have option A 5.25, option B 2.29, option C 2.68 and option D 5.52. So let's start by drawing out the situation that we have. So we have the ceiling and we have a two m long chain attached with a ball on the end of it. This chain again is two m and we know that this is released at an angle of 30 degrees. And so the initial velocity V knot is gonna be zero m per second. Now, this chain again, it makes an angle of 30 degrees with the vertical. We have the ground or the base line somewhere down here. No, we wanna find the top speed of this bow. When is that gonna occur? Well, we call that the top speed will occur when the angle is zero degrees. OK? When this ball is at the very bottom completely vertically. Yeah. So let's draw that situation. OK. So again, we have the ceiling, we have the two m long chain, we have the ball attached to it and this is where we're gonna get V max that maximum speed that we're looking for. Now, we're gonna consider that this height of the ball at the maximum height is a height zero. If this is height zero, then the height of the ball when it makes a 30 degree angle. OK. Or the distance from this height zero line two, the height of the ball initially is gonna be up and that's what we're gonna call it. All right. So we have some height, we have some speed in this problem. Let's think about our conservation of mechanical energy in conservation of mechanical energy. We know that the initial mechanical energy which is knot plus U knot, the initial kinetic energy plus the initial potential energy is equal to K F plus the U F, the final kinetic energy plus the final potential energy we're called that kinetic energy is given by one half M V squared. So each of our K terms we can replace with those variables, our potential energy. In this case, we don't have any springs acting here. So the potential energy is just gonna be the potential energy due to gravity. And we're called that, that's given by M G H. And so our equation is gonna become one half M V knot squared plus M G H knot is equal to one half M V F squared plus M G H F. Let's simplify where we can, we know the initial speed is zero. So the kinetic energy initially is going to be zero as well. That first term on the left hand side goes away, we know that the final height H F is gonna be zero because that's how we've defined our height. So the gravitational potential energy on the right hand side is gonna be zero. And that last term on the right hand side goes away, we can also divide both sides by the max M and that is going to divide out. And so what that tells us is that the result we get actually doesn't depend on that mass. It doesn't matter what the mass is, we're gonna get the same result. So what we're left with here is G multiplied by each knot is equal to one half V F squared. We want to solve for V F, we know the acceleration due to gravity G. But what are the H not? How can we actually calculate? Not, let's go ahead and do that. So we've said that H knot is a distance from this height equals zero line up to where we have our ball initially. We're gonna say that the rest of the height is a distance D, OK. And you can see that D is gonna be one side of our triangle where the hypotenuse is that two m long chain? Now, what we can notice is that two m, the entire length from the ceiling down to our H equals zero m line is gonna be equal to that distance D from our triangle plus that initial height H knot. So if we wanna find H knot, well, it's gonna be equal to two m minus uh distance D. How do we find that distance? Do? Well, we have our triangle, let's draw that out on its own. And we have a triangle with a hypotenuse of two m. It makes an angle of 30 degrees with the vertical and that vertical side has length D OK. So in order to calculate D, it's gonna be related to the angle through cosine because it's the adjacent side. So we have that cosine of 30 degrees. It's going to be equal to the adjacent side. D divided by the hypotenuse of two m, which tells us that that distance D is gonna be equal to two m multiplied by cosine of 30. All right. So we have a distance. So now we can write H knot our height, it's gonna be two m minus two m multiplied by cosine of 30. Substituting this into our equation. We have G which is 9.8 m per second squared, multiplied by that initial height. H nought again, two m minus D, which is two m multiplied by cosine of 30 degrees. And this is gonna be equal to one half V F squared. Let's give ourselves some more room here. We have that V F square and let me rewrite V F. We know that V F is actually gonna be the maximum speed. So let's write it as V max squared. Since V max is what we're looking for, we're gonna multiply both sides by two. In order to simplify, we get two multiplied by 9.8 m per second squared, we can factor two m out from each term in knot. So we have two m and that's multiplied by one minus cosine of 30 degrees. All right. Now we're gonna take the square root. V max is gonna be equal to the square root of everything we had before. OK. So we have two multiplied by 9. m per second squared, multiplied by two m multiplied by one minus cosine of 30 degrees. And when we take the square root, we're gonna get both the positive and the negative route that's gonna be equal to plus or minus 2. m per second. Now, what we have to do is interpret that and figure out whether we should be taking the positive or the negative route. In this case, we're asked just to find the speed. OK? So the direction doesn't matter, we want to just have the magnitude of that value. And so we are just going to take the positive route of 2.29 m per second. And that is gonna be the top speed of that ball that is hanging from that chain. If we compare this to our answer choices, we can see that this corresponds with answer choice B 2. m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

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A 20 kg child is on a swing that hangs from 3.0-m-long chains. What is her maximum speed if she swings out to a 45 degree angle?

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