A playground merry-go-round has radius 2.40 m and moment of inert...

Question

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg•m^2 about a vertical axle through its center, and it turns with negligible friction. A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest (b) How much work did the child do on the merry-go-round?

Accepted Answer

Welcome back everybody. We are taking a look at a flywheel that is pivoted on a vertical axle here and we are told a couple of different things. We are told that the moment of inertia generated by the flywheel is 2200 kg meter squared. We're told that it has a radius of 1.8 m. Now, an engineer who is studying the flywheel, he applies a tangential force of 35 newtons to it and he does that for a period of eight seconds here. Now we are told that the flywheel initially starts out at rest meaning that its angular velocity is going to start out at zero. And we are tasked with finding how much work is done by the engineer. Now for in order to figure out work, we're gonna have to use this formula, we're going to have to use that the work is equal to tau times our change in theta. Now we don't have either of these. So we're gonna have to calculate both of them. And we're gonna use some cinematics and and some other formulas as well. But let's start here. We know that our tau is equal to the moment of inertia, times our angular acceleration. And we also know that this is equal to our radius times our force are tangential force times the sine of the angle feta. Now, what is this data in this case? In this case it is simply just the angle between The centripetal forces there and the axis of rotation and for us that is just a 90° angle and just to visualize it real quick. You have a flywheel like this. You have an axle like this. This is your angle that we're looking at right here. Okay, so here's what I'm going to do. I am going to solve for angular acceleration because then through that we will be able to solve both for our delta theta using cinematics and then go back and solve for tau as well. I'm gonna divide both sides of this equation here by our moment of inertia, canceling that out on the left and we get that our angular acceleration is equal to our radius times are tangential force times the sine of our angle Theta all over the moment of inertia. So let's go ahead and plug in some values and find this angular acceleration we have that are angular acceleration is equal to our radius of 1.8 times are tangential force of 35 newtons times the sine of our angle theta, which we discussed to be 90 degrees all over our moment of inertia, which is 2200. When you plug this into your calculator, you get that are angular acceleration is 0. radiance per second. Great. So now what I'm going to do is I'm going to solve for our overall change in theta. And we have a formula for this kid, a magic formula. We have that our change in theta is equal to our initial angular velocity times time plus one half hour angular acceleration, times our times squared. And what is this going to equal? Well, we know that Our initial angular velocity is zero. So this first term will just be zero and then we will add one half times the angular acceleration of 0.028, six times our time of eight seconds squared. Which when you plug this into your calculator, you get that. Our angular acceleration is 0.915 radiance per second squared. Great. So now let's go ahead and use this relation that are tao. Let's see if I can't clear the the little scratch marks here that are tau is equal to our moment of inertia. Times our angular acceleration. So we have that tau is equal to i alpha is equal to 2200 times 0.286. Which when you plug this into your calculator, you get 63 newton meters. Great. So now we have both our tao and we have our change in our angle data and we can go ahead and compute the work done by the engineer. So our work is times delta theta. So we have 63 times 0.915. Which when you plug into your calculator gives us a final answer of 57. jewels of work done by the engineer corresponding to our final answer. Choice of d. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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