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) 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, what is its angular speed after this 15.0-s interval?

Video transcript

Hi, everyone. Let's take a look at this practice problem dealing with rotational motion. This problem says a rotating disc with a radius of 2.5 m and a moment of inertia of 2200 kg meters squared about its central vertical axis is initially at rest. A person applies a tangential force of 20 mutants to the edge of the disk for 12 seconds. Calculate the angular velocity of the disk after this 12 2nd period. We're given four possible choices as our answers. Choice A is 0.27 radiance per second. Choice B is 0.34 radiance per second. Choice C is 0.56 radiance per second and choice D is 0.72 radiance per second. Now we're asked to calculate the angle of velocity and we're given a time here. So the first thing that comes to mind is to use our rotational kin kinematic equations recall that our angular velocity omega is going to be equal to alpha T plus omega knot. Here, alpha is our angular acceleration is time and a mega knot is our initial angular velocity. Now, as given the initial angle of velocity, we're told it's at rest, we're also given the time, but we don't have the angular acceleration. So I need to find a different way to calculate that. And since I was given a force and the moment of inertia, we're going to actually use our equation that relates the net torque moment of inertia and angular acceleration together. So it's basically the new second law version of rotational motion to recall that our net torque Sigma Tau is equal to I multiplied by alpha. Here tell is our torque I is the moment of inertia and alpha is our angular acceleration. So I was given the moment of inertia, but I also need to calculate the torque. And so recall our definition of torque Tau is going to be equal to RF sign the or R here is the distance from the point of rotation to the uh force F is the force and theta is the angle between the vectors R and F. So I'm gonna plug that in or our net torque since that's the only torque that's acting on this disc will have RF nine. The and that's gonna be equal to I multiplied by alpha. So I'm gonna solve this for alpha, we have alpha is equal to RF line data divided by I. So I can now take that and plug it back into our kinematic equation. We'll have OMEGA equal to the quantity of RF nine the divided by I that quantity is multiplied by T plus omega knot. Now, if I look at the quantities I was given on the right hand side, those are all known quantities that was given in the problem so I can plug in their values. So I have Omega is equal to or R. We're told that um to R is the distance from the force to the axis rotation. We're told that the force is being applied at the edge of the disk. That means that the R here, in this case is just by radius, which is the 2.5 m that gets multiplied by the force which is 20 newtons and that gets multiplied by the sign. And in this case, we're told the person is applying the force tangentially to the edge of the disk. That means the angle between the R and F vector. In this case is 90 degrees that product gets divided by the moment of inertia which we were given that is the 2200 kilogram meters squared. And that fraction gets multiplied by the time which we were told was 12 seconds. And then we add to that our initial um angular velocity, which we're told it starts from rest. Our initial angular velocity is zero radiance per second. So I just have values on the right hand side of my equation now, so I can plug those into my calculator. And it turns out that Omega is equal to 0.27 radiance per second. And here I just kept two significant figures. And if you're cur curious where the radiance came from, remember that we're dealing with an angular quantity. And if you look at the units um from our previous expression, then you'd actually get units of one over seconds. But since we're dealing with an angular velo uh the angular velocity, here, we have that hidden units of radiance. So now I have my numerical value um I can compare that to the answer choices and this corresponds to choice a so quick little recap. Here. We started off by using our um formulas for the rotational kinematics. We're looking specifically for the angular velocity one and we were still missing the angular acceleration of that equation. So we had to use our net torque is equal to, I multiplied by alpha equation in order to find the angular acceleration. Once we did that, we could plug it in and calculate our final angular velocity. So I hope that this has been useful. And I'll see you in the next video.