(II) A person of mass 75 kg stands at the center of a rotating...

Question

(II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia 920. kg·m² . The platform rotates without friction with angular velocity 0.95 rad/s . The person walks radially to the edge of the platform.
(b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person’s walk.

(b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person’s walk.

Accepted Answer

Hey, everyone in this problem, we're told that a 42 kg dog is at the center of a large disk of radius 6 m. Initially, the disc rotates horizontally with an angular velocity of 0.82 radiance per second about a vertical axis through its center. Afterward, the dog walks directly to the edge of the disk. And we're given that the moment of inertia of the disc is 870 kg meters squared. And we're asked to determine the total rotational kinetic energy of the system with the dog initially on the disc at the center first. And then we're gonna do the same thing for when the dog walks to the edge of the disk. And we're told to assume that the disc rotates without any friction. We're given four answer choices in this problem A through D all of jewels and each of those answer choices contain a different value for the rotation of kinetic energy initially. And finally, we're gonna come back to these answer choices as we work through the problem. The first thing we're gonna do here is draw out what we have going on. OK. So if we think about this disc, it's rotating horizontally. So we're gonna look at it from a top down view. And so this is kind of a top down view if we were over top of it, and this disc is gonna rotate about its center. We know that initially it has this speed angular velocity story of 0.82 radiance per second. OK. So we can say that omega not the initial angular velocity is 0.82 radiance for a second, we know our disk has a radius of 6 m. Now, initially, we're gonna have our dog at the center. Yeah. So this is our dog initially. And then he's gonna walk out to the edge and at some final point, he's gonna be on the edge of that disk. All right. So we're asked to find rotational kinetic energy. And let's recall that a rotational kinetic energy krot this is just one half multiplied by the moment of inertia. I multiplied by omega squared. Now, the moment of inertia of the disk we're given and we're given that the moment of inertia eye for the disk is 870 kg meters squared. So let's start with the initial, OK. And when the dog is at the center, when the dog is at the center, and our rotational kinetic energy is just gonna be one half multiplied by that initial moment of inertia multiplied by the initial angular speed squared. And we have, oh my God not, we were given that in the problem. And we think about I knot, this moment of inertia and the moment of inertia, I knot is gonna be made up of the moment of inertia of the disc which we know was the moment of inertia of the dog. Now, the dog we can treat as a point mass. OK. So the moment of inertia is gonna be 870 kilogram meters squared. The moment of inertia of a point mass, we can look this up in a table in our textbook, that's gonna be the mass multiplied by the distance squared with that distance D is the distance of that point mass to the axis of rotation. Now, in this case, the dog is at the center of the disc, the disc is rotating about its center. And so the distance between the dog, the axis of rotation is just zero. And so the dog is not going to contribute to the moment of inertia initially. And so this term goes to zero. So if we think about our rotational kinetic energy, then it's just gonna be equal to one half multiplied by 870 kg meters squared, multiplied by 0.82 radiance per second, all squared, we work this out on our calculator and we get that our initial rotational kinetic energy is gonna be about 292.494 joules. If we round this to two significant um digits. Like the answer choices have done, we can write this as approximately 2.9 multiplied by 10 to the exponent to joules. OK. So this is the answer for the first part of that question. If we take a look at the answer choices we were given, we can eliminate some of them already. So option A and option B have that this rotational kinetic energy initially is 2.9 multiplied by 10 to the exponent two joules just like we found. Option C has 3.6 multiplied by 10 to the exponent two. And option D has 5.9 multiplied by 10 to the exponent two. So option C and option D, we can eliminate, we're left with option A and B. Now we can do that final kinetic energy to figure out which one it's going to be. So if we think about this final rotational kinetic energy, the process is gonna be similar, we're gonna have to do a little bit of extra work here. OK. So our final rotational kinetic energy we call it KF, this is gonna be one half I and let's call it K rotational app. So it's super clear what we're finding um I final omega final squared. Now this final moment of inertia, we can calculate it in a similar way to what we had done. And so this final moment of inertia, it's just gonna be the moment of inertia of the disc. Again, we know that plus the moment of inertia for the dog. OK. So we get 870 kg meters squared plus the mass of the dog multiplied by its distance to the rotational axis. Now, because the dog is at the edge of the disk, the distance between the dog and the rotational axis that is at the center of the disc is gonna be the same as the radius. OK. So the value D is just gonna be the radius of that disk. So we have that this will be 870 kg meters squared plus the mass of the dog, 42 kg multiplied by the radius of them um disc 6 m squared. And when we work this out, we get to this um moment of inertia in this final case, is gonna be 2382 kg meters squared. OK. So we have this much larger moment of inertia now. OK. So we have I, but what about Omega? We weren't given that final speed and when the dog walks out to the edge of that disk, it's going to change how fast that disk is spinning. So what we can do and we're told that we can assume the disc rotates without any friction, we can assume we have a conservation of momentum. And because we have this conservation of angular momentum, they were spinning. So we're dealing with angular momentum. That means our initial angular momentum L not will be equal to our final momentum angular momentum. Sorry LF. And you know, we're called that angular momentum is given by the moment of inertia eye multiplied by the angular velocity omega. So we get I knot, Omega knot is equal to if omega f. Now we have everything in this equation except for omega F. So we can just substitute in and solve for Omega. OK. Our initial moment of inertia at 870 kilogram meter squared multiplied by our initial angular velocity, 0.82 gradients per second. This is gonna be equal to that final moment of inertia which we just found 2382 kg meters squared multiplied by Omega F dividing both sides by 2382. Be it that our final angular velocity is approximately equal to 0.2995 radiance per second. All right. So what we found is that when the dog walks to the edge of this disk, the mo moment of inertia is going to increase, but the speed the angular velocity is going to decrease. OK. So let's see what that does for our rotational kinetic energy. A rotational kinetic energy now is going to be equal to one half multiplied by 2382 kg meters squared multiplied by 0.2995 radiance per second all squared. And when we work this out. We get that, that rotational kinetic energy is about 106.833 jewels. If we round to two significant digits, we write this in scientific notation like those answer choices, we can write this as 1.1 multiplied by 10 to the exponent two joules. So that is the final answer for the second part of this question. We can now go up and compare what we found. We had already narrowed it down to option A or option B. And now we can see that the correct answer is going to be option A. The total rotational kinetic energy initially is 2.9 multiplied by 10 to the exponent two joules. And the total rotational kinetic energy finally is 1.1 multiplied by 10 to the exponent two joules. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Rotational Kinetic Energy

Moment of Inertia

Conservation of Angular Momentum

Watch next