Problem

Three children are riding on the edge of a merry-…

03:15

Problem 39 Hard Difficulty

A playground merry-go-round has a mass of 120 $\mathrm{kg}$ and a radius of 1.80 $\mathrm{m}$ and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0 -kg child
gets onto it by grabbing its outer edge? The child is initially at rest.

Answer

$a ^ { \prime } = 0.366 \mathrm { mm } / \mathrm { s }$

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Physics 101 Mechanics

College Physics for AP® Courses

Chapter 10

Rotational Motion and Angular Momentum

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Video Transcript

So we're going to say of the angular moment of the initial angular momentum equals the final angular momentum. This is gonna be the initial moment of inertia times. The initial angular velocity equals the final moment of inertia times the final angular velocity. Now we can say that the initial angular velocity is equaling 0.5 00 revolutions per second, multiplied by two pi radiance for every rather for every revolution. And this is equaling. Essentially pie. Ah, radiance per second. And so we can say that then, Ah, the initial moment of inertia is simply m r square. The massive the merry go round times the radius of the merry go round square divided by two. This is, ah, modeling the merry go round as a disc. And then for the final ing moment of inertia, this would be m r squared over two plus treating them the child is a point mass. This would be m r squared. So the massive child times the distance between the child and the axis of rotation squared. This is, let's say, actually, m lower case R squared and capital are denotes the radius of the merry go round itself. Now we can see that the final angular velocity would be equaling the initial moment of inertia times the initial angular velocity divided by the final moment of inertia. And this is gonna be equaling. We can say r squared over two times the initial angular velocity. This would be divided by M r. Squared over two plus lower case and lower case R squared And so we can solve the final angular velocity would be equaling. This would be 120 kilograms. The mass of the merry go round times the radius of the merry go round squared multiplied. Rather, this would be divided by two multiplied by pi radiance. 1st 2nd and then this would be divided by the same thing. 120 kilograms multiplied by 1.80 meters. Quantity squared. This is divided by two plus the mass of the child of 22 kilograms. And he's standing on the edge of the merry go round, so this would be multiplied by 1.80 meters quantity squared, close parentheses close the denominator and we find that the final angular velocity is equaling 2.30 radiance per second. this would be our final answer makes sense because the final angular velocity should be less than the initial angular velocity. Given that the final moment of inertia is greater than the initial moment of inertia. So 2.30 radiance per second, that is that makes sense as well. That is the end of the solution. Thank you for what?

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