Mr.p: Good morning. Bobby, please read the problem and Billy, please translate.
"Flipping Physics" Intro
Bobby: A 25 kilogram child is sitting on the edge of a merry-go-round.
The merry-go-round has a mass of 255 kilograms and is rotating at 2.0 radians per second.
Billy: The mass of the child is 25 kilograms, the mass of the merry-go-round is 255 kilograms,
and the angular velocity of the merry-go-round is 2.0 radians per second. Bobby?
Bobby: The child crawls to the middle of the merry-go-round.
What is the final angular speed of the merry-go-round?
Billy: Well, the child goes from the outside edge to the center of the merry-go-round
and the final angular velocity of the merry-go-round equals question mark.
Bobby: You may make the following estimations:
the child is a point particle, the merry-go-round is a solid disc, and has an axle with negligible friction,
and the rotational inertia of a disc about its axis equals
one half the mass of the disc times the radius of the disc squared.
Billy: Okay. The child is a point particle,
the coefficient of friction of the axle of the merry-go-round is roughly zero,
and we have that equation for the rotational inertia of the disc about its axis.
Mr.p: Just so you know, I am using the letter c for child and the letter w for wheel or merry-go-round.
Okay, as the child crawls from the outside edge of the merry-go-round to its center,
assuming the axle of the merry-go-round is the axis of rotation,
the angular momentum of the child and merry-go-round system is conserved. Bo, please explain why that is.
Bo: Sure. Because net torque equals change in angular momentum over change in time,
we know angular momentum is conserved when the net external torque acting on the system equals zero.
There is no friction on the axle of the merry-go-round, so,
because it equals zero, friction on the axle does not cause an external torque on the merry-go-round.
As the child crawls toward its center, there is a force of static friction from the merry-go-round on the child and,
and according to Newton's third law,
an equal but opposite force of static friction from the child on the merry-go-round.
Because the axis of rotation is at the center of the merry-go-round,
those two forces caused equal but opposite torques on the child and merry-go-round system,
which means that two torques are internal to the system and cancel one another out.
So the net torque on the system is zero
and angular momentum, about the axle of the merry-go-round, is conserved.
Mr.p: Very nice Bo. Now there is no direction given in this problem and we are solving for angular speed.
So we are going to drop the vector symbol from the equations
and solve for the magnitude of the final angular velocity.
Therefore, conservation of angular momentum for this problem means
the sum of the initial angular momenta of the child and wheel
equals the sum of the final momenta of the child and wheel.
We know a rigid object with shape
has an angular momentum equal to its rotational inertia times its angular velocity.
It is also true that, when a point particle is moving in a circle,
the point particle has an angular momentum also equal to its rotational inertia times its angular velocity.
For now, let's assume this to be true, however, we will prove this to be true in a later lesson.
That means we can substitute rotational inertia times angular velocity
in for each angular momentum in the equation.
But notice we do not need subscripts for initial and final for the rotational inertia of the wheel
because its rotational inertia does not change.
However, we do need initial and final subscripts for the rotational inertia of the child,
because the location of the child changes
and therefore, the rotational inertia of the child changes during the problem.
Bobby, what is the equation for the rotational inertia of a system of particles?
Bobby: The rotational inertia of a system of particles
equals the mass of the particle times the square of the distance the particle is from the axis of rotation.
Oh, actually it's the sum of that term, one for each particle in the system.
Bo: Right, but we are treating the child as a single point particle, therefore, the rotational inertia of the child
equals just the mass of the child times the square of the distance the child is from the axis of rotation.
Billy: And, because they are moving together,
the angular velocities of the child and the wheel are going to be the same.
So we can substitute in final and initial angular velocities into the equation.
Bo: Let's use capital R for the radius of the wheel.
Bobby: That means the initial distance the child is from the axis of rotation
is the radius of the wheel or capital R
Bo: And the final "r" value for the child is zero, because the child is at the center of the wheel.
BBB: Everybody brought capital R squared to the party!
"Everybody brought mass" lyric
Mr.p: Then we can multiply through by 2, factor out initial velocity, and divide by the mass of the wheel.
That means the final angular velocity of the system equals
the initial angular velocity of the system times the quantity 2 times the mass and the child
plus the mass of the wheel all divided by the mass of the wheel.
Therefore, the final angular velocity of the system equals
2 times the quantity 2 times 25 plus 255 all divided by 255 which equals 2.39216
or 2.4 radians per second with 2 significant digits.
Okay, let's just take a step back for a moment and see if it makes sense that the angular velocity of the system
increased when the child moved to the center of the merry-go-round.
Bo, what are your thoughts on that?
Bo: I'm not ... I don't
Bobby: I think we need to go back to the conservation of angular momentum equation
near the beginning of the solution.
Bo: Okay, angular momentum is conserved.
The rotational inertia of the wheel remains unchanged throughout the whole event.
But as the child moves in toward the axis of rotation, the rotational inertia of the child decreases,
therefore, in order to maintain a constant angular momentum of the system,
the angular velocity of the system has to increase.
Billy: Right. The angular momentum of the system does not change,
so if the rotational inertia of the system decreases, the angular velocity of the system must increase.
That totally makes sense.
Bo: Yeah. Mr.p: Very nice.
Now, I want to return to the fact that we are treating the child as a point particle in this problem.
This means our child has zero rotational inertia when sitting at the center of the merry-go-round.
Hopefully you realize this will not be true in real life because the child has nonzero size and
therefore will have rotational inertia, even when sitting directly at the axis of rotation.
So this is a simplified solution, but it certainly helps with understanding.
And if you find a frictionless merry-go-round, please let me know.
You know what, one more thing. Let's determine the change in kinetic energy of the system.
There is no translational motion of the system, so all of the kinetic energy is rotational kinetic energy.
The change in kinetic energy of the system equals
the final kinetic energies of both objects minus the initial kinetic energies of both objects.
We can substitute in one half rotational inertia times angular velocity squared
for every rotational inertia term.
We already showed the final rotational inertia of the child equals zero.
The rotational inertia of the merry-go-round equals
one half times the mass of the wheel times the square of capital R, the radius of the wheel.
And we already showed that the rotational inertia of the child initial equals
the mass of the child times, again, the square of capital R, the radius of the wheel.
We can factor out capital R squared, multiply through by both of the one halves, and plug in numbers to show
that the change in kinetic energy of the child and merry-go-round system equals 60 times capital R squared,
or 6.0 times 10 to the first because we need 2 significant digits.
But I want to know why the change in kinetic energy of the system is positive.
Bobby: Uh... What?
Billy: Why is the change in kinetic energy of the system positive?
Bobby: Okay?
Bo: Well, the kid has to do work to pull their body into the center of the merry-go-round.
I mean, have y'all ever tried this?
You actually have to apply a pretty large force to move your body to the center of the merry-go-round.
I can totally see that the kid has to do work on their body and therefore the system to move to the center
of the merry-go-round.
Ryan: This is hard.
Bobby: But what does work have to do with kinetic energy?
Billy: Oh, net work equals change in kinetic energy.
The change in kinetic energy we just determined equals the net work on the child and merry-go-round system.
Mr.p just solved for the change in kinetic energy of the system,
and therefore also the net work done on the system.
The net work is positive because the child had
to do work on the system to increase the kinetic energy of the system.
Bobby: Angular momentum was conserved, but kinetic energy was increased. Cool.
Bo: But why is there a capital R in our answer?
Billy: The the radius of the merry-go-round was never defined in the problem.
That just shows that a larger wheel radius will mean a larger change in kinetic energy
and larger net work done on the system.
Bo: Right. Thanks. Billy: You are welcome.
Mr.p: Billy, Bobby, Bo, really good job today.
Thank you for learning with me today. I enjoyed learning with you.