What Happens to Angular Speed When You Move on a Merry-Go-Round?

[Bambic87]

I worked on this problem in class and i couldn't finish it i did half of it but i couldnt' figure out some of it.
A 65 kg student is spinning on a playground merry-go-round that has a mass of 5.25 x 10^2 and a radius of 2.00 m. she walks from the edge of the merry-go-round tward the center. if the angular speed of the merry-go-rounf is initially .20 rad/sec, what is it's angular speed when the student reaches a point .50 m from the center? (assume that the merry-go-round is a solid disk and the person is a point source)

 

[BobG]

You find your moment of inertia by summing. The moment of inertia for a spinning disk winds up being:

$$I_d=\frac{mr^2}{2}$$

The student could be considered a point mass, his moment of inertia is:

$$I_s=mr^2$$

You add the two together to get your total moment of inertia.

Angular momentum is equal to:

$$H=I\omega$$
with omega being you angular velocity (or angular speed)

Angular momentum stays constant unless an external force acts on it. The student walking closer to the center is not an external force. The student and merry-go-round are part of the same system.

Recalculate the student's new moment of inertia in his new position. Add it to the merry-go-round's moment of inertia.

Divide the angular momentum by the new moment of inertia to get your new angular velocity.

Thinking of what happens when a spinning ice skater brings their arms and legs in tight gives you a common sense check as to whether your answer is at least feasible.

 

[Bambic87]

Thank you!