A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center? Consider the merry-go-round is a solid 34 kg cylinder of radius of 1.6 m.
Answer in units of rad/s.

Hi there!

We can use the conservation of angular momentum to solve.

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

Begin by converting rev/sec to rad sec:

According to the above and the given information, we can write an equation and solve for ωf.

f is prop to m1xm2 double m1 double force

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