A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center

The given question is incomplete. The complete question is as follows.

A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0.6 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.

Explanation:

The given data is as follows.

    Initial angular velocity () = 0.14 rev/s

                        = 0.87 rad/s

Mass of man (M) = 84 kg,     Mass of cylinder (m) = 25 kg

Radius of the cylinder (r) = 2.4 m

Let be the initial distance of man from the axis of rotation = 2.0 m

Final distance of man from the axis of rotation () = 0.6 m

Formula to calculate the moment of inertia of cylinder is as follows.

        I =  

And, moment of inertia of man =

Therefore, initial moment of inertia of the system is as follows.

                =

                = 241.92 + 100

                = 341.92

Now, final moment of inertia of the system will be calculated as follows.

                     =

                     = 241.92 + 9

                     = 250.92

Now, according to the law of conservation of angular momentum is as follows.

      Initial angular momentum = Final angular momentum

or,  

                     =

                     = 1.18 rad/s

Thus, we can conclude that the new angular speed is 1.18 rad/s.

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