Part one
a merry-go-round rotates at the rate of
0.19 rev/s with an 97 kg man standing at
a point 2.5 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 63 kg
cylinder of radius of 2.5 m.
answer in units of rad/s

part two
what is the change in kinetic energy due to
this movement?
answer in units of j.

The  new angular speed when the  man walks to a point 0 m from the center  is 4.85 rad/s.

The change in kinetic energy is 1746 J.

Given data:

The rate of revolution of merry go-round is, n = 0.19 rev/s.

The mass of man is, M = 97 kg.

The distance of man from axis of rotation is, d = 2.5 m.

The mass of merry-go round is, m = 63 kg.

The radius of cylinder is, r = 2.5 m.

(1)

The problem can be solved by applying the law of conservation of angular momentum: in fact, the total angular momentum of the system must be conserved,

Here, is the new angular speed. and where the angular momentum of the man is zero because he is at the axis of rotation.

Solving as,

Thus, we can conclude that the  new angular speed when the  man walks to a point 0 m from the center  is 4.85 rad/s.

(2)

The expression for the change in kinetic energy is,

Kinetic energy change = Final kinetic energy - Initial kinetic energy

Thus, we can conclude that the change in kinetic energy is 1746 J.

Learn more about the Angular momentum here:

link

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

where is the circuit diagram ?