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
) = 0.14 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
 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
) = 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.