Answer:
The new angular speed of the merry-go-round is 8.31 rev/min.
Explanation:
Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):
[tex]L_f=L_i [/tex] (1)
The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:
[tex]L_i=L_{mi}=I\omega_i [/tex] (2)
with I the moment of inertia and ωi the initial angular speed of the merry-go-round
The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):
[tex]L_f=L_{mf}+L{cf}=I\omega_f+L{cf} [/tex] (3)
The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:
[tex]L{cf}=mRv_f [/tex] (4)
with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:
[tex]v_f=\omega_f R [/tex] (5)
(note that the angular speed is the same as the merry-go-round)
using (5) on (4), and (4) on (3):
[tex]L_f=I\omega_f+m\omega_f R^2 [/tex] (6)
By (5) and (2) on (1):
[tex]I\omega_f+m\omega_f R^2=I\omega_i [/tex]
Solving for ωf (12.0 rev/min = 1.26 rad/s):
[tex]\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2} [/tex]
[tex]\omega_f=0.87\frac{rad}{s}=8.31 \frac{rev}{min} [/tex]
