Answer:
The new angular velocity of the merry-go-round is 18.388 revolutions per minute.
Explanation:
The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:
[tex]\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f}[/tex] (1)
Where:
[tex]M[/tex] - Mass of the merry-go-round, in kilograms.
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the three children, in kilograms.
[tex]R[/tex] - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.
[tex]\dot n_{o}[/tex], [tex]\dot n_{f}[/tex] - Initial and final angular speed, in revolutions per minute.
If we know that [tex]M = 182\,kg[/tex], [tex]m_{1} = 17.4\,kg[/tex], [tex]m_{2} = 28.5\,kg[/tex], [tex]m_{3} = 32.8\,kg[/tex], [tex]R = 1.60\,m[/tex] and [tex]\dot n_{o} = 15.3\,\frac{rev}{min}[/tex], then the final angular speed of the system is:
[tex]\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)[/tex]
[tex]\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right][/tex]
[tex]\dot n_{f} = 18.388\,\frac{rev}{min}[/tex]
The new angular velocity of the merry-go-round is 18.388 revolutions per minute.
