Answer:
[tex]\dot n = 6.042\,rpm[/tex]
Explanation:
The final angle speed of the merry-go-round is determined with the help of the Principle of Angular Momentum Conservation:
[tex](270\,kg\cdot m^{2})\cdot \left(8\,rpm\right) = [270\,kg\cdot m^{2}+(27\,kg)\cdot (1.80\,m)^{2}]\cdot \dot n[/tex]
[tex]\dot n = 6.042\,rpm[/tex]
