Answer:0.304 rev/s
Explanation:
Given
radius [tex]r=1.8 m[/tex]
mass of Merry go round [tex]M=120 kg[/tex]
[tex]N=0.4 rev/s[/tex]
[tex]\omega _0=2\pi N=2.51 rad/s[/tex]
mass of child [tex]m=37.5 kg[/tex]
Initial moment of Inertia of system
assuming all the mass is at periphery
[tex]I_0=Mr^2[/tex]
Final Moment of Inertia
[tex]I=Mr^2+mr^2[/tex]
[tex]I=(M+m)r^2[/tex]
Conserving angular momentum
[tex]I_0\omega _0=I\omega [/tex]
[tex]\omega =\frac{I_0}{I}\times \omega _0[/tex]
[tex]\omega =\frac{Mr^2}{(M+m)r^2}\times \omega _0[/tex]
[tex]\omega =\frac{120}{120+37.5}\times 2.51[/tex]
[tex]\omega =1.91 rad/s[/tex]
[tex]N=\frac{1.91}{2\pi }=0.304\ rev/s[/tex]
