Answer:
0.1 rev/s
Explanation:
M = mass of the merry go round = 200 kg
R = radius of merry go round = 6 m
[tex]I_{o}[/tex] = Moment of inertia of merry go round = (0.5) MR² = (0.5) (200) (6)² = 3600 kgm²
m = mass of the man = 100 kg
[tex]I_{f}[/tex] = Moment of inertia of merry go round when man sits on it at the edge = (0.5) MR² + mR² = (0.5) (200) (6)² + (100) (6)² = 7200 kgm²
[tex]w_{o}[/tex] = initial Angular speed of merry-go-round before man sit = 0.2 rev/s
[tex]w_{f}[/tex] = Angular speed of merry-go-round after man sit = ?
Using conservation of angular momentum
[tex]I_{o}[/tex] [tex]w_{o}[/tex] = [tex]I_{f}[/tex] [tex]w_{f}[/tex]
(3600) (0.2) = (7200) [tex]w_{f}[/tex]
[tex]w_{f}[/tex] = 0.1 rev/s