Suppose that a merry-go-round, which can be approximated as a disk, has no one on it, but it is rotating about a central vertical axis at 0.2 revolutions per second. If a 100kg man quickly sits down on the edge of it, what will be its new speed? (A disk has a moment of inertia I=(1/2)mR2 , mass of merry-go-round = 200kg , radius of merry-goround=6m)

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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

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