Answer:
[tex]\omega=4.7916\ rad.s^{-1}[/tex]
Explanation:
Given:
- moment of inertia of the skater with extended arms, [tex]I'=2.5\ kg.m^2[/tex]
- moment of inertia of the the skater with pulled-in arms, [tex]I=1.2\ kg.m^2[/tex]
- angular velocity of the skater with extended arms, [tex]\omega'=2.3\ rad.s^{-1}[/tex]
Using the law of conservation of angular momentum:
[tex]I'.\omega'=I.\omega[/tex]
[tex]2.5\times 2.3=1.2\times \omega[/tex]
[tex]\omega=4.7916\ rad.s^{-1}[/tex]
