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kherson [118]
3 years ago
7

An ice skater has a moment of inertia of 2.5 kg.m2 when her arms are extended and a moment of inertia of 1.2 kg.m2 when her arms

are pulled in close to her body. If she goes into a spin with her arms extended and has an initial angular velocity of 2.3 rad/s, what is her angular velocity when she pulls her arms in close to her body? Answer in units of rad/
Physics
1 answer:
Damm [24]3 years ago
6 0

Answer:

\omega=4.7916\ rad.s^{-1}

Explanation:

Given:

  • moment of inertia of the skater with extended arms, I'=2.5\ kg.m^2
  • moment of inertia of the the skater with pulled-in arms, I=1.2\ kg.m^2
  • angular velocity of the skater with extended arms, \omega'=2.3\ rad.s^{-1}

<u>Using the law of conservation of angular momentum:</u>

I'.\omega'=I.\omega

2.5\times 2.3=1.2\times \omega

\omega=4.7916\ rad.s^{-1}

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