Question

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/

Answer 1

`\omega=4.7916\ rad.s^(-1)`

Step-by-step 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)`

Using the law of conservation of angular momentum:

`I'.\omega'=I.\omega`

`2.5* 2.3=1.2* \omega`

`\omega=4.7916\ rad.s^(-1)`