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

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, $\omega'=2.3\ rad.s^{-1}$

Using the law of conservation of angular momentum:

$I'.\omega'=I.\omega$

$2.5\times 2.3=1.2\times \omega$

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