Question

A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extends her arms, and her angular speed drops to 7.0 rad/s. What is her moment of inertia now

Answer 1

Answer:

When her hands extends, her momen of inertia is $4.28\ kg-m^2$.

Explanation:

Given that,

Initial angular speed, $\omega_i=8\ rad/s$

Initial moment of inertia, $I_1=100\ kg-m^2$

Final angular speed, $\omega_f=7\ rad/s$

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

$I_1\omega_1=I_2\omega_2$

$I_2$ is final moment of inertia

$I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2$

So, when her hands extends, her momen of inertia is $4.28\ kg-m^2$. Hence, this is the required solution.