Answer:
When her hands extends, her momen of inertia is [tex]4.28\ kg-m^2[/tex].
Explanation:
Given that,
Initial angular speed, [tex]\omega_i=8\ rad/s[/tex]
Initial moment of inertia, [tex]I_1=100\ kg-m^2[/tex]
Final angular speed, [tex]\omega_f=7\ rad/s[/tex]
Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :
[tex]I_1\omega_1=I_2\omega_2[/tex]
[tex]I_2[/tex] is final moment of inertia
[tex]I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2[/tex]
So, when her hands extends, her momen of inertia is [tex]4.28\ kg-m^2[/tex]. Hence, this is the required solution.
