Answer:
d. I0 /4
Explanation:
Given that:
Initial angular velocity of the skater = [tex]\omega_o[/tex]
After pulling arms, angular velocity changes to = [tex]4 \omega _o[/tex]
Provided that no external torque is acting on skater;
Then:
[tex]\tau _{ext} = 0[/tex]
The torque [tex]\tau_{ext} = \dfrac{dL}{dt}[/tex]
where;
[tex]\tau _{ext} = 0[/tex]
It implies that:
[tex]\dfrac{dL}{dt} =0[/tex]
where;
L = constant
[tex]L_{initial} = L_{final}[/tex]
So;
[tex]I_i\omega_i = I_f \omega _f[/tex]
[tex]I_f =\dfrac{I_i\omega_i}{ \omega _f}[/tex]
[tex]I_f =\dfrac{I_o\omega_o}{ 4\omega _o}[/tex]
[tex]\mathbf{ I_f =\dfrac{I_o}{ 4}}[/tex]
