Answer:
[tex]L=11.3\ kg-m^2/s[/tex]
Explanation:
Given that,
Angular speed of a skater, [tex]\omega=3\ rot/s=18.84\ rad/s[/tex]
The moment of inertia of the skater, I = 0.6 kg-m²
We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :
[tex]L=I\omega[/tex]
Substitute all the values,
[tex]L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s[/tex]
So, its angular momentum is equal to [tex]11.3\ kg-m^2/s[/tex].
