Answer:
the resulting angular speed after she pulls her hand inwards in (rad/s) is 27.02 rad/s
Explanation:
Given that :
the initial angular speed [tex]\omega_1 = 1.9 \ \ rev/s[/tex]
Initial rotational inertia [tex]I_1 = 12.00 \ \ kg.m^2[/tex]
Final angular speed [tex]\omega_2 = ???[/tex]
Final rotational inertia [tex]I_2 = 5.3 \ \ kg.m^2[/tex]
According to conservation of momentum :
Initial momentum = final momentum
[tex]I_1 \omega_1 = I_2 \omega_2[/tex]
[tex]\omega_2 = \frac{I_1 \omega_1}{I_2}[/tex]
[tex]\omega_2 = \frac{12.00*1.9}{5.3}[/tex]
[tex]\omega_2 = 4,3[/tex] [tex]rev/s[/tex]
To rad/s ; we have:
[tex]1 \ rev/s = 2 \pi \ \ rad/s[/tex]
[tex]\omega_2 = 4.3 * 2 \pi \ \ rad/s[/tex]
[tex]\omega_2 = 27.02 \ \ rad/s[/tex]
Therefore the resulting angular speed after she pulls her hand inwards in (rad/s) = 27.02 rad/s
