Answer:
2.8 rad/s
Explanation:
In absence of external forces, the total angular momentum of the system must be conserved.
The angular momentum when the arms of the student are extended horizontally is given by:
[tex]L_1 = (I_0 + 2I)\omega_1[/tex]
where:
[tex]I_0=3.8 kg m^2[/tex] is the moment of inertia of the student+stool
[tex]I=mr^2[/tex] is the moment of inertia of each mass, where:
m = 2.6 kg is one mass
r = 0.71 m is the distance of each mass from the rotation axis
[tex]\omega_1=1.8 rad/s[/tex] is the initial angular velocity
So we have
[tex]L_1=(I_0+2mr^2)\omega_1[/tex]
When the student pulls the weights to a distance of r' = 0.23 m, the angular momentum is:
[tex]L_2=(I_0+2I')\omega_2[/tex]
where:
[tex]I'=mr'^2[/tex] is the new moment of inertia of each mass, with
r' = 0.23 m
Since the angular momentum must be constant, we have:
[tex]L_1=L_2\\(I_0+2mr^2)\omega_1 = (I_0+2mr'^2)\omega_2[/tex]
and solving for [tex]\omega_2[/tex], we find the final angular velocity:
[tex]\omega_2 = \frac{I_o+2mr^2}{I_0+2mr'^2}\omega_1=\frac{3.8+2(2.6)(0.71)^2}{3.8+2(2.6)(0.23)^2}(1.8)=2.8 rad/s[/tex]
