(a) 2.75 rev/min
The moment of inertia of the rod rotating about its center is:
[tex]I_R=\frac{1}{12}ML^2[/tex]
where
[tex]M=3.30\cdot 10^{-2} kg[/tex] is its mass
L = 0.450 m is its length
Substituting,
[tex]I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2[/tex]
The moment of inertia of the two rings at the beginning is
[tex]I_r = 2mr^2[/tex]
where
m = 0.200 kg is the mass of each ring
[tex]r=5.20\cdot 10^{-2} m[/tex] is their distance from the center of the rod
Substituting,
[tex]I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2[/tex]
So the total moment of inertia at the beginning is
[tex]I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2[/tex]
The initial angular velocity of the system is
[tex]\omega_1 = 35.0 rev/min[/tex]
The angular momentum must be conserved, so we can write:
[tex]L=I_1 \omega_1 = I_2 \omega_2[/tex] (1)
where [tex]I_2[/tex] is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is
[tex]r=\frac{0.450 m}{2}=0.225 m[/tex]
so the moment of inertia of the rings is
[tex]I_r=2(0.200)(0.225)^2=0.0203 kg m^2[/tex]
and the total moment of inertia is
[tex]I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2[/tex]
Substituting into (1), we find the final angular speed:
[tex]\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min[/tex]
(b) 103.0 rev/min
When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:
[tex]I_2 = I_R = 5.57\cdot 10^{-4}kg m^2[/tex]
So using again equation of conservation of the angular momentum:
[tex]L=I_1 \omega_1 = I_2 \omega_2[/tex]
We find the new final angular speed:
[tex]\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min[/tex]
