Calculating Angular Speed in a Rotating System with Sliding Rings

[lina29]

A uniform rod of mass 2.95×10−2 kg and length 0.440 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.240 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.70×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

A-What is the angular speed of the system at the instant when the rings reach the ends of the rod?

B-What is the angular speed of the rod after the rings leave it?

Using these equations I get
I1 = (1/12)(m1)L² + 2(m2)r²=.00154
I2 = (1/12)(m1)L² + 2(m2)(L/2)²=.02371
ω2 = ω1(I1/I2)
I know the answer for a and b are the same I'm just confused how I use ω2 = ω1(I1/I2) to figure out the answer

 

[AlexChandler]

I know the answer for a and b are the same I'm just confused how I use ω2 = ω1(I1/I2) to figure out the answer

Are you sure the answers are the same?

 

[lina29]

Yes. I figured it out though. I realized that I was given w1 and w2 = 35*[.00154/.02371] = 2.27 rpm.

Thanks though!