A uniform rod of mass 2.60 x 10^-2 kg and length 0.440m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.150kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.7 x 10^-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0rev/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 when the rings are at the end of the rod?
b) What is the angular speed after rings leave the rod?
I got part A simply enough, by applying I1w1=I2w2.
Generic expression for I (not specifying whether the rings are the end or not)
I = (1/12)ML^2 +2md^2
Now going onto part B, where the rings leave the rod, wouldn't the moment of inertia decrease to merely
I = (1/12)ML^2 because "d" = 0?
Naturally when moment of inertia decreases in a system with no net external torque (angular momentum is conserved), the angular speed should increase right? but for some reason it stays the same.