A uniform rod of mass 2.70×10 2 kg and length 0.440 m rotates in a horizontal pl

Question

A uniform rod of mass 2.70×102 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 0.210 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.70×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 26.0 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.

Explanation / Answer

The rod:
I1 = 1/12*m*L^2 <--- This doesn't change as the rings slide
I1 = 4.356*10^-4 kg*m^2

The rings:
The trick to the rings is that their moments of inertia change. Let's list the intial moments of inertia. We treat them as point masses.
I_point mass = m*r^2

Since they have the same mass and initially they have the same radii
I2i = I3i = m*ri^2 = .210 kg * (4.70*10^-2 m)^2 = 4.64*10^-4 kg*m^2

The final moment of interia for the masses before the fly off:
With rf = 0.440 m /2 = 0.220 m
I2f = I3f = m*rf^2 = .0101 kg*m^2

Conservation of Angular Momentum for (3) objects
I1*w1i + I2*w2i + I3*w3i = I1*w1f + I2*w2f + I3*w3f

Since I2 and I3 change:
I1*w1i + I2i*w2i + I3i*w3i = I1*w1f + I2f*w2f + I3f*w3f

Initially and finally they are rotating with the same speed:
w1i = w2i = w3i = wi and w1f = w2f = w3f = wf
wi*(I1 + I2i + I3i) = wf*(I1 + I2f + I3f)

Solve for wf
wf = wi * (I1 + I2i + I3i) / (I1 + I2f + I3f)

Now plug in numbers
wf = 26.0 rpm * ( 4.356*10^-4 kg*m^2 + 2*(4.64*10^-4 kg*m^2)) / (4.356*10^-4 kg*m^2 + 2*(.010))
wf = 1.734 rpm

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