10.88) A uniform rod of mass 3.00×102 kg and length 0.440 m rotates in a horizon
ID: 1555210 • Letter: 1
Question
10.88) A uniform rod of mass 3.00×102 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 0.230 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.30×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 33.0 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?
Explanation / Answer
The angular speed at both cnditions are same. So answer for A and B is same.
The rod:
I1 = 1/12*m*L^2 <--- This doesn't change as the rings slide
I1 = 4.84*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 = 0.230 kg * (5.30*10^-2 m)^2 = 6.46*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 = 11.13 * 10^-3 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 = 33.0 rpm * (4.84*10^-4 kg*m^2 + 2*(6.46*10^-4 kg*m^2)) / (4.84*10^-4 kg*m^2 + 2*(11.13 * 10^-3 kg*m^2))
wf = 1.776*10-3 / 0.045
= 0.03946 radian/s
= 1.0467*10-4 rpm
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