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Home Q & A Board Science-Related Homework Help Physics
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ilocanababy ilocanababy
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11 years ago
a cylinder with moment of inertia I1 rotates about a vertical, frictionless axle with angular speed wi . a second cylinder, this one having moment of inertia I2 and initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed (wf)

1) What is wf?

2) Show that the Kinetic energy of the system decreases in this interaction and calculate the ratio of the final to initilal rotational energy.
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#1
11 years ago
Use conservation of angular momentum in first part.
(I1)(wi)=(I1+I2)(wf)
for second,first calculate initial energy of the firstcylinder and then finally calculate the energy of both cylinders using formula:
K.e=1/2(I)(w^2)
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bugmenomorebugmenomore
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#2
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11 years ago
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#3
11 years ago
This problem is tricky, because it is NOT a matter of conservation of angular momentum! Why not? Because when the second cylinder is next to the first one, the friction between them should not only communicate rotation from the first to the second, but should also cause the second to revolve around the first. It doesn't because it's held in place (not stated): because that holding requires a force and thus a torque, the changes in the angular momentum of the two cylinders are each negative.

1) What is wf?
Initial angular momentum is:
I1 * wi
Suppose this is pointing out at us, for definiteness.

The second cylinder is held in contact with the first, just above. The friction force acts at the surface of the lower (first) cylinder: it points towards the right. Since the cylinders end up each rotating with the same angular speed wf, they must have the same radius: call it R.

Looking at the first cylinder, it experiences force +F at radius R for time t, at the top surface. So the total impulse in angular momentum is
-RFt,
so the new angular momentum is I1*wi - RFt. But this must also be I1*wf.

The second cylinder experiences force -F at radius R for time t; at the bottom surface, so it's total impulse in angular momentum is also
-RFt,
so the new angular momentum is -RFt. But this must also be -I2*wf .
Therefore,
RFt = I2*wf

and
I1*wf = I1*wi - RFT = I1*wi - I2*wf
I1*wf + I2*wf = I1*wi

wf = wi*I1/(I1+I2)

2) Kinetic Energy

Originally, the KE was
I1*wi^2/2

The new KE is

KE-new = (I1+I2)wf^2/2
               = (i1+I2)(I1/(I1+I2))^2 * wi^2/2
               = ((i1+I2)/I1)(I1/(I1+I2))^2 * (I1*wi^2)/2
               = (I1/(I1+I2))*(I1*wi^2)/2
               = (I1/(I1+I2))* KE-original

So, KE-new/KE-original = I1/(I1+I2)
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