Rotating Discs

Rotating Discs

A Uniform 2.00 kg circular disk of radius 20.0 cm is rotating clockwise about an axis through its center...?

with an angular speed 30.0 revolutions per second. A second uniform 1.50 kg circular disk of radius 15.0 cm that is not rotating is dropped onto the first disk so that the axis of rotation of the first disk passes through the center of the second disk. What is the final angular speed of the two discs when they are rotating together?

I had this on a test and could not for the life of me figure out how to figure it out.

In this problem some energy is converted into heat by friction, but angular momentum is conserved. The initial angular momentum is I1 w1, where I1 is the moment of inertia of the first disk and w1 its angular velocity measure in radians per second. Afterwards the angular momentum is (I1+I2)w, where w is the final angular velocity measured in radians per second. Thus:

I1 w1 =(I1+I2)w

Solve for w:

w=I1 w1/(I1+I2) = w1/(1+I2/I1)

Note that w1= 2 pi f1, where f1 =30 rev/s, and w=2pi f. Thus

f= f1/(1+I2/I1)

Now, the moment of inertia of a disk is MR^2/2, where M is its mass and R its radius. Thus:
I2/I1=(M2 R2^2)/(M1 R1^2)=(1.5* 15^2)/(2*20^2)=1.6875
Hence the final angular velocity in revs per second is
f=30/(1+1.6875)=11.16 revolutions per second.

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To verify that you can not use the law of conservation of energy, compute the amount of energy converted into heat, as follows:

Q=initial KE -final KE
= (1/2) I1 w1^2 - 1/2(I1+I2)w^2

Since I1 = 2*.20^2/2=.02
I2=1.5* .15^2/2=.0169
Q = (1/2) [.02 (2 pi 30)^2- .0369(2 pi 11.16)^2]=

2 pi^2 [.02 (30)^2- .0369(11.16)^2]=260 J.

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