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I have a small circular ring placed on a circular disc. The outer radius of the ring is same as the radius of the disc. I have worked out the difference of Moment of Inertia of the combined shape to be = 1/2 M(ring) {R1^2 + R2^2 }. Is this correct?

Worked out as follows :-

For disc--> I(1) = I(disk) = 1/2 * M(disc) * R(disk)^2

For ring--> I(ring) = 1/2 * M(ring) * {R1^2 + R2^2}

R1= inner radius of ring R2 = outer radius of ring

Total MOI of combined shape --> I(2) = I(disk) + I(ring)

What I want to find is I(2) - I(1) = 1/2 * M(ring) {R1^2 + R2^2}

Is this method correct??

Note: 1. This is for the Torsion Pendulum experiment 2. The minor shapes like nuts and bolts are neglected.

BEFORE COMBINING AFTER COMBINING

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  • $\begingroup$ Isn't I(2) - I(1) just the ring? $\endgroup$ – Jiminion Feb 25 '15 at 6:19
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I(ring) = Idisk(R2) - Idisk(R1).

The trick is figuring out the mass.

Mass of R2-sized disk would be MR2 = M*(pi*R2*R2)/((pi*R2*R2)-(pi*R1*R1))

Mass of R1-sized disk would be MR1 = MR2*(pi*R1*R1)/(pi*R2*R2)

So I(ring) = 1/2MR2*(R2*R2) - 1/2MR1*(R1*R1)

       = 1/2(  (M*(pi*R2*R2*R2*R2) - M*(pi*R1*R1*R1*R1))/((pi*R2*R2) - (pi*R1*R1)) ) )

I guess the pis can come out:

       = 1/2*M*(R2^4-R1^4)/(R2^2 - R1^2)

Ugh, this means:

       = 1/2*M*(R2^2+R1^2)*(R2^2-R1^2)/(R2^2 - R1^2)
       = 1/2*M*(R2^2+R1^2)

So, yep.

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    $\begingroup$ Yup. Jiminion, you are right. I2 - I1 is just I(ring). Anyway it seems you also arrived at the same conclusion. So guess I can proceed with that equation. $\endgroup$ – HANZEL HARRY FERNANDEZ Feb 26 '15 at 17:30

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