Moment of Inertia for Three Point Masses

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SUMMARY

The discussion centers on calculating the moment of inertia for three point masses located at A=(-a, -a), B=(a, -a), and C=(0, a) with respect to an axis along the z-axis through the origin. The formula used is I = m((rA)² + (rB)² + (rC)²), leading to the calculation of I = 5ma². However, there is confusion as the expected answer is 4.12ma², prompting participants to question the source of the discrepancy.

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atlantic
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Homework Statement



Three particles of mass m are placed at A=(-a, -a), B=(a, -a) and C=(0, a)

Find the moment of inertia for an axis along the z-axis through the origin



Homework Equations



I = m((rA)2 + (rB)2 + (rC)2)



The Attempt at a Solution


I calculate that:
(rA)2 = (sqrt[(a)2 + (a)2])2 = 2a2
(rB)2 = (sqrt[(-a)2 + (a)2])2 = 2a2
(rC)2 = (a)2 = a2

So that:
I = m(a2 + 2a2 + 2a2) = 5ma2

But the answer is supposed to be 4.12ma2. What am I doing wrong?
 
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atlantic said:
So that:
I = m(a2 + 2a2 + 2a2) = 5ma2
Looks good to me.

But the answer is supposed to be 4.12ma2. What am I doing wrong?
I don't think you're doing anything wrong. What book are you using?
 

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