Two points are on a disk that rotates about an axis perpendicular to the plane

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SUMMARY

The discussion centers on calculating the moment of inertia for two unequal masses, m and 2m, attached to a thin bar rotating about an axis. Initially, the moment of inertia is calculated as I = 6md² when mass m is at a distance of 2d and mass 2m at d. After interchanging the masses, the moment of inertia becomes I' = 9md². The ratio of the new moment of inertia to the original is 3/2, clarifying that the initial assumption of I being doubled was incorrect.

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Two unequal masses m and 2m are attached to a thin bar of negligible mass that rotates about an axis perpendicular to the bar. When m is a distance 2d from the axis and 2m is a distance d from the axis, the moment of inertia of this combination is I. If the masses are now interchanged, the moment of inertia will be?

This is the work I tried:

The mases are m and 2m
The distance of the mass m from the axis is 2d
The distance of the mass 2m from the axis is d

The moment of inertia of the masse before exchanged is

I = m(2d)^2 + 2m(d)^2

I = 6md^2

The moment of inertia of the masse After exchanged is

The distance of the mass m from the axis is d
The distance of the mass 2m from the axis is 2d

I = m(d)^2 + 2m(2d)^2

I = 8md^2

I thought that after the masses were interchanged, the I would be 2I, the difference between before and after. The answer was wrong. I think the answer might be (2/3)I but I'm not sure. help is much appreciated
 
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There's an error in your last calculation... I'=m(d)^2+2m(2d)^2=m(d)^2+8m(d)^2=9m(d)^2
Meaning that the ratio between I' (after the masse were interchanged) and I is 9/6=3/2.
But I'm not sure I've understood your problem...
 

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