# Why is the moment of inertia of a rod not the same as that of a disk?

Take a disk and cut it to the centre like you'd do with scissors, and bring the cut sides to each other taking along all the mass in the middle. you'd have a rod. This should work because $R$ for the individual particles don't change when you're squeezing (moment of inertia of two particles of mass m on opposite sides of diameter of a circle is $2m R^2$, like one particle of mass $2m$).

This works for a ring and a point mass, why not for a disk?

## 1 Answer

Your analysis is flawed in one important way:

When you cut the disc to the centre and collapse it like closing a fan, the "rod" created is not uniform, in terms of mass distribution, along its length.

The part of the rod close to the centre of the disc contains the mass on a short circular line from the former disc. The other end of the rod contains more mass formerly on a much larger circle.

• oh that's right, but then why is the MI of a disk the same as half a disk, quarter disk, 1/8th disk etc? – Vrisk Oct 16 '17 at 12:42
• Assuming that the fractional discs are all pie-shaped pieces of various sizes, the distribution of mass radially outward is the same for all pieces, irrespective of size... – DJohnM Oct 16 '17 at 16:43
• uh, hey, another question if you don't mind - I can use the same logic for a sharply bent rod (at centre) and say the MI (in location of bent perpendicular to plane of rod) same regardless of the angle of bent. I can keep making this angle smaller and then suddenly it poofs and the moment of inertia changes? – Vrisk Oct 18 '17 at 12:52
• There is no change in the moment of inertia in the situation described... – DJohnM Oct 18 '17 at 18:59
• won't i get a single straight rod when the angle bw the rods is 0? that'll be mr^2/3 as opposed to mr^2/12 right? – Vrisk Oct 19 '17 at 8:27