# Moment of Inertia of a Disk

## Homework Statement

A disk of radius R has an initial mass M. Then a hole of radius (1/4) is drilled, with its edge at the disk center.

Find the new rotational inertia about the central axis. Hint: Find the rotational inertia of the missing piece, and subtract it from that of the whole disk. You’ll find the parallel-axis theorem helpful.

## Homework Equations

I = Icm + Md2

Moment of Inertia of a Disk: 1/2 MR2

## The Attempt at a Solution

Based on the formula for mass of the disk, let M' be mass of the drilled out smaller disk.

M = 2πr2
Since all factors are constant here except r, where the new radius is 1/4 of the original, thus (1/4)2 = (1/16) M

so M' = (1/16)M

Plugging in values to find the moment of inertia of the drilled out disk piece:

I = Icm + Md2

I = (1/2)(1/16M)(R/4)2 + (M/16)(R/4)2

This yields I = (3/512) MR2

The new moment of inertia is (1/2)MR2 - (3/512)MR2

= (253/512) MR2

The problem is the correct answer for this question according to the quiz was:

(1/2)MR - (1/256)MR2

which is different from what I got. Not sure where I went wrong. Anything helps!

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