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## 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.

Express your answer in terms of the variables M and R.

## Homework Equations

I = I

_{cm}+ Md

^{2}

Moment of Inertia of a Disk: 1/2 MR

^{2}

## 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 = 2πr

^{2}hρ

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 = I

_{cm}+ Md

^{2}

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

^{2}+ (M/16)(R/4)

^{2}

This yields I = (3/512) MR

^{2}

The new moment of inertia is (1/2)MR

^{2}- (3/512)MR

^{2}

= (253/512) MR

^{2}

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

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

^{2}

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