How Do You Calculate the Moment of Inertia for a Disk with a Central Hole?

[Jefffff]

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 + 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 = 2πr²hρ
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 + 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!

 

[Orodruin]

It seems to me that the answer from the quiz fails to account for the hole’s centre of mass moment of inertia.

 

[Jefffff]

It seems to me that the answer from the quiz fails to account for the hole’s centre of mass moment of inertia.

Yes, this is what seems to be the case. Thanks for checking the work! I'll submit a request to the instructor.