What is the Moment of Inertia of a Disk with a Hole?

[ArticMage]

Homework Statement

A uniform circular disk has radius 35 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 8.75 cm is cut out of it. The center of the hole is a distance 13.125 cm from the center of the disk. Find the moment of inertia of the modified disk about the Z-axis.


So first the mass of the circle with the hole is 328.125 g and the small circle is 21.875g . Then what I did was first found the MI of the disk without the hole.

.5*M*R^2=214375 g*cm^2

Then I tried finding the cm after the hole was made and used the parallel axis theorem.

The cm is (-.875cm,0) so the -.875 becomes the distance from the original cm and then

214375 + M(-.875)2=214626 which isn't right

Then I tried the MI of the whole disk - the MI of the hole which also was wrong.

So if anyone can help that would be great.

 

[mukundpa]

It is
MI of the whole disk about z axis - the MI of the hole (disk) about z axis.

 

[ArticMage]

Ok I see what I did wrong.

I did it MI(whole)-MI(small) but the MI(small) was just the MI of the removed piece itself, not as part of the whole. This makes sense, thanks.