Disk with Hole Parallel Axis Thm problem

[engus18]

Greetings everyone, been having some problems with this question,
A uniform circular disk has radius 36 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 7.2 cm is cut out of it. The center of the hole is a distance 10.8 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.

Homework Equations

I=1/2mr^2
I=Icm+Md^2

The Attempt at a Solution

Well, I know I need to find the MoI of the disk, and then find the MoI of the removed disk, and subtract. Have had problems with finding the MoI of the removed disk, here's my work:

Large disk MoI: 1/2(350g)(36cm^2) = 226800g/cm

Small disk mass: 70g, found from setting up a ratio 36/350=7.2/x
Small disk MoI: I=Icm+ Md^2
I=1/2mr^2 + Md^2
I=1/2(70g)(7.2cm)^2 + (70g)(10.8cm)^2
I=9979.2 gcm^2

Then I subtract 9979.2 from the 226800, still doesn't get me the right answer, what am I doing wrong?

Thanks

 

[alphysicist]

Hi engus18,

Greetings everyone, been having some problems with this question,
A uniform circular disk has radius 36 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 7.2 cm is cut out of it. The center of the hole is a distance 10.8 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.

Homework Equations

I=1/2mr^2
I=Icm+Md^2

The Attempt at a Solution

Well, I know I need to find the MoI of the disk, and then find the MoI of the removed disk, and subtract. Have had problems with finding the MoI of the removed disk, here's my work:

Large disk MoI: 1/2(350g)(36cm^2) = 226800g/cm

Small disk mass: 70g, found from setting up a ratio 36/350=7.2/x

I don't believe this ratio is correct. Think about what happens if you make a disk with double the radius. By what factor does its mass increase (assuming everything else is the same)?