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)?

 

[kerimek]

for any help!

Dear student,

It seems like you are on the right track in your solution. However, there are a few errors in your calculations.

Firstly, the moment of inertia of the large disk should be calculated using the parallel axis theorem, as the disk is not rotating about its center of mass. This will give you a different value for the moment of inertia.

Secondly, when finding the moment of inertia of the small disk, you should use the formula I=1/2mr^2, as the small disk is rotating about its center of mass. Using the parallel axis theorem here would give you an incorrect value.

Once you have corrected these errors, you should be able to subtract the moment of inertia of the small disk from the moment of inertia of the large disk to get the moment of inertia of the modified disk.

I hope this helps you in solving the problem. Good luck!