What is the moment of inertia of a disk with a hole about the Z-axis?

[ledphones]

Homework Statement

A uniform circular disk has radius 39 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 9.75 cm is cut out of it. The center of the hole is a distance 14.625 cm from the center of the disk. Find the moment of inertia of the modified disk about the Z-axis.

a) answer is in g cm^2

HELP: Think of the disk as the sum of the modified disk plus the cutout part.
HELP: Using the hint, apply the definition of moment of inertia. That is, consider the whole disk (disk without hole) to be composed of two constituent objects (modified disk plus cutout part). Then think about the answers to the following questions and try to figure out how answering these questions leads you to solve the problem:
How is the moment of inertia of the disk related to the moments of inertia of the modified disk and the cutout part?
What is the moment of inertia of the disk about the origin?
What is the mass of the cutout part?
What is the moment of inertia of the cutout part about the center of the cutout part?
What is the moment of inertia of the cutout part about the origin?

Homework Equations

1/2MR^2
Parallel-Axis Theorem (I=Icm+M(h)^2) where Icm is I subscript center of mass

The Attempt at a Solution

I found the mass to be .0781299743 grams per cm^2 from 350/((39^2)*pi-(9.75^2)*pi)

I then multiplied this by (39^2)*pi and (9.75^2)*pi to get the mass of the disk without a hole as 373.33 grams and the mass of the hole to be 23.33 grams.

Next I found the inertia of the disk w/o the hole
I=.5MR^2
.5(373.33)39^2 = 283917.47

I then found the inertia of the hole with the parallel-axis theorem
.5M(r)^2+M(h)^2
.5(23.33)(9.75)^2+(23.33)(14.625)^2 = 6098.97

I then subtracted the inertia of the disk w/o the hole and the inertia of the hole. which should give me the disk w/o the hole
(28317.47-6098.97)=277818.5

This was wrong please help!

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I also tried the disk without the hole but multiplied the hole's h by the mass of the disk without the hole and the mass of the disk.

.5(23.33)(9.75)^2+(373.33)(14.625)^2

.5(23.33)(9.75)^2+(350)(14.625)^2

attaining 202956.7789and 207946.85 respectively

 

[Yuqing]

You may find it easier to think of the disk as composed of two parts. First is the whole disk without any holes. Second is a smaller mass in the middle of negative mass to compensate for the hole.

edit: looking a bit more closely at what you've written, your density term seems to be off. Can you tell me exactly how you got that?

 

[ledphones]

I took the mass of the large disk without a hole found the area (39*39*pi) and subtracted the area of the hole (9.75*9.75*pi) and then i divided the total mass by this to find the density.

 

[Yuqing]

Ah, you can't really do that. The mass given is the mass of the disk without the hole. So to find the density, you divide the mass by the volume of the disk without the hole. You don't need to subtract away anything.

 

[ledphones]

Thank you so much! That was exactly what I was doing wrong. Mass was the mass of the disk with the hole. Not the disk without the hole.