Moment of inertia (disk with off center hole)

In summary, you need to find the mass of the hole in order to use the first MoI equation. Then use the parallel axis theorem to find the moment of inertia about an axis not through the centre of mass.
  • #1
Imperitor
19
0
I have a disk with some thickness to it and I need its moment of inertia.
So this is the formula with r1=0
37f7b2c6aaaf32e1972af5e228064928.png


Now there is a "circular hole of diameter 'd' at a distance of 'r' from the geometric center of the disk." So I'm thinking that I should subtract the MoI of the hole from the disk. Here is a picture if you need it.

http://img99.imageshack.us/img99/782/50817033.jpg

I could use the same MoI equation as I did for the disk on the hole but that doesn't account for the hole being off centered. Any ideas??

I need to eventually equate this to a small mass rotating around a center point, so if there is a more direct approach please tell me. Thanks in advance for the help.
 
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  • #2
Welcome to PF!

Hi Imperitor ! Welcome to PF! :smile:
Imperitor said:
I could use the same MoI equation as I did for the disk

That's right! :smile:

And then use the parallel axis theorem to find the moment of inertia about an axis not through the centre of mass. :wink:
 
  • #3
Thanks. I really should have known that...

I'm really enjoying this forum. It's a great resource for an engineering student.
 
  • #4
Wait a minute... hit another snag. To use that theorem I need to know what the mass of the missing hole is. I only know the mass of the disk as is (with the hole in it). Also the first formula I gave could only be used if I knew what the mass of the disk was with the hole filled in. This is driving me nuts. I can't even start working on the question I'm doing without getting this MoI.
 
  • #5
just got up… :zzz:
Imperitor said:
Wait a minute... hit another snag. To use that theorem I need to know what the mass of the missing hole is. I only know the mass of the disk as is (with the hole in it). Also the first formula I gave could only be used if I knew what the mass of the disk was with the hole filled in. This is driving me nuts. I can't even start working on the question I'm doing without getting this MoI.

Easy-peasy …

divide the given mass by the area, to give you the density :wink:

then multiply by each area separately! :smile:
 
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Likes harmyder

Question 1: What is moment of inertia?

Moment of inertia is a physical property of a rotating object that measures its resistance to changes in its rotational motion.

Question 2: How is moment of inertia calculated for a disk with an off center hole?

The moment of inertia for a disk with an off center hole is calculated by using the parallel axis theorem, which states that the moment of inertia of a body about an axis is equal to the moment of inertia of the body about a parallel axis through its center of mass plus the product of the mass of the body and the square of the distance between the two axes.

Question 3: Why does a disk with an off center hole have a different moment of inertia compared to a solid disk?

A disk with an off center hole has a different moment of inertia because the mass of the object is distributed differently. The removed material from the off center hole affects the distribution of mass and thus changes the moment of inertia.

Question 4: How does the moment of inertia affect the rotation of a disk with an off center hole?

The moment of inertia affects the rotation of a disk with an off center hole by determining how much torque is needed to accelerate or decelerate the object. A larger moment of inertia means a greater resistance to changes in rotational motion.

Question 5: What are some real-world applications of moment of inertia for a disk with an off center hole?

Moment of inertia for a disk with an off center hole is important in various engineering and scientific fields such as calculating the stability and oscillations of rotating machines, determining the energy required for an object to spin, and analyzing the behavior of celestial bodies in orbit.

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