Find the moment of inertia of this disk

In summary, to find the mass of the cutout in terms of the overall mass, you can use the disk's density to calculate the mass per unit area of the disk, and then use that to calculate the mass of the cutout based on its area.
  • #1
234jazzy2
11
0

Homework Statement


A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R0 and mass M. It then has a circular hole of radius R1 drilled into it. The hole's center is a distance h from the center of the disk.

Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C
Express your answer in terms of the variables M, R0, R1, and h.

How do i convert mass of the cutout in terms of overall mass?

Homework Equations


I = (1/2)mr^2

The Attempt at a Solution


I = (1/2)M(R0)^2 - I cutout
I cutout = (1/2)m(R1)^2 + mh^2

Thanks
 
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  • #2
Hello @234jazzy2 ,

Welcome to Physics Forums! :welcome:

234jazzy2 said:

Homework Statement


A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R0 and mass M. It then has a circular hole of radius R1 drilled into it. The hole's center is a distance h from the center of the disk.

Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C
Express your answer in terms of the variables M, R0, R1, and h.

How do i convert mass of the cutout in terms of overall mass?
Are you inquiring about how to find the mass of the cutout?

Before the hole was drilled, the overall disk had a particular area that you can calculate. And the overall disk has a known mass, M. With those you can calculate the disk's mass per unit area. That's sort of like the density of the disk.

Then calculate the area of the drilled out hole section. Since you know the area of the cutout and the disk's mass per unit area ("density" like characteristic) you should be able to determine the mass of the cutout.

Homework Equations


I = (1/2)mr^2

The Attempt at a Solution


I = (1/2)M(R0)^2 - I cutout
I cutout = (1/2)m(R1)^2 + mh^2

It looks like you are on the right track so far. :smile:
 
  • #3
collinsmark said:
Hello @234jazzy2 ,

Welcome to Physics Forums! :welcome:Are you inquiring about how to find the mass of the cutout?

Before the hole was drilled, the overall disk had a particular area that you can calculate. And the overall disk has a known mass, M. With those you can calculate the disk's mass per unit area. That's sort of like the density of the disk.

Then calculate the area of the drilled out hole section. Since you know the area of the cutout and the disk's mass per unit area ("density" like characteristic) you should be able to determine the mass of the cutout.
It looks like you are on the right track so far. :smile:

Thanks! Didn't know you could think of it in terms of density.
 

1. What is the moment of inertia of a disk?

The moment of inertia of a disk is a measure of its resistance to changes in rotational motion. It is a mathematical representation of the distribution of mass around the axis of rotation.

2. How do you find the moment of inertia of a disk?

The moment of inertia of a disk can be calculated using the formula I = 1/2 * m * r^2, where I is the moment of inertia, m is the mass of the disk, and r is the radius of the disk.

3. What are the units for moment of inertia?

The units for moment of inertia depend on the units used for mass and distance. In the SI system, the moment of inertia is typically measured in kilograms per square meter (kg/m^2).

4. How does the moment of inertia of a disk differ from that of a solid cylinder?

The moment of inertia of a disk is half that of a solid cylinder with the same mass and radius. This is because the mass of a disk is concentrated at the edges, while the mass of a solid cylinder is evenly distributed throughout its volume.

5. Are there any real-world applications of finding the moment of inertia of a disk?

Yes, the moment of inertia of a disk is used in engineering and physics to analyze the motion of rotating objects. It is also important in fields such as robotics, where precise control of rotational motion is required.

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