• Support PF! Buy your school textbooks, materials and every day products Here!

Moment of inertia of a disc with a hole

  • Thread starter iva
  • Start date
  • #1
iva
21
0
I think i have this right, but would like to confirm i did the right thing:

Homework Statement



I have a lamina disc radius R with a disc of radius r cut out of it from its centre. Total mass of the actual object is M, Find moment of inertia when lamina rotates about axis perpendicular to the lamina through its centre


Homework Equations



dm = density x area = density x circumference x width = p 2 pi r dr

Inertia = mr2 = Integration sign ( dIr)



The Attempt at a Solution



I chose to integrate over thin concentric rings with width dr, so that mass of each was

dm = density x area = density x circumference x width = p 2 pi r dr

So that I for each ring about the axis is

dIr = dm r2 = 2 p pi r3 dr

We integrate over all rings that make up this disk ie from radius r to R

So, I = (integral) dIr
= (integral)R to r ( 2 p pi r3 dr)
= 2 p pi (integral)R to r (r3 dr)
= 2 p pi (r4/4)]R to r
= p pi /2 (R4 - r4)

Since the area of this disk with the cutout is pi R2 - pi r2, its density is

p = Mass / area = mass / (pi R2 - pi r2)

Substitute this in the equation for I giving:

I = M pi (R4 - r4) / 2 (pi R2 - pi r2)

since (R4-r4) = (R2+r2)(R2-r2) we can cancel so

I = M (R2 + r2)/2

Thanks!
 

Answers and Replies

  • #3
iva
21
0
thank you !!!
 

Related Threads on Moment of inertia of a disc with a hole

Replies
13
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
6K
Replies
5
Views
17K
  • Last Post
Replies
3
Views
14K
Replies
9
Views
3K
Replies
5
Views
5K
Replies
4
Views
2K
Top