Moment of Inertia of a Solid Cylinder With a Wedge Removed

  • Thread starter cedoty1989
  • Start date
  • #1
cedoty1989
2
0
Homework Statement:
Imagine a solid cylinder height h able to rotate vertically around z-axis (centered at x=0 y=0). There is a wedge cut out so that when looking down with z hat pointing out of the page there is an angle 2a formed such that their is symmetry with respect to reflection over the x axis (See picture). The question is to calculate the moment of inertia.
Relevant Equations:
I=∫∫∫ dm r^2 -> Cylindrical Coords: (r from 0 to R) (z from zero to h) (theta from -Pi + a to Pi-a)

Uniform density -> dm=(M/V)dV

dV = (dr)r(dtheta)(dz)

The rest of the equations are in the picture...
CamScanner 11-23-2020 21.50-1.jpg
 

Attachments

  • CamScanner 11-23-2020 21.50-1.jpg
    CamScanner 11-23-2020 21.50-1.jpg
    62.5 KB · Views: 64

Answers and Replies

  • #2
cedoty1989
2
0
I know there should be an alpha in there somewhere but I cannot see where I'm going wrong. Thank you in advance.
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,216
8,528
The question is to calculate the moment of inertia.
About what axis? z axis through origin or through mass centre?
Your r22+z2 doesn't make sense for either.
 

Suggested for: Moment of Inertia of a Solid Cylinder With a Wedge Removed

Replies
3
Views
335
Replies
21
Views
494
Replies
4
Views
467
Replies
5
Views
636
Replies
13
Views
525
Replies
4
Views
531
  • Last Post
Replies
3
Views
551
Replies
8
Views
580
Top