Moment of inertia of a rod with varying density

Click For Summary
SUMMARY

The discussion focuses on calculating the moment of inertia of a thin rod with a varying density between 4 lb/ft and 24 lb/ft. The mass of the rod is determined to be 140 lbs using the equation dm = (2x+4) dx integrated from 0 to 10 ft. The x-coordinate of the centroid is calculated as 130/21 ft. The moment of inertia calculation initially presented challenges, but it was clarified that the variable x must consistently represent the distance from the center of mass, leading to the correct answer of 6.92m when the limits are adjusted appropriately.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with the concept of moment of inertia
  • Knowledge of centroids and their calculation
  • Experience with variable density problems in physics
NEXT STEPS
  • Study the derivation of moment of inertia for varying density objects
  • Learn about the integration of variable density functions in calculus
  • Explore the application of centroids in engineering mechanics
  • Review examples from Mary L Boas' "Mathematical Methods" for further practice
USEFUL FOR

This discussion is beneficial for physics students, engineering students, and professionals involved in mechanics and structural analysis, particularly those dealing with variable density materials.

bigevil
Messages
77
Reaction score
0

Homework Statement



A thin rod is 10 ft long and has a density which varies uniformly between 4 and 24 lb/ft. Find:
a) the mass
b) the x-coordinate of centroid
c) Moment of inertia about an axis perpendicular to the rod
d) Moment of inertia about an axis perpendicular to the rod passing thru the heavy end.

The Attempt at a Solution



a) isn't difficult. I have got dm = (2x+4) dx, then integrate between 0 and 10 to get 140 lbs.

b) From \int \bar{x} dm = \int x dm, integrate and use earlier results to get 130/21.

c) I'm stuck at this one. I assume this means that the axis of MI passes through the centre of mass, so I set up the integration limits from 80/21 to -130/21 (rod is 10 foot). Then, taking I = \int x^2 dm = \int_{-130/21}^{80/21} 2x^3 + 4x^2 dx = 1.7(140) = 1.7m (where m=140 lbs). But the answer (this question is from Mary L Boas' mathematical methods book) is 6.92m.

Am I missing something here? I am only calculating for the x-coordinate because the rod is "thin".
 
Physics news on Phys.org
I haven't actually calculated anything, but the problem might be that in your integral you are making "x" refer to two different things.

In your expression for dm, x is referring to distance from the light end of the rod. But since you are finding moment of inertia and passing the axis through the center of mass, the "x^2" in the integral is referring to distance from the center of mass. The two reference points are not the same.

Maybe if you replaced x^2 by (130/21 - x)^2, it would end up working.

Oh, checked it and it does work. But you have to change the limits to 0 to 10, since I made x the distance from the light end.
 
Oh yeah... thanks Darksun... :)
 

Similar threads

Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
1K
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
8
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K