Question about the inertia matrix of a bar

  • Thread starter Thread starter influx
  • Start date Start date
  • Tags Tags
    Inertia Matrix
Click For Summary
SUMMARY

The discussion centers on the moment of inertia of a bar, specifically addressing the confusion surrounding the inertia matrix and its constants. The moment of inertia for a bar around any axis perpendicular to it is defined as mℓ²/12, while the moment of inertia about a parallel axis is zero. The participants clarify that the problem assumes the rod is infinitely thin, which justifies the use of mℓ²/12 in the inertia matrix. Additionally, the angle in the mark scheme is noted as -β due to the orientation of the rotation.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with the parallel axis theorem
  • Knowledge of rigid body dynamics
  • Ability to interpret engineering diagrams and notation
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about the parallel axis theorem and its applications
  • Explore the implications of assuming an "infinitely thin" rod in physics problems
  • Review engineering mechanics focusing on rotational dynamics
USEFUL FOR

Students in engineering and physics, particularly those studying dynamics and mechanics, as well as educators looking to clarify concepts related to moment of inertia and rotational motion.

influx
Messages
162
Reaction score
1

Homework Statement


756c9e.png


Homework Equations


ml2/12

The Attempt at a Solution


5bf3a7.png
So according to my databook:

3267d3.png


The axis x'-x' in the question corresponds to axis ZZ in the databook image above. That means in terms of radius, the moment of inertia about axis x'-x' is mr2/2. So in light of this, why is the constant outside the matrix ml2/12? Why not mr2/2?
 
Last edited:
Physics news on Phys.org
The moment of inertia for a bar around any axis perpendicular to it is ##m\ell^2/12##.
 
Orodruin said:
The moment of inertia for a bar around any axis perpendicular to it is ##m\ell^2/12##.
I understand that but the x'-x' axis in the question is parallel to the bar and not perpendicular?

Thanks
 
influx said:
I understand that but the x'-x' axis in the question is parallel to the bar and not perpendicular?

Thanks
Exactly, and the moment of inertia relative to that axis is zero, as shown in the solution.
 
Orodruin said:
Exactly, and the moment of inertia relative to that axis is zero, as shown in the solution.

If the moment of inertia relative to the parallel axis is zero, then why is the moment of inertia about axis ZZ (in my data book) not 0? This axis is parallel to the cylinder?
 
influx said:
If the moment of inertia relative to the parallel axis is zero, then why is the moment of inertia about axis ZZ (in my data book) not 0? This axis is parallel to the cylinder?
Because the problem you are solving is idealising the rod as infinitely thin - i.e., having negligible diameter.
 
  • Like
Likes   Reactions: influx
Orodruin said:
Because the problem you are solving is idealising the rod as infinitely thin - i.e., having negligible diameter.

Ah that makes sense! This question is from an exam paper set by my university's engineering department a few years ago, is there any way to determine that this rod is infinitely thin from the wording of the question? Does the ''uniform slender'' part suggest this?

Also, I was wondering why (according to the mark scheme) the angle is -β? I mean the diagram shows a counter clock wise rotation so shouldn't the angle be β?
 
I
influx said:
Ah that makes sense! This question is from an exam paper set by my university's engineering department a few years ago, is there any way to determine that this rod is infinitely thin from the wording of the question? Does the ''uniform slender'' part suggest this?

Also, I was wondering why (according to the mark scheme) the angle is -β? I mean the diagram shows a counter clock wise rotation so shouldn't the angle be β?

I was wondering whether anyone could help me out with this?
 
influx said:
Ah that makes sense! This question is from an exam paper set by my university's engineering department a few years ago, is there any way to determine that this rod is infinitely thin from the wording of the question? Does the ''uniform slender'' part suggest this?

Also, I was wondering why (according to the mark scheme) the angle is -β? I mean the diagram shows a counter clock wise rotation so shouldn't the angle be β?
So you need a negative angle to counter that offset. The y-axis is pointing into the screen so a positive rotation would typically be clockwise.
 

Similar threads

Replies
8
Views
14K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K