Question about the inertia matrix of a bar

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The discussion centers on the moment of inertia of a bar and its relation to different axes. The moment of inertia for a bar around an axis parallel to it is zero, while the inertia around a perpendicular axis is given by ml²/12. Participants clarify that the problem assumes the rod is infinitely thin, which justifies the use of this formula. The term "uniform slender" in the question suggests the rod's negligible diameter. Additionally, there is confusion regarding the angle of rotation, with clarification that a negative angle is necessary to account for the defined coordinate system.
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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?
 
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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.
 
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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.
 

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