What is the Moment of Inertia for a Rod?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a rod system, with a focus on understanding the implications of different axes of rotation and the distribution of mass. Participants are examining the setup and parameters involved in the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the treatment of the rod as a point mass and the confusion between mass per rod and total mass. There are attempts to calculate the moment of inertia using different methods and interpretations of the problem setup.

Discussion Status

The discussion is ongoing, with participants providing guidance on the calculations and clarifying misunderstandings. There is recognition of errors in the initial approach, and multiple interpretations of the problem are being explored.

Contextual Notes

Some participants note a misunderstanding of the question's requirements, particularly regarding the axis of rotation and the treatment of the rods in the system. This has led to confusion about the correct method for calculating the moment of inertia.

coldblood
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Hi friends,
Please help me in solving this problem, I'll appreciate the help.

The problem is as:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1467427_1461727597387679_1141225220_n.jpg

Attempt -

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1472752_1461727710721001_1917071548_n.jpg

Thank you all in advance.
 
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You treated the rod furthest from the axis as a point mass at its centre.
Also, I think you have confused the mass per rod with the total mass.
 
haruspex said:
You treated the rod furthest from the axis as a point mass at its centre.
Also, I think you have confused the mass per rod with the total mass.

So what will with the farthest rod Should I take its moment of inertia as ml2/12(about center) + m(l√3/2)2

If I do so,

The total M.I. is found to be (3/2) ml2

Which gives radius of giration as l√(3/2)

But the answer is l/√2
 
coldblood said:
So what will with the farthest rod Should I take its moment of inertia as ml2/12(about center) + m(l√3/2)2
Yes.
If I do so,

The total M.I. is found to be (3/2) ml2

Which gives radius of giration as l√(3/2)

But the answer is l/√2
That's the other error I mentioned. If each rod has mass m then the entire structure has mass 3m. If its radius of gyration is k then its moment of inertia will be 3mk2.
 
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Well, I did a mistake in reading the question also. I thought that I have to calculate the moment of inertia of the system about the axis parallel to the plane of the structure.

If I have to find the M.I. of the system about an axis parallel, So what would the M.I of the farthest rod?
In that case should I treat it as a point mass situated at the center of the rod?
 
coldblood said:
Well, I did a mistake in reading the question also. I thought that I have to calculate the moment of inertia of the system about the axis parallel to the plane of the structure.

If I have to find the M.I. of the system about an axis parallel, So what would the M.I of the farthest rod?
In that case should I treat it as a point mass situated at the center of the rod?

Yes.
 
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Likes   Reactions: 1 person

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