Find Moment of Inertia of 4 Rods Constructing a Square: Explained

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To find the total moment of inertia of four rods forming a square when hung from a corner, first calculate the moment of inertia of one rod about its center, which is I = (1/12)ML^2. Then, use the parallel axis theorem to determine the distance from the center of the square to the corner for each rod. The correct approach involves calculating the moment of inertia for all four rods at their centers and then adjusting for their positions at the corner. The total moment of inertia is found to be 10ML^2/3 when the calculations are correctly applied. Understanding the distances involved and applying the theorem accurately is crucial for arriving at the correct answer.
makyol
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Homework Statement


When we put together 4 rod in case they will construct a square and when we hang it from the corner. How can we find its total moment of inertia?

Length of rods: L
Mass of rods: M



Homework Equations


Paralel Axis Theorem: I = I + Md^2


The Attempt at a Solution


As my book said answer is 10ML^2/3. Can you please explain it how can i calculate it? Thanks a lot.
 
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Find the moment of inertia of one rod about its centre. Find the distance to the centre of the square, then use the parallel axis theorem.
 
Does it make sense firstly finding all of four rods moment of inertia at its centre then taking it and finding the new moment of inertia secondly at the corner of square?
Actually i cannot understand precisely, can you please be more clear?
 
makyol said:
Does it make sense firstly finding all of four rods moment of inertia at its centre then taking it and finding the new moment of inertia secondly at the corner of square?
Actually i cannot understand precisely, can you please be more clear?

Sorry, I didn't see the hanging from the corner part.

I=(1/12)ML2.

You need to find the distances of the centres of each rod to the corner and then use the parallel axis theorem. Then just simply add them up.
 
rock.freak667 said:
Sorry, I didn't see the hanging from the corner part.

I=(1/12)ML2.

You need to find the distances of the centres of each rod to the corner and then use the parallel axis theorem. Then just simply add them up.

But again i cannot get the result that is given by book. So do i have to take perpendicular distance to the corner for the each rod? I mean for example; for top two rods it is easy:
I = (1/12)ML^2 + M(L/2)^2 but when it comes to bottom rods, it is being complicated, unfortunately. Again what distance i will use?
L or squareroot of M(5/4)L^2 ?
Still i could not get the right answer despite the deal with two cases.
If you have any suggestion, i will appreciate it. Thanks.
 
makyol said:
Again what distance i will use?
L or squareroot of M(5/4)L^2 ?
Md^2 = M(5/4)L^2
 
makyol said:
Does it make sense firstly finding all of four rods moment of inertia at its centre then taking it and finding the new moment of inertia secondly at the corner of square?
This approach would work. The moment of inertia about the center of the square would work out to be 4/3 ML^2. Then you can apply the parallel-axis theorem to the square as a whole. You just have to be a little careful with this step because the mass of the square is 4M, not just M.
 

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