Moment of inertia of a rod bent into a square

In summary: You are asking about the moment of inertia of a single object about an axis that is not the center of mass. If you want to find the moment of inertia of a two-rod system, you should use the parallel axis theorem.
  • #1
Eggue
13
2
Homework Statement
A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem
Relevant Equations
Parallel axis theorem, Moment of inertia of a rod about it's end
I tried to find the moment of inertia of 2 rods connected at the corners by adding up their moments of inertia:
[tex] \frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2 [/tex]
I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem
[tex] \frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2 [/tex]
However, when i try to solve for I_cm i get a negative value. I was going to multiply I_cm by 2 to get the final answer

I know the correct solution is finding the moment of inertia about the center of mass of each rod and using the parallel axis theorem but I'm wondering why my approach is wrong?
 
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  • #2
Eggue said:
I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem
[tex] \frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2 [/tex]
However, when i try to solve for I_cm i get a negative value. I was going to multiply I_cm by 2 to get the final answer
The "center of mass" you are using here appears to be the center of mass of the four rods (center of the square) rather than the center of mass of the two-rod assembly you are working with.

The parallel axis theorem requires you to use the correct center of mass position.
 
  • #3
jbriggs444 said:
The "center of mass" you are using here appears to be the center of mass of the four rods (center of the square) rather than the center of mass of the two-rod assembly.

The parallel axis theorem requires you to use the correct center of mass position.
Aren't they the same though?
 
  • #4
Eggue said:
Aren't they the same though?
Calculate them. The center of mass of the two-rod assembly will be on a line connecting the center of mass of the one rod with the center of mass of the other. The center of mass of the four-rod assembly will be on a line connecting opposite corners of the square (ends of the two rods).
 
  • #5
jbriggs444 said:
Calculate them. The center of mass of the two-rod assembly will be on a line connecting the center of mass of the one rod with the center of mass of the other. The center of mass of the four-rod assembly will be on a line connecting opposite corners of the square (ends of the two rods).
So will the distance to the center of mass be [itex] \frac{\sqrt{2}a}{4} [/itex]?
 
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  • #6
Eggue said:
So will the distance to the center of mass be [itex] \frac{\sqrt{2}a}{4} [/itex]?
Yes, that is what I get.
 
  • #7
jbriggs444 said:
Yes, that is what I get.

I did the working but my answer isn't correct
Working:
IMG20200305204921.jpg

Where I_p is the moment of inertia about the center of the square
 
  • #8
Eggue said:
I did the working but my answer isn't correct
That picture is not of great quality. But I shall try to make sense of it.

You start with $$\frac{1}{6}Ma^2 = I_{\text{cm}}+M(\frac{\sqrt{2}}{4}a)^2$$ But that equation is already wrong. You are only dealing with ##\frac{M}{2}##, not ##M##.
 
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  • #9
jbriggs444 said:
That picture is not of great quality. But I shall try to make sense of it.

You start with
16Ma2=Icm+M(√24a)216Ma2=Icm+M(24a)2​
But that equation is already wrong. You are only dealing with M2M2, not MM.
Haha sorry about the image quality. You're right, once i correct the error i get the correct answer. One more question though, Why does the center of mass of the 2 rods lie on the line connecting the center of mass of each rod? Shouldn't it lie halfway between both rods which would give a distance of √(a2)2+(a2)2=√2a2(a2)2+(a2)2=2a2
 
  • #10
Eggue said:
Haha sorry about the image quality. You're right, once i correct the error i get the correct answer. One more question though, Why does the center of mass of the 2 rods lie on the line connecting the center of mass of each rod? Shouldn't it lie halfway between both rods which would give a distance of [itex] \frac{\sqrt{2}a}{2}[/itex]
It does lie halfway between both rods. It lies halfway between their centers.

If you have a collection of objects, the center of mass of the collection is positioned at the [weighted] average position of the individual centers of mass.
 
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  • #11
jbriggs444 said:
It does lie halfway between both rods. It lies halfway between their centers.

If you have a collection of objects, the center of mass of the collection is positioned at the [weighted] average position of their individual centers of masses.
Ohhh. Thank you so much for the help I've been trying to figure out where I've gone wrong for 2 days
 
  • #12
But why do a 2 rod system? The 4 sides are identical in their relation to the axis. Find the moment of inertia of one side about that axis and multiply by four.
 
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  • #13
Cutter Ketch said:
But why do a 2 rod system? The 4 sides are identical in their relation to the axis. Find the moment of inertia of one side about that axis and multiply by four.
I did a problem before that asked to find the moment of inertia of the 2 rod system so i sort of had tunnel vision
 

Related to Moment of inertia of a rod bent into a square

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in rotational motion. It is dependent on the object's mass and distribution of mass around its axis of rotation.

2. How is moment of inertia calculated for a rod bent into a square?

The moment of inertia for a rod bent into a square can be calculated using the formula I = (mL^2)/12, where m is the mass of the rod and L is the length of one side of the square.

3. How does the moment of inertia change when the shape of the rod is bent into a square?

The moment of inertia increases when the shape of the rod is bent into a square. This is because the mass is distributed further from the axis of rotation, resulting in a larger moment of inertia.

4. What factors affect the moment of inertia of a rod bent into a square?

The moment of inertia of a rod bent into a square is affected by its mass, length, and distribution of mass along its length. The moment of inertia also increases as the length of the rod increases.

5. How is the moment of inertia of a rod bent into a square used in real-world applications?

The moment of inertia of a rod bent into a square is used in engineering and physics to analyze the rotational motion of objects. It is important in designing structures and machines that require stable and efficient rotational motion.

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