Moment of inertia of a rod bent into a square

[Eggue]

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:
\frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2
I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem
\frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2
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?

 

[jbriggs444]

I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem
\frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2
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.

 

[Eggue]

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?

 

[jbriggs444]

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).

 

[Eggue]

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 \frac{\sqrt{2}a}{4}?

 

[jbriggs444]

So will the distance to the center of mass be \frac{\sqrt{2}a}{4}?

Yes, that is what I get.

 

[Eggue]

Yes, that is what I get.

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

Where I_p is the moment of inertia about the center of the square

 

[jbriggs444]

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 $M$.

 

[Eggue]

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

 

[jbriggs444]

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 \frac{\sqrt{2}a}{2}

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.

 

[Eggue]

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

 

[Cutter Ketch]

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.

 

[Eggue]

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