How Is the Moment of Inertia of a Cube Calculated Incorrectly?

[iva]

Homework Statement

Find the moment of inertia of a cube, all edges length A, with mass M through the centre ( through mid point of 2 opposite sides)

Homework Equations

I = integral( r^2 dm
dm = density x volume where volume of a slice ( thin) square = AxAxdx = A^2 dx
= p A^2 dx

The Attempt at a Solution

Divide cube into thin square slices with rotation axis through the centre. Caulculate I for each AxA lamina by dividing each lamina into thin rodes each made of small mass elements with width dx, dm = pdx. For each rod I is

I = integral (r² dm)
= p (integral)A/2 to -A/2 (x² dx)
= p (x³/3)A/2 to -A/2
= p ( A³/24 + A³/24)
= pA³/12 where p = m/A so
I = mA²/12

The recagle is the sum of all the small masses over the bigger mass Mi so I for the square is

I = MA²/12

I know this is wrong though because the answer is MA²/6 but where did i go wrong, how did i get half of what it should be?

Thanks!

 

[Gokul43201]

You've found the MoI for a single rod, about a perpendicular axis through its own midpoint. But the square sheet is composed of rods (or strips) who do not all pass through the center of the square.

 

[iva]

I should have rather said that the axis go through the middle of the sheet so that all the rods go through the centre. Isn't my error futher down the line or is it here at the rods?

Thanks

 

[Gokul43201]

I should have rather said that the axis go through the middle of the sheet so that all the rods go through the centre.

I don't understand your construction. How do you divide up a square sheet into a series of strips (rods) that ALL pass through the center of the square?

 

[iva]

The axis is in the plane of the square lamina. I found the MoI for a rod then summed it up for the sheet. Then summed up the sheet for the cube.

Is this approach wrong?

 

[Gokul43201]

The axis is in the plane of the square lamina.

If the axis passes through the center of the cube, going through the centers of opposite faces, then clearly, only the central lamina contains the axis. All the other laminae that make up the cube are separated from this axis by some distance. But their MoI's can be found easily using the parallel axis theorem. But then the addition at the final step is no longer trivial.

 

[iva]

AAAAH I see, but now I'm stumped, i can't use the parallel or perpedicular axis theorems because like you've pointed out that MoI waas taken for a sheet on the face of the cube .
BUT can't i say that the sheet was taken from the centre of the cube? the laminas will be the same through out right?

 

[iva]

Just realized ( maybe ) Could the MoI on this lamina be seen as the Ix or Iy for the cube, ie i could use the perpedicular axes theorem by using this lamina result as the x-axis MoI, double it as it will be the same being a cube and that is where i would get MA^2/6 ?!

 

[Gokul43201]

Just realized ( maybe ) Could the MoI on this lamina be seen as the Ix or Iy for the cube, ie i could use the perpedicular axes theorem by using this lamina result as the x-axis MoI, double it as it will be the same being a cube and that is where i would get MA^2/6 ?!

You're pretty close but not exactly correct in the details. The MoI you found for the lamina can not be seen as the Ix or Iy of the cube. But it most definitely is the Ix (or Iy) of the square lamina. From the perpendicular axis theorem, you can find Iz for the square lamina. And now you can extrapolate from the square lamina to the cube, about that same axis. Do you understand why that works?

 

[iva]

Oh i see, I jumped a step because i knew the result, is it because for the lamina that Iz would be going through the COM of the cube, then you just sum up all the little masses of the slices geting 1 big mass for the cube?

 

[Gokul43201]

Yes, that's basically correct. In essence, you can add up elements to generate a larger object, so long as all the elements are positioned identically with respect to the axis of rotation.

In this problem, that was true of the rods making up the square sheet, with respect to an axis lying along the plane of the sheet and parallel to an edge. It is also true of the sheets making up the cube, with respect to an axis perpendicular to the plane of the sheets. But it would not be true of the sheets making up the cube, with respect to an axis lying parallel to the plane of the sheets. That's why you had to use the perpendicular axis theorem at that stage: to give you a more suitable axis.

 

[iva]

Because a suitable axis would have to be one that goes through the COM right?

Thank you so much your help was great, having the answer infront of me led me to not think about each step