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Moment of Inertia of a Slab

  • Thread starter jumbogala
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Homework Statement


A slab has width a, length a, and thickness a/4. What is the moment of inertia about its symmetry axis?

Use the one parallel to the large face of the slab.


Homework Equations





The Attempt at a Solution


The answer is supposedly (ma2)/12 + m(a/4)2/12.

I can't see why so I tried to verify it using integration, but that's not working either. Basically I tried to find the moment of inertia of a thin plate, dimensions a x a. I found that to be ma2/12 like I should.

Then I wanted to basically take a stack of thin plates to make one thick plate of thickness a/4. But I can't figure out what the calculus for that would look like. Help?
 

Answers and Replies

  • #2
SammyS
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Which symmetry axis? It has a couple of them
 
  • #3
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If you took a notebook sitting on a table, do the symmetry axis parallel to the table and through the center of the the notebook (horizontal). Since it has dimensions a x a it doesn't matter which direction the axis points (ie. in the x or y direction) because they will be equal.

I just do not want the axis that points up from the table (vertical).

I found something on the internet that shows how to do this, but they did not use a "stack". Is it possible to do it by stacking up a bunch of thin slabs?
 
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SammyS
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The corner to corner diagonal is also a symmetry axis, since it's square.
 
  • #5
SammyS
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...

I can't see why so I tried to verify it using integration, but that's not working either. Basically I tried to find the moment of inertia of a thin plate, dimensions a x a. I found that to be ma2/12 like I should.

Then I wanted to basically take a stack of thin plates to make one thick plate of thickness a/4. But I can't figure out what the calculus for that would look like. Help?
It can be done this way using the parallel axis theorem with integration.
 

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