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Moment of Inertia of Parallelepiped
By Nikky Pujari on Tuesday, October 02, 2001 - 12:59 pm:

I need to know how can we prove that the moment of inertia about the x-axis of a homogeneous rectangular parallelepiped having dimensions a,b,c with respect to a centroidal axis parallel to the edges of length a, isM( b2+c2)/12. M=mass of parallelepiped. thank u.

By Arun Iyer on Tuesday, October 02, 2001 - 07:09 pm:

this should be quite simple if you could prove that K2 = (b2+c2)/12.....
where K is the radius of gyration.

try it you will get it.....
love arun

By David Loeffler on Friday, October 05, 2001 - 05:48 pm:

What do you mean by "rectangular parallelepiped" in this context? This result is correct for a cuboid, which is (I suppose) a fully rectangular parallelepiped - is this the result you were after?
If so, let me know and I'll post the proof.
If your parallelepiped is not a cuboid, the result is still correct, but you have to consider b and c to be not the edge lengths but their components perpendicular to the edge of length a (if that makes sense).

By David Loeffler on Saturday, October 06, 2001 - 04:23 pm:


Perhaps I should explain exactly what a parallelepiped is.

It is a solid shape having six faces, all of them parallelograms. A picture is worth a thousand words:
Parallelepiped

As you can see, it looks something like a rather squashed cuboid.

A rectangular parallelepiped is one where all the angles are right angles. This happens when all the faces are rectangles, so the shape is simply a cuboid.

David

By Nikky Pujari on Sunday, October 07, 2001 - 10:17 am:

David , plz give me the proof for a cuboid ,i think that should give me my answer.thanx.

By David Loeffler on Sunday, October 07, 2001 - 04:48 pm:


Well, as for the cuboid:

We can imagine splitting the cuboid up into a very large number of tiny pieces.

Let's use some axes: x, y and z axes are parallel to the edges of length a, b and c respectively, with the origin at the centre of the shape. So the axis of rotation is the x coordinate axis.

Consider a small element at (x, y, z), a small cuboid whose dimensions are dx by dy by dz. If r is the density of the shape, this will have mass dm =r dx dy dz.

Also, its distance r from the axis of rotation is simply sqrt(y2+z2). So its moment of interia about the axis is r2 dm = (y2+z2)r dx dy dz.

Now, let's imagine adding all these bits together. Consider first a long thin rod running the whole length of the cuboid parallel to the x axis, consisting of a number of small pieces as described above. The total moment of inertia of all these bits is S (y2+z2)r dx dy dz) = (y2+z2)rdy dz S dx.
The last sum becomes, as we make the pieces smaller and more numerous, an integral : ò-a/2 a/2dx = a. (The limits are ±a/2 since these are the x-distances of the ends of the cuboid from the centre).

So, each thin rod has moment of inertia (y2+z2)ardydz. Suppose we now repeat the process by considering a set of such rods having the same z-value, forming a "slice" through the cuboid parallel to the x and y axes.
The moment of this slice will be a sum of terms (y2+z2)ardydz. We are now summing over y, so it becomes an integral of the form

ardz ò-b/2 b/2(y2+z2)dy
=ardz (bz2+1/12 b3)

We can now perform a final sum over all values of z from -c/2 to c/2, obtaining an integral

abr ò-c/2 c/2 (z2+1/12 b2) dz
= abr (1/12 c3 + 1/12 b2c)
=abcr (b2+c2)/12.

Now the first term is the mass of the cuboid, so its moment of inertia is 1/12 M(b2+c2).

David