Moment of inertia of triangular prism

[HHveluj]

Homework Statement

Find the moment of inertia of triangular prism (equilateral triangles with side 2a, parallel to xy-plane), mass M. It's centered at the origin with long side parallel to z-axis. Find moment of inertia about the z-axis, and without doing integrals explain two products of inertia (Ixy, Ixz, I suppose)

Homework Equations

The integral equation for Izz=int[(M/V)*(x^2+y^2)dV]
where V=volume, or in cylindrical coords. we can write (x^2+y^2)=r^2, where r=distance from z-axis

The Attempt at a Solution

I just can't figure out - if I do it in cartesian coords - what limits of x and y should I substitute? It doesn't look like it should be -a to a, since that wouldn't be geometrical center, and plus to that - it's a triangle, not square - so I should connect x to y somehow... Or should I do it in different coord. system?...

Thanks a lot in advance!
I understand that this is easy question, but I just can't figure it out for some reason...

 

[OlderDan]

I don't think you want to use cylindrical coordinates for this. Cartesian coordinates will do nicely. You need to know where the centroid of the tiangle is to put it at the origin. By symmetry, it is at the intersection of the bisectors of the angles. I assume you will want one side parallel to the x-axis (or y axis; your choice). The limits of integration are determined from the equations of the boundary lines forming the triangle. Draw the trianngle in the x-y plane and write the equations of the lines.

 

[m2003]

I would like to first clarify the question by restating it in more specific terms. The moment of inertia of a triangular prism is being sought, with the prism having equilateral triangular cross-sections with side length 2a and a mass of M. The prism is centered at the origin, with the long side parallel to the z-axis. The requested moment of inertia is with respect to the z-axis, and the problem also asks for an explanation of the two products of inertia, Ixy and Ixz.

To solve this problem, we can use the integral equation for moment of inertia about the z-axis, which is given as Izz = int[(M/V)*(x^2+y^2)dV], where V is the volume of the prism. In cylindrical coordinates, we can rewrite (x^2+y^2) as r^2, where r is the distance from the z-axis. With this in mind, the limits of integration for x and y should be from 0 to a, since the prism is centered at the origin and has a side length of 2a.

As for the products of inertia, Ixy and Ixz, these represent the off-diagonal terms in the moment of inertia tensor and can be calculated using the same integral equation. Ixy represents the product of inertia between the x and y axes, and Ixz represents the product of inertia between the x and z axes. These products of inertia can be visualized as the distribution of mass around the x and y axes, and the x and z axes, respectively. Without doing the actual integrals, we can determine that these products of inertia will be non-zero due to the asymmetry of the triangular prism, which will result in a distribution of mass that is not symmetrical about these axes. Therefore, to fully describe the moment of inertia of a triangular prism, we must also consider these products of inertia.

I hope this explanation helps in understanding the problem and how to approach it. It is important to always clarify the question and break it down into smaller, more specific parts in order to find the appropriate solution.