# Moment of inertia tensor

Find the tensor of inertia about the center of mass of a flat rigid body in the shape of 45 degrees right triangle with uniform mass density. What are the principal axes. What is the moment of inertia with respect to the axis perpendicular to the plane of the triangle through the right angle vertex of the triangle.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachment.

The components of the moment of inertia tensor are:

(1.1)

For a thin lamina, z=0, so this integral becomes a surface integral:

(1.2)

We will construct the system where the triangles hypotenuse is aligned with the x-axis, the altitude to the hypotenuse is aligned with the y-axis and the z-axis is perpendicular to the plane and is passing through the intersection of the hypotenuse and the altitude.

The equation of the left side is and the right side is

For a right triangle of uniform mass density the integrals become:

And:

The off-diagonal elements are:

The tensor is:

(1.3)

The tensor is already diagonal, so the original axes of rotation we set are the principal axes, and any axes system that is parallel to that is also principal axes system.

For a rotation axis that is perpendicular to the plane (parallel to the z-axis) and passes through the right angle vertex we will use the parallel axes theorem:

(1.4)

Where h is the length of the altitude and M is the total mass of the triangle.

We get:

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