Question

A thin, flat, uniform metal triangle lies in the \(x y\) plane with its corners at \((1,0,0),(0,1,0),\) and the origin. Its surface density (mass/area) is \(\sigma=24\). (Distances and masses are measured in unspecified units, and the number 24 was chosen to make the answer come out nicely.) (a) Find the triangle's inertia tensor I. (b) What are its principal moments and the corresponding axes?

Short answer

Inertia tensor: \(I_{xx} = 2, I_{yy} = 2, I_{xy} = -0.5\). Principal moments: \(1.5, 2.5\). Axes found via eigenvectors.

Step-by-step solution

Understanding the Triangle's Shape and Properties

The triangle is defined by its vertices at the points \((1, 0, 0)\), \((0, 1, 0)\), and the origin \((0, 0, 0)\). This is a right triangle lying in the \(xy\)-plane. Its base and height are of length 1 unit each.

Calculating the Area of the Triangle

The area of a right triangle is given by \(\text{area} = \frac{1}{2} \times \text{base} \times \text{height}\). Substituting the lengths, we calculate \(\text{area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\) square units.

Establishing the Mass Distribution

The surface density \(\sigma\) is given as 24 (units of mass per unit area). Therefore, the total mass \(m\) of the triangle must be \(m = \sigma \times \text{area} = 24 \times \frac{1}{2} = 12\).

Forming the Inertia Tensor Basis

The inertia tensor for a two-dimensional distribution in the \(xy\)-plane can be written in a matrix form \(I = \begin{bmatrix} I_{xx} & I_{xy} & 0 \ I_{yx} & I_{yy} & 0 \ 0 & 0 & 0 \end{bmatrix}\). We aim to find the components \(I_{xx}\), \(I_{yy}\), and \(I_{xy}\).

Calculating the Inertia Tensor Elements

Using known properties of the center of mass and symmetric distribution, set up integrals to solve for inertia tensor elements. For a triangle with uniform planar density, \(I_{zz} = \frac{1}{6}m(a^2+b^2)\) where \(a\) and \(b\) are legs of the triangle. Here, \(a = b = 1\).

Finding \(I_{xx}\) and \(I_{yy}\)

We compute \(I_{xx} = \frac{1}{6}m\) and \(I_{yy} = \frac{1}{6}m\) according to distribution symmetry and depth (or height) parameters. Substituting \(m = 12\), we determine \(I_{xx} = I_{yy} = 2\).

Calculating \(I_{xy}\) and Principal Moments

Based on the uniformity and symmetry of the triangle, \(I_{xy} = -\frac{1}{24}m\), leading to \(I_{xy} = -0.5\) after substitution of \(m\). Principal moments (eigenvalues) are obtained by diagonalizing the tensor.

Identifying Principal Axes

The principal axes are found using eigenvectors associated with the eigenvalues from the diagonalization step, corresponding to the zero-off-diagonal form of the inertia tensor.

Key concepts

Principal Moments

Principal moments are essential when analyzing the inertia properties of a shape. They tell us the resistance of an object to being rotated about different axes. In the context of our triangle, finding the principal moments involves calculating the eigenvalues of the inertia tensor.

The inertia tensor for a 2D object like our metal triangle is a matrix that embodies how mass is distributed about various axes in the plane. The principal moments emerge from this tensor when it is diagonalized, which means transforming the tensor into a form where all off-diagonal elements are zero.

• The diagonal elements then represent the principal moments.
• These values signify the object's inertia about specific, orthogonal axes called principal axes.
• For a uniformly dense right triangle, symmetry helps simplify calculating these moments.

This results in eigenvalues that signify how difficult it is to set the triangle rotating about these axes compared to others. Thus, principal moments are not merely mathematical abstractions. They offer practical insights into the rotational dynamics of bodies.

Surface Density

Surface density is a measure of how much mass is distributed over an area. For our metal triangle, this is represented by the symbol \( \sigma \), and it's given as 24 units of mass per unit area.

Understanding surface density is crucial for several reasons:

• It allows for the straightforward calculation of total mass when the area is known.
• It's essential for determining how the mass contributes to inertia.
• Knowing surface density simplifies evaluating how objects behave under physical forces.

In our exercise, the surface density helps derive total mass from the triangle's area: \( m = \sigma \times \text{area} \). This relationship allows us to proceed with inertia calculations, highlighting why understanding surface density can be so critical in physics and engineering contexts.

Uniform Metal Triangle

A uniform metal triangle, like the one in our exercise, has consistent mass distribution over its entire area. This uniformity implies every square unit of the triangle carries the same mass.

The triangle we are considering is defined as having:

• Vertices at known coordinates forming a right triangle.
• Fixed surface density provided in the problem.
• Symmetry that aids in simplifying calculations.

Because of this uniformity and symmetry, calculations for properties like inertia tensor components become more manageable. The uniform distribution allows for easier integration using geometric properties, which can help compress and simplify the math required for finding mass-related properties.
This is particularly advantageous in theoretical problems like ours, where uniformity simplifies assumptions, allowing a focus on core concepts like inertia tensors and principal axes.