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Chapter 10: Problem 37

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

Expert verified
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

01

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.
02

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.
03

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 \(m = \sigma \times \text{area} = 24 \times \frac{1}{2} = 12\).
04

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 \(I_{xy}\).
05

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, \(a = b = 1\).
06

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 \(I_{xx} = I_{yy} = 2\).
07

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.
08

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

To illustrate the result (10.18), that the total KE of a body is just the rotational KE relative to any point that is instantaneously at rest, do the following: Write down the KE of a uniform wheel (mass M) rolling with speed \(v\) along a flat road, as the sum of the energies of the CM motion and the rotation about the CM. Now write it as the energy of the rotation about the instantaneous point of contact with the road and show that you get the same answer. (The energy of rotation is \(\frac{1}{2} I \omega^{2} .\) The moment of inertia of a uniform wheel about its center is \(I=\frac{1}{2} M R^{2} .\) That about a point on the rim is \(I^{\prime}=\frac{3}{2} M R^{2}\).)

Suppose that you have found three independent principal axes (directions \(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\) ) and corresponding principal moments \(\lambda_{1}, \lambda_{2}, \lambda_{3}\) of a rigid body whose moment of inertia tensor \(\mathbf{I}\) (not diagonal) you had calculated. (You may assume, what is actually fairly easy to prove, that all of the quantities concerned are real.) (a) Prove that if \(\lambda_{i} \neq \lambda_{j}\) then it is automatically the case that \(\mathbf{e}_{i} \cdot \mathbf{e}_{j}=0\) (It may help to introduce a notation that distinguishes between vectors and matrices. For example, you could use an underline to indicate a matrix, so that \(\underline{\mathbf{a}}\) is the \(3 \times 1\) matrix that represents the vector a, and the vector scalar product a \(\cdot \mathbf{b}\) is the same as the matrix product \(\tilde{\mathbf{a}} \mathbf{b}\) or \(\underline{\mathbf{b}}\) a. Then consider the number \(\tilde{\mathbf{e}}_{i} \mathbf{I} \mathbf{e}_{j},\) which can be evaluated in two ways using the fact that both \(\mathbf{e}_{i}\) and \(\mathbf{e}_{j}\) are eigenvectors of I.) (b) Use the result of part (a) to show that if the three principal moments are all different, then the directions of three principal axes are uniquely determined. (c) Prove that if two of the principal moments are equal, \(\lambda_{1}=\lambda_{2}\) say, then any direction in the plane of \(\mathbf{e}_{1}\) and \(\mathbf{e}_{2}\) is also a principal axis with the same principal moment. In other words, when \(\lambda_{1}=\lambda_{2}\) the corresponding principal axes are not uniquely determined. (d) Prove that if all three principal moments are equal, then any axis is a principal axis with the same principal moment.

Here is a good exercise in vector identities and matrices, leading to some important general results: (a) For a rigid body made up of particles of mass \(m_{\alpha},\) spinning about an axis through the origin with angular velocity \(\omega,\) prove that its total kinetic energy can be written as $$ T=\frac{1}{2} \sum m_{\alpha}\left[\left(\omega r_{\alpha}\right)^{2}-\left(\omega \cdot \mathbf{r}_{\alpha}\right)^{2}\right] $$ Remember that \(\mathbf{v}_{\alpha}=\boldsymbol{\omega} \times \mathbf{r}_{\alpha} .\) You may find the following vector identity useful: For any two vectors a and b, $$ (\mathbf{a} \times \mathbf{b})^{2}=a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2} $$ (If you use the identity, please prove it.) (b) Prove that the angular momentum \(\mathbf{L}\) of the body can be written as $$ \mathbf{L}=\sum m_{\alpha}\left[\omega r_{\alpha}^{2}-\mathbf{r}_{\alpha}\left(\omega \cdot \mathbf{r}_{\alpha}\right)\right] $$ For this you will need the so-called \(B A C-C A B\) rule, that \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})=\mathbf{B}(\mathbf{A} \cdot \mathbf{C})-\mathbf{C}(\mathbf{A} \cdot \mathbf{B})\) (c) Combine the results of parts (a) and (b) to prove that $$ T=\frac{1}{2} \omega \cdot \mathbf{L}=\frac{1}{2} \tilde{\omega} \mathbf{I} \omega $$ Prove both equalities. The last expression is a matrix product; \(\omega\) denotes the \(3 \times 1\) column of numbers \(\omega_{x}, \omega_{y}, \omega_{z},\) the tilde on \(\tilde{\omega}\) denotes the matrix transpose (in this case a row), and I is the moment of inertia tensor. This result is actually quite important; it corresponds to the much more obvious result that for a particle, \(T=\frac{1}{2} \mathbf{v} \cdot \mathbf{p} .\) (d) Show that with respect to the principal axes, \(T=\frac{1}{2}\left(\lambda_{1} \omega_{1}^{2}+\lambda_{2} \omega_{2}^{2}+\lambda_{3} \omega_{3}^{2}\right)\) as in Equation (10.68).

Show that the inertia tensor is additive, in this sense: Suppose a body \(A\) is made up of two parts \(B\) and \(C\). (For instance, a hammer is made up of a wooden handle wedged into a metal head.) Then \(\mathbf{I}_{A}=\mathbf{I}_{B}+\mathbf{I}_{C} .\) Similarly, if \(A\) can be thought of as the result of removing \(C\) from \(B\) (as a hollow spherical shell is the result of removing a small sphere from inside a larger sphere), then \(\mathbf{I}_{A}=\mathbf{I}_{B}-\mathbf{I}_{C}\).

A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal \(z\) axis, being free to swing in the \(x y\) plane ( \(x\) horizontal, \(y\) vertically down). Its mass is \(m,\) its \(\mathrm{CM}\) is a distance \(a\) from the pivot, and its moment of inertia (about the \(z\) axis) is \(I\). (a) Write down the equation of motion \(\dot{L}_{z}=\Gamma_{z}\) and, assuming the motion is confined to small angles (measured from the downward vertical), find the period of this compound pendulum. ("Compound pendulum" is traditionally used to mean any pendulum whose mass is distributed \(-\) as contrasted with a "simple pendulum," whose mass is concentrated at a single point on a massless arm.) (b) What is the length of the "equivalent" simple pendulum, that is, the simple pendulum with the same period?
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