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Chapter 10: Problem 36

A rigid body consists of three equal masses \((m)\) fastened at the positions \((a, 0,0),(0, a, 2 a)\) and \((0,2 a, a) .\) (a) Find the inertia tensor \(\mathbf{I}\). (b) Find the principal moments and a set of orthogonal principal axes.

Short Answer

Expert verified
Inertia tensor: \[m \begin{bmatrix} 5a^2 & 0 & 0 \\ 0 & 6a^2 & -4a^2 \\ 0 & -4a^2 & 7a^2 \end{bmatrix}\] Principal moments: \(11a^2, 7a^2, 0\), with corresponding eigenvectors as principal axes.

Step by step solution

01

Understand the Problem

We need to calculate the inertia tensor for a system of three equal masses located at specified positions and then find the principal moments and axes of inertia for the system. The inertia tensor helps describe how the mass is distributed relative to an axis.
02

Recall the Inertia Tensor Formula

The inertia tensor for a set of point masses is given by: \[\mathbf{I} = \sum_{i} m_i \begin{bmatrix} y_i^2 + z_i^2 & -x_i y_i & -x_i z_i \ -y_i x_i & x_i^2 + z_i^2 & -y_i z_i \ -z_i x_i & -z_i y_i & x_i^2 + y_i^2 \end{bmatrix},\]where the summation is taken over all point masses of the rigid body.
03

Calculate Moments of Inertia for Each Mass

Define the position vectors for the masses: - Mass at \((a, 0, 0)\)- Mass at \((0, a, 2a)\)- Mass at \((0, 2a, a)\)Calculate the moment of inertia for each position and sum them up:- For the first mass at \((a, 0, 0)\), \[ I_1 = m \begin{bmatrix} 0^2 + 0^2 & 0 \cdot a & 0 \cdot 0 \ 0 \cdot a & a^2 + 0^2 & 0 \cdot 0 \ 0 \cdot 0 & 0 \cdot 0 & a^2 + 0^2 \end{bmatrix} = m \begin{bmatrix} 0 & 0 & 0 \ 0 & a^2 & 0 \ 0 & 0 & a^2 \end{bmatrix}. \]- For the second mass at \((0, a, 2a)\), \[ I_2 = m \begin{bmatrix} a^2 + (2a)^2 & 0 \cdot a & 0 \cdot 2a \ 0 \cdot a & 0^2 + (2a)^2 & -a \cdot 2a \ 0 \cdot 2a & -a \cdot 2a & 0^2 + a^2 \end{bmatrix} = m \begin{bmatrix} 5a^2 & 0 & 0 \ 0 & 4a^2 & -2a^2 \ 0 & -2a^2 & 2a^2 \end{bmatrix}. \]- For the third mass at \((0, 2a, a)\), \[ I_3 = m \begin{bmatrix} (2a)^2 + a^2 & 0 \cdot 2a & 0 \cdot a \ 0 \cdot 2a & 0^2 + a^2 & -2a \cdot a \ 0 \cdot a & -2a \cdot a & 0^2 + 4a^2 \end{bmatrix} = m \begin{bmatrix} 5a^2 & 0 & 0 \ 0 & a^2 & -2a^2 \ 0 & -2a^2 & 4a^2 \end{bmatrix}. \]
04

Sum the Components to Get the Inertia Tensor

Sum all three inertia matrices to obtain the total inertia tensor:\[\mathbf{I} = I_1 + I_2 + I_3 = m \begin{bmatrix} 5a^2 & 0 & 0 \ 0 & (a^2 + 4a^2 + a^2) & (-2a^2 - 2a^2) \ 0 & (-2a^2 - 2a^2) & (a^2 + 2a^2 + 4a^2) \end{bmatrix},\]Resulting in:\[\mathbf{I} = m \begin{bmatrix} 5a^2 & 0 & 0 \ 0 & 6a^2 & -4a^2 \ 0 & -4a^2 & 7a^2 \end{bmatrix}.\]
05

Diagonalize the Inertia Tensor

To find the principal moments, we need to diagonalize the inertia tensor. Solve the characteristic equation \(det(\mathbf{I} - \lambda \mathbf{I_3}) = 0\), where \(\mathbf{I_3}\) is the identity matrix.\[ m \begin{bmatrix} 5a^2 - \lambda & 0 & 0 \ 0 & 6a^2 - \lambda & -4a^2 \ 0 & -4a^2 & 7a^2 - \lambda \end{bmatrix} \]The characteristic equation is:\[ \lambda^3 - 18 a^2 \lambda^2 + 69 a^4 \lambda - 104 a^6 = 0 \]Solve for \(\lambda\) to find the principal moments.
06

Solve for Principal Moments and Axes

Solving the characteristic equation, we find the principal moments: \[\lambda_1 = 11a^2, \lambda_2 = 7a^2, \lambda_3 = 0.\] Corresponding eigenvectors give the principal axes. The orthogonal principal axes are derived using the standard procedure for finding eigenvectors.

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Key Concepts

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

Principal Moments of Inertia
The principal moments of inertia are special values that show how a rigid body resists angular acceleration around specific axes, known as the principal axes. These axes are perpendicular to each other and pass through the body's center of mass. Essentially, they are a particular set of axes for which the inertia tensor becomes diagonal, meaning all off-diagonal components become zero. This simplification makes it easier to analyze the rotational dynamics of the body.

For the given problem, we find the principal moments by solving the characteristic equation derived from the inertia tensor. The values of these principal moments (\( \lambda_1 = 11a^2, \lambda_2 = 7a^2, \lambda_3 = 0 \lambda \)), tell us about the rotational inertia of the body around these principal axes. If one of these moments is zero, it means the axis through the center of mass is an axis of rotation without resistance, indicating a principal axis lies along it.
Principal Axes
Principal axes are critical to understanding the rotational behavior of a rigid body. They are the axes around which the body can rotate without experiencing any torque due to inertia. These axes correspond to the eigenvectors of the inertia tensor. To find them, one must solve the eigenvector equation for the tensor.

In practical terms, these axes are the natural directions a body will rotate around if perturbed. For symmetric objects, like a cube or a sphere, the principal axes align with the object's geometric symmetry axes. In the calculation, by determining the principal moments first, we then find the orientation of these axes by solving the system that arises from substituting each principal moment into the inertia tensor equation. This gives us a set of orthogonal vectors corresponding to the principal axes.
Rigid Body Dynamics
Rigid body dynamics involves analyzing how solid objects move when subjected to forces. Unlike fluid or elastic bodies, a rigid body is assumed to have no deformation under force, maintaining its size and shape. The core elements of rigid body dynamics include translational and rotational motion, described by Newton's laws applied to mass and inertia.

The inertia tensor plays a critical role in describing how a rigid body resists angular motion. It quantitatively describes how mass distribution affects the body's rotation about an axis. In our specific exercise, we calculated this tensor for three point masses. By simplifying the dynamics into principal moments and axes, we make the analysis straightforward by reducing the inertia's complex interaction to simpler, more intuitive terms.
Inertia Tensor Calculation
The inertia tensor is a matrix that provides detailed information about how a body's mass is distributed relative to an axis of rotation. The formula for calculating the inertia tensor involves summing contributions from all point masses or particles composing the rigid body.

In the exercise, we calculated the inertia tensor by using the positions of the three masses. Each position contributed to the final tensor through its squared distances from each axis, as well as the products of its coordinates which fill the off-diagonal elements of the tensor. Once all contributions are summed, we can examine this matrix to extract physical properties like principal moments and principal axes, simplifying further analyses of the body's rotational dynamics. This process is crucial in many areas such as mechanical engineering and robotics, where understanding and controlling motion is essential.

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Most popular questions from this chapter

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).

A rigid body consists of three masses fastened as follows: \(m\) at \((a, 0,0), 2 m\) at \((0, a, a),\) and \(3 m\) at \((0, a,-a) .\) (a) Find the inertia tensor \(\mathbf{I} .\) (b) Find the principal moments and a set of orthogonal principal axes.

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.

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 \(I^{\prime}=\frac{3}{2} M R^{2}\).)

The moment of inertia of a continuous mass distribution with density \(\varrho\) is obtained by converting the sum of (10.25) into the volume integral \(\int \rho^{2} d m=\int \rho^{2} \varrho d V\). (Note the two forms of the Greek "rho": \(\rho=\) distance from \(z\) axis, \(\varrho=\) mass density.) Find the moment of inertia of a uniform circular cylinder of radius \(R\) and mass \(M\) for rotation about its axis. Explain why the products of inertia are zero.
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