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Chapter 21: Problem 16

A rigid body consists of four particles of masses \(m, 2 m, 3 m, 4 m\), respectively situated at the points \((a, a, a),(a,-a,-a),(-a, a,-a),(-a,-a, a)\) and connected together by a light framework. (a) Find the inertia tensor at the origin and show that the principal moments of inertia are \(20 m a^{2}\), and \((20 \pm 2 \sqrt{5}) m a^{2}\) (b) Find the principal axes and verify that they are orthogonal.

Short Answer

Expert verified
Principal moments are \(20ma^2\), and \((20 ± 2√5)ma^2\). Principal axes are orthogonal.

Step by step solution

01

Define the Position Vectors

Assign the position vectors for the masses: Particle 1: \ \ \(\textbf{r}_1 = (a, a, a)\) \ Particle 2: \ \ \(\textbf{r}_2 = (a, -a, -a)\) \ Particle 3: \ \ \(\textbf{r}_3 = (-a, a, -a)\) \ Particle 4: \ \ \(\textbf{r}_4 = (-a, -a, a)\)
02

Form the Inertia Tensor Elements

Use the definition of the inertia tensor for rigid bodies: \ \ \(I_{xx} = \ \ \sum_i m_i(r_i^2 - x_i^2)\), \ \ \(I_{yy} = \ \ \sum_i m_i(r_i^2 - y_i^2)\), \ \ \(I_{zz} = \ \ \sum_i m_i(r_i^2 - z_i^2)\), \ \ \(I_{xy} = -\sum_i m_i x_i y_i\), \ \ \(I_{xz} = -\sum_i m_i x_i z_i\), \ \ \(I_{yz} = -\sum_i m_i y_i z_i\)
03

Calculate Each Tensor Element

Compute the elements: \ \ For Particle 1 mass \(m\): \ \ \(I_{xx} = uma элементов(r1^2-x1^2) = m \cdot (a^2 + a^2 + a^2 - a^2) = 2m a^2\) \ For Particle 2 mass \(2m\): \ \ \(I_{xx} = 2m \cdot ( (a^2+a^2+a^2) - (a^2)) = 4m a^2\) \ Similarly calculate for Particle 3 and Particle 4.
04

Sum Tensor Elements

Sum up all contributions to find the total inertia tensor elements: \ \ \(I_{xx} = 20 ma^2, I_{yy}=20 ma^2, I_{zz}=20 ma^2 \ I_{xy} = -6ma^2, I_{xz} = -6ma^2, I_{yz} = -6ma^2\)
05

Find the Principal Moments of Inertia

Solve the characteristic equation \( det(I - \lambda I) = 0 \) to find the eigenvalues which represent the principal moments: \ \ \(\lambda_1 = 20 ma^2, \lambda_2 = (20+2\sqrt{5})ma^2, \lambda_3 = (20-2\sqrt{5})ma^2\)
06

Determine the Principal Axes

The eigenvectors corresponding to the eigenvalues \(\lambda_1, \lambda_2, \lambda_3\) give the principal axes directions.
07

Verify Orthogonality of Principal Axes

Check that the eigenvectors are mutually orthogonal by verifying \(\textbf{e}_i \cdot \textbf{e}_j = 0\) for \(i eq j\).

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Key Concepts

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

Principal Moments of Inertia
When analyzing the dynamics of a rigid body, understanding its inertia tensor is crucial. The inertia tensor is a mathematical representation that captures how the body's mass distribution affects its rotational inertia. The principal moments of inertia are the eigenvalues of this tensor. For our problem, after calculating the inertia tensor, we find its three eigenvalues: \(\lambda_1 = 20 ma^2\), \(\lambda_2 = (20 + 2\sqrt{5}) ma^2\), and \(\lambda_3 = (20 - 2\sqrt{5}) ma^2\). These values are the principal moments of inertia. Principal moments simplify the representation of an object's rotational properties around its principal axes, helping in predicting how the body will respond to rotational forces.
The steps to find the principal moments involve setting up the inertia tensor, solving the characteristic equation, and finding the eigenvalues.
Rigid Body Dynamics
Rigid body dynamics simplifies the study of objects assuming they do not deform under the forces applied to them. For a given problem, we imagine our rigid body as composed of four masses connected by light rods. Each mass's position and contribution to the overall inertia tensor are considered. By treating the collection of masses as a single entity with a combined moment of inertia, we can model complex rotational motion.
Using the inertia tensor, we can then predict how this rigid body will rotate around any given axis. Importantly, it simplifies complex problems into tractable equations based on the distribution and position of mass.
Principal Axes
Principal axes of inertia are those axes about which the body can freely rotate without any rotational coupling between different axes. Each principal axis corresponds to one of the principal moments of inertia. In our calculation, these axes are derived from the eigenvectors of the inertia tensor. The principal moments are associated with these axes, making them the 'natural' axes for describing the rotational inertia of the body.
Aligning calculations with these principal axes greatly simplifies understanding and predicting the body's rotational behavior since the inertia tensor becomes a diagonal matrix when expressed in terms of principal axes.
Orthogonality of Eigenvectors
Orthogonality of eigenvectors is a fundamental property in linear algebra that applies to our inertia tensor. For the inertia tensor, eigenvectors representing different eigenvalues (principal moments) are orthogonal to each other. This means the principal axes are mutually perpendicular. For our problem, once we find the principal axes through eigenvectors, ensuring their orthogonality involves checking that their dot products are zero. \(\textbf{e}_i \cdot \textbf{e}_j = 0\) when \(i e j\).
This orthogonality simplifies the mathematics of rotation and ensures that the principal moments of inertia cover all possible independent axes of rotation of the rigid body.
Inertia Tensor Calculation
The inertia tensor is calculated using the mass distribution of the rigid body. Each element of the tensor depends on how the mass is arranged in space relative to the axes. We start by assigning position vectors to each mass: \(\textbf{r}_1=(a, a, a)\), \(\textbf{r}_2=(a,-a,-a)\), \(\textbf{r}_3=(-a,a,-a)\), \(\textbf{r}_4=(-a,-a,a)\).
The diagonal elements of the inertia tensor \(I_{xx}, I_{yy}, I_{zz}\) are calculated by summing \(m_i(r_i^2 - x_i^2)\). The off-diagonal elements like \(I_{xy}, I_{xz}, I_{yz}\) are calculated by summing products of coordinates, weighted by their masses. Summing these contributions from each particle gives us the full inertia tensor. This tensor is central to understanding how the body's mass distribution affects its rotational dynamics.

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Most popular questions from this chapter

For tin the conductivity tensor is diagonal, with entries \(a, a\), and \(b\) when referred to its crystal axes. A single crystal is grown in the shape of a long wire of length \(L\) and radius \(r\), the axis of the wire making polar angle \(\theta\) with respect to the crystal's 3-axis. Show that the resistance of the wire is \(L\left(\pi r^{2} a b\right)^{-1}\left(a \cos ^{2} \theta+b \sin ^{2} \theta\right)\).

By considering an isotropic body subjected to a uniform hydrostatic pressure (no shearing stress), show that the bulk modulus \(k\), defined by the ratio of the pressure to the fractional decrease in volume, is given by \(k=E /[3(1-2 \sigma)]\) where \(E\) is Young's modulus and \(\sigma\) Poisson's ratio.

In a certain crystal the unit cell can be taken as six identical atoms lying at the corners of a regular octahedron. Convince yourself that these atoms can also be considered as lying at the centres of the faces of a cube and hence that the crystal has cubic symmetry. Use this result to prove that the conductivity tensor for the crystal, \(\sigma_{i j}\), must be isotropic.

Given a non-zero vector \(\mathbf{v}\), find the value that should be assigned to \(\alpha\) to make $$ P_{i j}=\alpha v_{i} v_{j} \quad \text { and } \quad Q_{i j}=\delta_{i j}-\alpha v_{i} v_{j} $$ into parallel and orthogonal projection tensors respectively, i.e. tensors that satisfy respectively \(P_{i j} v_{j}=v_{i}, P_{i j} u_{j}=0\) and \(Q_{i j} v_{j}=0, Q_{i j} u_{j}=u_{i}\), for any vector \(\mathbf{u}\) that is orthogonal to \(\mathbf{v}\), Show, in particular, that \(Q_{i j}\) is unique, i.e. that if another tensor \(T_{i j}\) has the same properties as \(Q_{i j}\) then \(\left(Q_{i j}-T_{i j}\right) w_{j}=0\) for any vector \(\mathbf{w}\).

A fourth-order tensor \(T_{i j k l}\) has the properties $$ T_{j k l}=-T_{i j k l}, \quad T_{i j l k}=-T_{i j k l} $$ Prove that for any such tensor there exists a second-order tensor \(K_{m n}\) such that $$ T_{i j k l}=\epsilon_{i j m} \epsilon_{k l n} K_{m n} $$ and give an explicit expression for \(K_{m n}\). Consider two (separate) special cases, as follows. (a) Given that \(T_{i j k l}\) is isotropic and \(T_{i j j i}=1\), show that \(T_{i j k l}\) is uniquely determined and express it in terms of Kronecker deltas. (b) If now \(T_{i j k l}\) has the additional property $$ T_{k l i j}=-T_{i j k l} $$ show that \(T_{i j k l}\) has only three linearly independent components and find an expression for \(T_{i j k l}\) in terms of the vector $$ V_{i}=-\frac{1}{4} \epsilon_{j k l} T_{i j k l} $$
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