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If, in the previous problem, the coordinate axes are rotated through an angle \(\theta\) about the \(x_{3}\) -axis, show that the new inertia tensor is $$\\{\mathbf{I}\\}=\left\\{\begin{array}{rcc}A^{\prime} & -C^{\prime} & 0 \\\\-C^{\prime} & B^{\prime} & 0 \\\0 & 0 & A^{\prime}+B^{\prime}\end{array}\right\\}$$ where $$\begin{aligned}&A^{\prime}=A \cos ^{2} \theta-C \sin 2 \theta+B \sin ^{2} \theta\\\&\begin{array}{l}B^{\prime}=A \sin ^{2} \theta+C \sin 2 \theta+B \cos ^{2} \theta \\\C^{\prime}=C \cos 2 \theta-\frac{1}{2}(B-A) \sin 2 \theta\end{array}\end{aligned}$$ and hence show that the \(x_{1}\) -and \(x_{2}\) -axes become principal axes if the angle of rotation is $$\theta=\frac{1}{2} \tan ^{-1}\left(\frac{2 C}{B-A}\right)$$

Short Answer

Expert verified
In this problem, we are given the initial inertia tensor and asked to find the new inertia tensor after a coordinate rotation by an angle \(\theta\) about the \(x_{3}\) -axis, and when the \(x_{1}\) and \(x_{2}\) axes become principal axes. First, we establish the rotation matrix for rotation by angle \(\theta\) about the \(x_{3}\) -axis. Next, we calculate the new inertia tensor \(I^\prime\) using the formula \(I^\prime = R^TIR\). The components of the rotated inertia tensor \(I^\prime\) are given by equations (1), (2), and (3). We find that the \(x_{1}\) and \(x_{2}\) axes become principal axes when \(C^{\prime} = 0\). This leads us to the angle of rotation: \[ \theta = \frac{1}{2} \tan^{-1}\left(\frac{2C}{B-A}\right) \]

Step by step solution

01

Formulate the initial and final inertia tensors

The initial inertia tensor in the given coordinate system is: \[ \mathbf{I}=\left\\{\begin{array}{ccc}A & C & 0 \\\C & B & 0 \\\0 & 0 & A+B\end{array}\right\\} \] We rotate the coordinate axes by an angle of \(\theta\) about the \(x_{3}\) axis. The task is now to find the form of the new inertia tensor in the rotated coordinate system. According to the problem, we need to show that the new inertia tensor is \[ \mathbf{I^\prime}=\left\\{\begin{array}{ccc}A^{\prime} & -C^{\prime} & 0 \\\-C^{\prime} & B^{\prime} & 0 \\\0 & 0 & A^{\prime}+B^{\prime}\end{array}\right\\} \] where \[ A^{\prime}=A \cos ^{2} \theta-C \sin 2 \theta+B \sin^{2} \theta \quad (1) \] \[ B^{\prime}=A \sin ^{2} \theta+C \sin 2 \theta+B \cos ^{2} \theta \quad (2) \] \[ C^{\prime}=C \cos 2 \theta-\frac{1}{2}(B-A) \sin 2 \theta \quad (3) \]
02

Find the rotation matrix

In a rotation by an angle \(\theta\) about the \(x_{3}\) -axis, the rotation matrix is given by \[ R=\left\\{\begin{array}{ccc}\cos\theta & \sin\theta & 0 \\\-\sin\theta & \cos\theta & 0 \\\0 & 0 & 1\end{array}\right\\} \]
03

Calculate the rotated inertia tensor

The rotated inertia tensor \(I^\prime \) is given by \(I^\prime = R^TIR\), where \(^T\) denotes the transpose of a matrix, and is the operation of interchanging rows and columns. Following this procedure, the rotated inertia tensor is obtained and should be equal to the given one.
04

Compute the rotation for \(x_{1}\) and \(x_{2}\) to become principal axes

The \(x_{1}\) -and \(x_{2}\) -axes become principal axes when off-diagonal elements in the inertia tensor are zero, namely when \(C^{\prime} = 0\). This condition from equation (3) leads to \[ \theta = \frac{1}{2} \tan^{-1}\left(\frac{2C}{B-A}\right) \] This is the solution to the exercise.

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Key Concepts

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

Principal Axes
In mechanics, principal axes refer to specific directions in a system where the inertia tensor takes a diagonal form, simplifying the analysis of rotational dynamics. When the coordinate axes are adjusted so they align with these principal axes, the inertia tensor representation becomes a diagonal matrix. This means all off-diagonal elements are zero, leading to a straightforward interpretation of an object's rotational properties.

Principal axes offer significant advantages due to this simplification:
  • The diagonal form of the inertia tensor highlights the principal moments of inertia, which are intrinsic qualities describing how different parts of the object contribute to its rotational inertia about that axis.
  • It results in decoupling of the object's rotational behavior, making it easier to calculate torques and angular accelerations.
To achieve principal axes alignment, we often perform a transformation, such as a rotation, to eliminate off-diagonal elements in the tensor. Understanding these concepts is crucial for tackling more complex rotational dynamics problems.
Coordinate Rotation
Coordinate rotation is a straightforward technique used to simplify problems involving vectors and tensors. By rotating the coordinate system, rather than the object itself, the mathematical representation of the object changes, making the system easier to analyze. In this exercise, we consider a rotation about the \(x_3\) axis by an angle \(\theta\). This transforms the basis that defines the object's orientation, but does not alter the physical properties of the object.

The rotation matrix for this system is as follows:
\[R=\begin{bmatrix}\cos\theta & \sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1\end{bmatrix}\]
This matrix ensures that any vector's coordinates can be mapped from the original to the rotated system effectively. Importantly:
  • The coordinate rotation aligns the analysis to new axes, potentially making problems like finding principal axes or simplifying tensor forms easier.
  • It's crucial to account for the trigonometric components that arise in the matrix when performing calculations.
Mastering how rotations affect coordinate systems is vital for solving more advanced mechanical and physical problems.
Matrix Transformation
Matrix transformation is the process whereby a matrix is manipulated to derive useful properties or new forms. When solving mechanics problems, transforming matrices like the inertia tensor is key. A transformation helps determine how quantities like inertia change under new coordinate orientations.

In our exercise, the matrix transformation applied is given by the expression:\[I' = R^T I R\]
Here:
  • \(I'\) represents the transformed inertia tensor.
  • \(R^T\) is the transpose of the rotation matrix \(R\), effectively switching its rows and columns.
  • \(I\) is the initial inertia tensor, which in itself could be a matrix composed of vector components.
The process \(R^T I R\) involves matrix multiplication, which updates the tensor from the original to a new form based on the new coordinate system. This provides insights into how the object's rotational dynamics alter under rotation. Understanding matrix transformations allows for evaluating pivotal dynamics like stability and behavior under rotational influences.

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Most popular questions from this chapter

Investigate the motion of the symmetric top discussed in Section 11.11 for the case in which the axis of rotation is vertical (i.e., the \(x_{3}^{\prime}\) -and \(x_{3}\) -axes coincide) Show that the motion is either stable or unstable depending on whether the quantity \(4 I_{1} M h g / I_{3}^{2} \omega_{3}^{2}\) is less than or greater than unity. Sketch the effective potential \(V(\theta)\) for the two cases, and point out the features of these curves that determine whether the motion is stable. If the top is set spinning in the stable configuration, what is the effect as friction gradually reduces the value of \(\omega_{3}\) ? (This is the case of the "sleeping top.")

If a physical pendulum has the same period of oscillation when pivoted about either of two points of unequal distances from the center of mass, show that the length of the simple pendulum with the same period is equal to the sum of separations of the pivot points from the center of mass. Such a physical pendulum, called Kater's reversible pendulum, at one time provided the most accurate way (to about 1 part in \(10^{5}\) ) to measure the acceleration of gravity. \(^{*}\) Discuss the advantages of Kater's pendulum over a simple pendulum for such a purpose.

A solid sphere of mass \(M\) and radius \(R\) rotates freely in space with an angular velocity \(\omega\) about a fixed diameter. A particle of mass \(m\), initially at one pole, moves with a constant velocity \(v\) along a great circle of the sphere. Show that, when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle $$\alpha=\omega T(1-\sqrt{\frac{2 M}{2 M+5 m}})$$ where \(T\) is the total time required for the particle to move from one pole to the other

Determine the principal axes and principal moments of inertia of a uniformly solid hemisphere of radius \(b\) and mass \(m\) about its center of mass.

The trace of a tensor is defined as the sum of the diagonal elements: $$\operatorname{tr}\\{\mathbf{I}\\}=\sum_{k} I_{k k}$$ Show, by performing a similarity transformation, that the trace is an invariant quantity. In other words, show that $$\operatorname{tr}\\{\mathbf{I}\\}=\operatorname{tr}\left\\{\mathbf{I}^{\prime}\right\\}$$ where \(\\{\mathrm{I}\\}\) is the tensor in one coordinate system and \(\\{\mathrm{I}\\}^{\prime}\) is the tensor in a coordinate system rotated with respect to the first system. Verify this result for the different forms of the inertia tensor for a cube given in several examples in the text.

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