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Gegeben ist der Tr盲gheitstensor $$ J=\left(\begin{array}{ccc} 24 & -4 & -8 \\ -4 & 60 & -2 \\ -8 & -2 & 60 \end{array}\right)\left[\mathrm{kg} \mathrm{m}^{2}\right] $$ Bestimmen Sie die Menge aller Winkelgeschwindigkeiten \(\omega\) bez眉glich derer die Rotationsenergie \(T_{0}=\frac{1}{2} \omega^{T} J \omega=\) \(1.0 \frac{\mathrm{kg} \mathrm{m}^{2}}{\mathrm{~s}^{2}}\) ist.

Short Answer

Expert verified
Answer: The set of all possible angular velocities given the provided inertia tensor and rotation energy is {蠅鈧, 蠅鈧, 蠅鈧儅, where 蠅鈧 = [2/鈭10, -1/鈭10, 1/鈭10], 蠅鈧 = [1/(2鈭14), 1/(2鈭14), 1/鈭14], and 蠅鈧 = [1/(2鈭34), -1/鈭34, -1/(2鈭34)].

Step by step solution

01

Find the eigenvalues and eigenvectors of the inertia tensor

First, to find the eigenvalues of the inertia tensor J, we need to solve the characteristic equation \(\det(J-\lambda I)=0\), where \(I\) is the identity matrix, and \(\lambda\) is the eigenvalue. Solving for 位, we get: $$ \left|\left(\begin{array}{ccc} 24-\lambda & -4 & -8 \\ -4 & 60-\lambda & -2 \\ -8 & -2 & 60-\lambda \end{array}\right)\right|=0. $$ Computing the determinant on the left side we get the characteristic polynomial: $$ (\lambda-20)(\lambda-56)(\lambda-68)=0. $$ Thus, the eigenvalues are 位鈧=20, 位鈧=56, and 位鈧=68. Now, we need to find the eigenvectors corresponding to each of these eigenvalues. For eigenvalue 位鈧=20, we solve the system (J-20I)v = 0: $$ \left(\begin{array}{ccc} 4 & -4 & -8 \\ -4 & 40 & -2 \\ -8 & -2 & 40 \end{array}\right)\left(\begin{array}{c} v_{x} \\ v_{y} \\ v_{z} \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right). $$ Solving this system of equations, we get the eigenvector v鈧 associated with 位鈧=20: v鈧=[2,-1,1]. Similarly, by solving the respective systems of equations, we get the eigenvectors for 位鈧=56 and 位鈧=68: v鈧=[1,1,2] and v鈧=[1,-2,-1].
02

Use the principal axes of inertia to find the angular velocities

Using the principal axes system of coordinates and the energy equation, we can calculate the total energy for a specific angular velocity: $$ T_0 = \frac{1}{2}\omega^T\cdot J \cdot\omega = \frac{1}{2}(\omega_x^{2}\lambda_{1}+\omega_y^{2}\lambda_{2}+\omega_z^{2}\lambda_{3}). $$ We are given T鈧=1 kg鈰卪虏/s虏. Plugging in the values of the eigenvalues and rearranging for the set of possible angular velocities, we get: $$ 1 = \frac{1}{2}(20\omega_x^{2}+56\omega_y^{2}+68\omega_z^{2}). $$ Now, in order to find the angular velocities, we can normalize the principal axes eigenvectors (v鈧, v鈧, and v鈧) and multiply them by their corresponding eigenvalues: $$ \omega = \sqrt{\frac{2}{\lambda_{i}}}v_{i}. $$ This can be done for all three eigenvectors v鈧, v鈧, and v鈧.
03

Find the set of possible angular velocities

To find the set of possible angular velocities, we plug in the normalized eigenvectors v鈧, v鈧, and v鈧 into the equation for 蠅 obtained in Step 2: For v鈧, $$ \omega_{1}=\sqrt{\frac{2}{20}}\left[2,-1,1\right] = \left[\frac{2}{\sqrt{10}}, -\frac{1}{\sqrt{10}}, \frac{1}{\sqrt{10}}\right]. $$ For v鈧, $$ \omega_{2}=\sqrt{\frac{2}{56}}\left[1,1,2\right] = \left[\frac{1}{2\sqrt{14}}, \frac{1}{2\sqrt{14}}, \frac{1}{\sqrt{14}}\right]. $$ For v鈧, $$ \omega_{3}=\sqrt{\frac{2}{68}}\left[1,-2,-1\right] = \left[\frac{1}{2\sqrt{34}}, -\frac{1}{\sqrt{34}}, -\frac{1}{2\sqrt{34}}\right]. $$ Now, the set of all possible angular velocities 蠅 will be the union of these three sets of angular velocities: $$ \omega = \{ \omega_{1}, \omega_{2}, \omega_{3} \}. $$

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Key Concepts

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

Eigenvalues and Eigenvectors
The eigenvalues and eigenvectors are fundamental concepts in linear algebra, especially when dealing with matrices. In the context of physics, particularly rotational dynamics, understanding them is crucial for analyzing the behavior of objects. Consider the inertia tensor, often a symmetric matrix representing how mass is distributed in a rigid body. To find eigenvalues, we look for scalars \( \lambda \) that satisfy the equation \( \det(J-\lambda I)=0 \), where \( J \) is the inertia tensor and \( I \) is the identity matrix.

Eigenvalues represent the principal moments of inertia, key in understanding how an object rotates around different axes. After finding \( \lambda_1 = 20, \lambda_2 = 56, \lambda_3 = 68 \), we proceeded to find the corresponding eigenvectors, which indicate the principal axes of rotation. Solving\( (J-\lambda I)\mathbf{v}=0 \) for each \( \lambda \), confers a direction vector, \( \mathbf{v} \), that does not change direction under the transformation. The determination of \( \mathbf{v}_1 = [2, -1, 1] \), \( \mathbf{v}_2 = [1, 1, 2] \), and \( \mathbf{v}_3 = [1, -2, -1] \) allows us to align our rotational frame with these significant directions.

Thus, by using eigenvalues and eigenvectors, one simplifies the complexity of rotational dynamics, making calculations and predictions about rotational behavior more manageable.
Rotational Energy
Rotational energy is the kinetic energy an object possesses due to its rotation. In the case of rigid bodies, it is a valuable measure to understand how the mass distribution affects motion. For an object rotating around its principal axes, the rotational energy \( T_0 \) is expressed as \( T_0 = \frac{1}{2} \omega^T \cdot J \cdot \omega \), where \( J \) is the inertia tensor and \( \omega \) is the angular velocity vector.

This rotational energy expression can be simplified when working with principal axes of inertia. By aligning our coordinate system along these axes, the product \( \omega^T \cdot J \cdot \omega \) becomes a sum of simpler terms: \( \frac{1}{2}(\omega_x^{2}\lambda_1 + \omega_y^{2}\lambda_2 + \omega_z^{2}\lambda_3) \). This transformation clarifies how each component of angular velocity contributes to the total energy. Here, \( \lambda_1, \lambda_2, \lambda_3 \) denote the eigenvalues of \( J \), revealing the incorporated mass distribution鈥檚 effect on energy.

Awareness of rotational energy assists in designing and analyzing objects, predicting their behavior under specific angular conditions. By knowing the energy derived from our derived angular velocities, one can ensure stability and efficient energy use in rotational mechanics.
Angular Velocity
Angular velocity \( \omega \) describes how quickly an object rotates about an axis and is represented by a vector quantity. This vector indicates both the rotational rate and the direction of rotation. When evaluating a rigid body's motion, it is often beneficial to consider the principal axes, as calculated from the inertia tensor's eigenvalues and eigenvectors.

To find the set of possible angular velocities that yield a rotational energy of \( 1 \text{ kg} \cdot \text{m}^2/\text{s}^2 \), we utilize each principal axis. The eigenvectors' normalization and multiplication by their respective eigenvalue-derived factors provide these velocities. For example, \( \omega_1=\sqrt{\frac{2}{20}}\left[2,-1,1\right] \) emerges from eigenvalue \( \lambda_1 = 20 \). Similarly, other angular velocities \( \omega_2 \) and \( \omega_3 \) derive from the remaining eigenvalues.

These calculations yield a set of angular velocities that fulfill the rotational energy requirement using different configurations. Understanding angular velocity in such a systematic way provides deeper insights into controlling rotational dynamics, crucial in engineering applications such as machinery design and even spacecraft orientation.

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Most popular questions from this chapter

Wir betrachten vier Populationen unterschiedlicher Arten \(a_{1}, a_{2}, a_{3}, a_{4}\). Vereinfacht nehmen wir an, dass bei einem Fortpflanzungszyklus, der f眉r alle vier Arten gleichzeitig stattfindet, die vier Arten mit einer gewissen H盲ufigkeit mutieren, aber es entstehen bei jedem solchen Zyklus wieder nur diese vier Arten. Mit \(f_{i j}\) bezeichnen wir die H盲ufigkeit, mit der \(a_{i}\) zu \(a_{j}\) mutiert. Die folgende Matrix gibt diese H盲ufigkeiten wieder \(-\) dabei gelte \(0 \leq t \leq 1:\) $$ \boldsymbol{F}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1-t & t & t \\ 0 & t & 1-t & t \\ 0 & t & t & 1-t \end{array}\right) $$ Gibt es nach hinreichend vielen Fortpflanzungszyklen eine Art, die dominiert?

Geben Sie die Eigenwerte und Eigenvektoren der folgenden Matrizen an: (a) \(\boldsymbol{A}=\left(\begin{array}{cc}3 & -1 \\ 1 & 1\end{array}\right) \in \mathbb{R}^{2 \times 2}\) (b) \(\boldsymbol{B}=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right) \in \mathbb{C}^{2 \times 2}\) (c) \(\boldsymbol{C}=\left(\begin{array}{ll}a & b \\ b & d\end{array}\right) \in \mathbb{R}^{2 \times 2}\)

Haben die quadratischen \(n \times n\)-Matrizen \(A\) und \(A^{T}\) dieselben Eigenwerte? Haben diese gegebenenfalls auch dieselben algebraischen und geometrischen Vielfachheiten?

Haben 盲hnliche Matrizen dieselben Eigenwerte? Haben diese dann gegebenenfalls auch dieselben algebraischen und geometrischen Vielfachheiten?

Gegeben ist eine elastische Membran im \(\mathbb{R}^{2}\), die von der Einheitskreislinie \(x_{1}^{2}+x_{2}^{2}=1\) berandet wird. Bei ihrer (als lineare Abbildung angenommene) Verformung gehe der Punkt \(\left(\begin{array}{l}v_{1} \\ v_{2}\end{array}\right)\) in den Punkt \(\left(\begin{array}{l}5 v_{1}+3 v_{2} \\ 3 v_{1}+5 v_{2}\end{array}\right)\) 眉ber. (a) Welche Form und Lage hat die ausgedehnte Membran? (b) Welche Geraden durch den Ursprung werden auf sich abgebildet?

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