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Chapter 10: Q5P (page 508)

Point masses 1 at (1, 1, 1) and at (-1, 1, 1).

Short Answer

Expert verified

Answer:

Inertia tensor is $t = (\begin{matrix} 4 & 0 & 0 \\ 0 & 4 & - 2 \\ 0 & - 2 & 4 \end{matrix})$ .

Step by step solution

01

Given Information

The given information is mentioned below:

$m_{1} = m_{2} = 1 r_{1} = (1, 1, 1) r_{2} = (- 1, 1, 1)$

02

Definition of Inertia tensor.

Inertia tensor is the angular momentum of a rigid body determined by the product of its tensor and its rotation vector.

03

Find the eigenvalues.

Solve for eigenvalues.

$l_{i j} = (3 δ_{i j} - 1) + (3 δ_{i j} - {(- 1)}^{δ_{1 l} + δ_{1 l}}) l = (\begin{matrix} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{matrix}) + (\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}) = (\begin{matrix} 4 & 0 & 0 \\ 0 & 4 & - 2 \\ 0 & - 2 & 4 \end{matrix})$

04

Find principal axes and corresponding moment of inertia.

Find the moment of inertia.

$V_{1} = (1, 0, 0) V_{2} = (0, - 1, 1) V_{3} = (0, 1, 1)$

Simplify Further

$l_{1} = 4 l_{2} = 6 l_{3} = 2$

Hence, the inertia tensor is, $t = (\begin{matrix} 4 & 0 & 0 \\ 0 & 4 & - 2 \\ 0 & - 2 & 4 \end{matrix})$ .

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Most popular questions from this chapter

Write equations (2.12) out in detail and solve the three simultaneous equations (say by determinants) for $x_{1}, x_{2}, x_{3}$ in terms of $x_{1}', x_{2}', x_{3}'$ to verify equations (2.13) . Use your results in Problem 4.

For the point mass m we considered in (4.2) to (4.4), the velocity is $v = ω \times r$ so the kinetic energy is $T = \frac{1}{2} m v^{2} = \frac{1}{2} (ω \times r) \nabla \dot{} (ω \times r)$ .Show that T can be written in matrix notation as $T = \frac{1}{2} ω^{T} I ω$ where I is the inertia matrix, $ω$ is a column matrix, and is a row matrix with elements equal to the components of $ω$

P Derive the expression (9.11)for curl V in the following way. Show that $\nabla x_{1} = \frac{e_{1}}{h_{1}}$ and $\nabla \times (\nabla x_{1}) = \nabla \times (\frac{e_{1}}{h_{1}}) = 0$ . Write V in the form $V = \frac{e_{1}}{h_{1}} (h_{1} V_{1}) + \frac{e^{2}}{h_{2}} (h_{2} V_{2}) + \frac{e_{3}}{h_{3}} (h_{3} V_{3})$ and use vector identities from Chapter 6 to complete the derivation.

Parabolic.

Write the tensor transformation equations for $ε_{i j k} ε_{m n p}$ to show that this is a (rank 6) tensor (nota pseudo tensor). Hint:Write (6.1) for each $ε$ and multiply them, being careful not to re-use a pair of summation indices.

See all solutions

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