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Chapter 10: Q6P (page 508)

Point masses 1 at (1, 1, -2) and 2 at (1, 1, 1).

Short Answer

Expert verified

Answer:

Inertia tensor is $l = (\begin{matrix} 9 & - 3 & 0 \\ - 3 & 9 & 0 \\ 0 & 0 & 6 \end{matrix})$ .

Step by step solution

01

Given Information.

The given information is mentioned below:

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

02

Definition of Point mass.

The concept of a physical object with non-zero mass but explicitly and particularly minuscule volume or linear dimension is known as point mass.

03

Find the eigenvalues.

Solve for eigenvalues.

$l_{i j} = (6 δ_{i j} - {(- 2)}^{δ_{3 i} + δ_{3 i}}) + 2 (3 δ_{i j} - 1) l = (\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix}) - (\begin{matrix} 1 & 1 & - 2 \\ 1 & 1 & - 2 \\ - 2 & - 2 & 4 \end{matrix}) + 2 (\begin{matrix} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \end{matrix}) = (\begin{matrix} 5 & - 1 & 2 \\ - 1 & 5 & 2 \\ 2 & 2 & 2 \end{matrix}) + (\begin{matrix} 4 & - 2 & - 2 \\ - 2 & 4 & - 2 \\ - 2 & - 2 & 4 \end{matrix}) l = (\begin{matrix} 9 & - 3 & 0 \\ - 3 & 9 & 0 \\ 0 & 0 & 6 \end{matrix})$

04

Find principal axes and corresponding moment of inertia.

Find the moment of inertia.

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

Simplify further.

$l_{1} = 12 l_{2} = 6 l_{3} = 6$

Hence, inertia tensor is, $l = (\begin{matrix} 9 & - 3 & 0 \\ - 3 & 9 & 0 \\ 0 & 0 & 6 \end{matrix})$

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Most popular questions from this chapter

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.

Show that the nine quantities $T_{ij} = (\partial V_{i}) / (\partial x_{J})$ (which are the Cartesian components of $\nabla V$ where V is a vector) satisfy the transformation equations $(2.14)$ for a Cartesian $2^{nd}$ -rank tensor. Show that they do not satisfy the general tensor transformation equations as in $(10.12)$ . Hint: Differentiate $(10.9)$ or $(10.10)$ partially with respect to, say, $x'_{k}$ . You should get the expected terms [as in $(10.12)$ ] plus some extra terms; these extraneous terms show that $(\partial V_{i}) / (\partial x_{J})$ is not a tensor under general transformations. Comment: It is possible to express the components of $\nabla V$ correctly in general coordinate systems by taking into account the variation of the basis vectors in length and direction.

$\begin{array}{l} x = uv \\ y = u \sqrt{1 - v^{2}} \end{array}$

Observe that a simpler way to find the velocity $\frac{ds}{dt}$ in (8.10)is to divide the vectordsin (8.6)by. Complete the problem to find the acceleration in cylindrical coordinates.

In spherical coordinates $\nabla^{2} r, \nabla^{2} (r^{2}), \nabla^{2} (\frac{1}{r^{2}}), \nabla^{2} e^{i k r c o s θ}$ .

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