TEST BANK FOR Classical Dynamics of Particles and Systems 5th Ed By Stephen T. Thornton and Jerry B.
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Axes and lie in the plane. 1x′3x′13xx
The transformation equations are:
113cos45cos45xxx=°−°′
22xx=′
331cos45cos45xxx=°+°′
111122xx=−′ 3 x
22xx=′
311122xx=−′ 3 x
So the transformation matrix is:
1102201011022− 1
2 CHAPTER 1
1-2.
a)
x1ABCDαβγOEx2x3
From this diagram, we have
cosOEOAα=
cosOEOBβ= (1)
cosOEODγ=
Taking the square of each equation in (1) and adding, we find
2222coscoscosOAOBODαβγ++=++
2 2 2 OE (2)
But
22OAOBOC+= 2 (3)
and
22OCODOE+=2 (4)
Therefore,
222OAOBODOE++=2 (5)
Thus,
222coscoscos1αβγ++ = (6)
b)
x3AA′x1x2OEDCBθC′B′E′D′
First, we have the following trigonometric relation:
222cosOEOEOEOEEEθ 2 ′′+−= ′ (7)
MATRICES, VECTORS, AND VECTOR CALCULUS 3
But,
2222222coscoscoscoscoscosEEOBOBOAOAODODOEOEOEOEOEOEββαγγ′′′′=−+−+−′′=−+−′′′+−′
α
(8)
or,
22222222222coscoscoscoscoscos2coscoscoscoscoscos2coscoscoscoscoscosEEOEOEOEOEOEOEOEOEαβγαβααββγγ
γ
ααββγ′′=+++++′′′′−++′′′′=+−++′′′ γ
′ (9)
Comparing (9) with (7), we find
coscoscoscoscoscoscosθααββγγ=++′′ ′ (10)
1-3.
x1e3′x2x3Oe1e2e3Ae2′e1′e2e1e3
Denote the original axes by , , , and the corresponding unit vectors by e,, . Denote the new axes by , , and the corresponding unit vectors by 1x2x3x12e3e1x′2x′3x′1′e, 2′e, e. The effect of the rotation is ee, , e. Therefore, the transformation matrix is written as: 3′13′→21′ee→3→2′e
()()()()()()()()()111213212223313233cos,cos,cos,010cos,cos,cos,001100cos,cos,cos,′′′λ′′′==′′′eeeeeeeeeeeeeeeeee
1-4.
a) Let C = AB where A, B, and C are matrices. Then,
ijikkjkCA= B Σ (1)
()tjijkkikijkijkkCCABBA===ΣΣ
[Solved] TEST BANK FOR Classical Dynamics of Particles and Systems 5th Ed By Stephen T. Thornton and Jerry B.
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- Submitted On 12 Nov, 2021 07:15:33
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