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sol ass1 - ANALYTICAL MECHANICS CIVE 281 Solution to...

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1 ANALYTICAL MECHANICS, CIVE 281 Solution to Assignment No. 1 1. The motion of a particle is defined by the position vector ( ( j i r t t t A t t t A cos sin sin cos - + + = , where t is expressed in seconds. Determine the values of t for which the position vector and the acceleration vector (a) perpendicular, (b) parallel. (B. & J. 11.93) Solution: Position vector, ( ( j i r t t t A t t t A cos sin sin cos - + + = Velocity vector, v r = dt d ( ( j i t t t t A t t t t A cos sin cos sin cos sin - + + + + - = ( ( j i t t A t t A sin cos + = Acceleration vector, ( 29 ( 29 j i a v t t t A t t t A dt d sin cos cos sin + + + - = = (a) For the position vector, r and acceleration vector, a to be perpendicular, 0 = a r Definition of dot product of two vectors z z y y x x B A B A B A + + = B A ( ( [ ] ( ( [ ] 0 sin cos cos sin cos sin sin cos = + + + - - + + j i j i t t t A t t t A t t t A t t t A O P y x P 0 A r Fig. P11.93
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2 ( ( ( ( [ ] 0 sin cos cos sin cos sin sin cos 2 = + - + + - + t t t t t t t t t t t t A ( ( ( ( 0 sin cos cos sin cos sin sin cos = + - + + - + t t t t t t t t t t t t ( ( 0 cos sin sin cos cos sin cos sin cos sin cos sin 2 2 2 2 2 2 = - + - + + + - - t t t t t t t t t t t t t t t t t t 0 sin cos cos sin 2 2 2 2 2 2 = + - + - t t t t t t ( ( 0 cos 1 sin 1 2 2 2 2 = - + - t t t t ( ( 0 cos sin 1 2 2 2 = + - t t t Trigonometric pythagorean identity 1 cos sin 2 2 = + t t Therefore, 0 1 2 = - t 1 2 = t t = 1 s (b) For the position vector, r and acceleration vector, a to be perpendicular, 0 = × a r Definition of cross product of two vectors ( 29 ( 29 ( 29 x y y x z x x z y z z y z y x z y x B A B A ˆ B A B A ˆ B A B A ˆ B B B A A A ˆ ˆ ˆ - + - + - = = × k j i k j i B A ( ( [ ] ( ( [ ] 0 sin cos cos sin cos sin sin cos = + + + - × - + + j i j i t t t A t t t A t t t A t t t A ( ( ( ( [ ] 0 cos sin cos sin sin cos sin cos 2 2 = + - - - + + k ˆ t t t t t t A t t t t t t A ( [ ] 0 cos cos sin cos sin sin sin cos sin cos sin cos 2 2 2 2 2 2 2 = - + + - - + + + k ˆ t t t t t t t t t t t t t t t t t t A ( 0 cos 2 sin 2 2 2 = + k ˆ t t t t ( 0 cos sin 2 2 2 = + k ˆ t t t 0 2 = t t = 0
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3 2. At a general time t, a particle has an acceleration ( k j i a ˆ t ˆ t ˆ e t 3 cos 6
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