A1 9th Edition Solution

# A1 9th Edition Solution - ANALYTICAL MECHANICS CIVE 281...

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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. 11.95) ( ) ( ) j i r t t t A t t t A cos sin sin cos + + = Solution: Position vector, 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, ( ) ( ) 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. P11.95

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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 0 = × a r t = 1 s (b) For the position vector, r and acceleration vector, a to be perpendicular, Definition of cross product of two vectors ( ) ( ) ( ) 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 =
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A1 9th Edition Solution - ANALYTICAL MECHANICS CIVE 281...

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