14.5Ex44-46

14.5Ex44-46 - S E C T I O N 14.5 Motion in Three-Space (ET...

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SECTION 14.5 Motion in Three-Space (ET Section 13.5) 565 Substituting (1), (3) and (6) into (5) gives: a N N = h 2 , 0 i − 4 t 4 ( t 2 + 4 ) h 2 t , 4 i = h 2 , 0 i − t t 2 + 4 h 2 t , 4 i = * 2 2 t 2 t 2 + 4 , 4 t t 2 + 4 + = ¿ 8 t 2 + 4 , 4 t t 2 + 4 À = 1 t 2 + 4 h 8 , 4 t i (7) Since a N = k a × v k k v k , a N is positive, hence N is a unit vector in the direction of a N N . Hence, by (7): N = a N N k a N N k = 1 t 2 + 4 h 8 , 4 t i 1 t 2 + 4 q 8 2 + ( 4 t ) 2 = h 8 , 4 t i p 64 + 16 t 2 = 4 h 2 , t i 4 p 4 + t 2 = 1 p 4 + t 2 h 2 , t i (8) Combining (7) and (8) yields: a N N = 4 t 2 + 4 h 2 , t i = 4 p t 2 + 4 N a N = 4 p t 2 + 4 (9) (c) By (4) and (9) we obtain the following decomposition: a ( t ) = a T ( t ) T + a N ( t ) N = Ã 2 t p t 2 + 4 ! T + Ã 4 p t 2 + 4 ! N 43. Find the components a T and a N of the acceleration vector of a particle moving along a circular path of radius R = 100 cm with constant velocity v 0 = 5cm / s. SOLUTION Since the particle moves with constant speed, we have v 0 ( t ) = 0, hence: a T = v 0 ( t ) = 0 The normal component of the acceleration is a N = κ ( t )v( t ) 2 . The curvature of a circular path of radius R = 100 is ( t ) = 1 R = 1 100 , and the velocity is the constant value v( t ) = v 0 = 5. Hence, a N = 1 R v 2 0 = 25 100 = 0 . 25 cm / s 2 44. At time t 0 , a moving particle has velocity vector v = 2 i and acceleration vector a = 3 i + 18 k . Determine the curvature ( t 0 )
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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14.5Ex44-46 - S E C T I O N 14.5 Motion in Three-Space (ET...

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