problem15_57

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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15.57: a) 2 2 2 ) , ( ) , ( A y x z y x y = + The trajectory is a circle of radius . A . 0 ) 0 , 0 ( , ) 0 , 0 ( , 0 At = = = z A y t . ) 2 , 0 ( , 0 ) 2 , 0 ( , 2 At A ω π z ω π y ω π t - = = = . 0 ) 2 , 0 ( , ) , 0 ( , At = - = = ω π z A ω π y ω π t A ω π z ω π y ω π t + = = = ) 2 3 , 0 ( , 0 ) 2 3 , 0 ( , 2 3 At b) ) cos( , ) sin( ωt kx dt dz v ωt kx dt dy v z y - - = = - + = = , 2 2 v v v z y = + = so the speed is constant. k j r ˆ ˆ z y + = ) sin( ) cos( ) cos( ) ( sin 2 2 t ω kx t ω kx ω A t ω kx t ω kx ω A zv yv z y - - - - - = + = v r v v r so , 0 = is tangent to the circular path. c) ) sin( , ) cos( 2 2 t ω kx dt dv a t ω kx dt dv a z z y y - - = = - - = = 2 2 2 2 2 2 )] ( sin ) ( cos [ ω A t ω kx t ω kx ω A za ya z y - = - + - - = + = a r ra a A r - = = = a r
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Unformatted text preview: a a r and 180 so cos ° = = ⋅ φ ra is opposite in direction to a r ; is radially inward. , 2 2 2 A z y = + so the path is again circular, but the particle rotates in the opposite sense compared to part (a )....
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