HW4_solution - P1112 Fall 08 Solutions for HW 4 Problem 1:...

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Solutions for HW 4 Problem 1: Kinematics of Circular Motion (a) The object’s distance from the origin is given by the magnitude of its radial position vector R. R = [X(t) , Y(t)] = [Rcos( ω t) , Rsin( ω t)] R = ( R R ) (1/2) = (R 2 cos 2 ( ω t) + R 2 sin 2 ( ω t) ) (1/2) = ( R 2 ( cos 2 ( ω t) + sin 2 ( ω t) ) ) (1/2) = ( R 2 (1) ) (1/2) = R (b) We must show that the velocity V , the time derivative of R , is perpendicular to R for all time. In other words, V R = 0. R = [Rcos( ω t) , Rsin( ω t)] V = [-R ω sin( ω t) , R ω cos( ω t)] V R = -R 2 ω sin( ω t)cos( ω t) + R 2 ω sin( ω t)cos( ω t) = 0 (c) V = ( V V ) (1/2) = ( R 2 ω 2 ( sin 2 ( ω t) + cos 2 ( ω t) ) ) (1/2) = (R 2 ω 2 (1) ) (1/2) = R ω (d) The acceleration A is the time derivative of V . A = [-R ω 2 cos( ω t) , -R ω 2 sin( ω t)] = - ω 2 R We see that A is just a negative constant times R . They always point in opposite directions; R away from the origin and A towards the origin. A = (
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HW4_solution - P1112 Fall 08 Solutions for HW 4 Problem 1:...

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