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Solutions-Week5

Solutions-Week5 - Homework Week 5 Solutions 13.5.P1 If the...

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Homework Week 5 Solutions 13.5.P1 If the speed of the particle is constant, the tangential component, a T ( t ) = v ( t ), of the acceleration is zero. However, the normal component, a N ( t ) = κ ( t ) v ( t ) 2 is not necessarily zero, since the particle may change its direction. 13.5.P2 For a particle in uniform circular motion around a circle, the acceleration vector a ( t ) points towards the center of the circle, whereas v ( t ) is tangent to the circle. 13.5.P3 (a) The velocity vector points in the direction of motion, hence the velocities of the two objects point in the same direction. (b) The length of the velocity vector is the speed. Since the speeds are not necessarily equal, the velocity vectors may have different lengths. (c) The acceleration is determined by the tangential component v ( t ) and the normal component κ ( t ) v ( t ) 2 . Since v and v may be different for the two objects, the acceleration vectors may have different directions. 13.5.P4 If the speed is constant, v ( t ) = 0. Therefore, the acceleration vector has only the normal component: a ( t ) = a N ( t ) N ( t ) The velocity vector always points in the direction of motion. Since the vector N ( t ) is orthogonal to the direction of motion, the vectors a ( t ) and v ( t ) are orthogonal. 13.5.P5 Since a line has zero curvature, the normal component of the acceleration is zero, hence a ( t ) has only the tangential component. The velocity vector is always in the direction of motion, hence the acceleration and the velocity vectors are parallel to the line. We conclude that (b) is the correct statement. 13.5.P6 The acceleration vector is given by the following decomposition: a ( t ) = v ( t ) T ( t ) + κ ( t ) v ( t ) 2 N ( t ) (1) In our case v ( t ) = 4 is constant hence v ( t ) = 0. In addition, the curvature of a circle of radius 2 is κ ( t ) = 1 2 . Substituting v ( t ) = 4, v ( t ) = 0 and κ ( t ) = 1 2 in (1) gives: a ( t ) = 1 2 · 4 2 N ( t ) = 8 N ( t ) The length of the acceleration vector is, thus, a ( t ) = 8 cm / s 2 13.5.P7 The tangential acceleration a T and the normal acceleration a N are the following values: a T ( t ) = v ( t ); a N ( t ) = κ ( t ) v ( t ) 2 At the moment where both speedometers read 110 mph, the speeds of the two cars are v = 110 mph. Since the track is circular, the curvature κ ( t ) is constant, hence the normal accelerations of the two cars are equal at this moment. Statement (b) is correct. 13.5.5 Differentiating r ( θ ) = sin θ, cos θ, cos 3 θ gives: v ( θ ) = r ( θ ) = cos θ, - sin θ, - 3 sin 3 θ v π 3 = cos π 3 , - sin π 3 , - 3 sin π = 1 2 , - 3 2 , 0 a ( θ ) = r ( θ ) = - sin θ, - cos θ, - 9 cos 3 θ a π 3 = - sin π 3 , - cos π 3 , - 9 cos π = - 3 2 , - 1 2 , 9 1
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The speed is the magnitude of the velocity vector, that is: v π 3 = v π 3 = 1 2 2 + - 3 2 2 + 0 2 = 1 13.5.8 We compute the velocity and acceleration vectors at t = 0 and t = 1: v ( t ) = r ( t ) = - 2 t, - 1 v (0) = 0 , - 1 v (1) = - 2 , - 1 a ( t ) = v ( t ) = - 2 , 0 a (0) = a (1) = - 2 , 0 Below is a sketch of the path and the velocity and acceleration vectors at t = 0 and t = 1: x y t = 0 t = 1 t = 2 t = - 2 v (0) v (1) a (0) a (1) r ( t ) = 1 - t 2 , 1 - t , - 2 t 2 13.5.11 We find v ( t ) by integrating a ( t ): v ( t ) = t 0 a ( u ) du = t 0 u, 4 du = 1 2 u 2 , 4 u t 0 + v 0 = t 2 2 , 4 t + v 0 The initial condition gives: v (0) = 0 , 0 + v 0 = 1 3 , - 2 v 0 = 1 3 , - 2 Hence, v ( t ) = t 2 2 , 4 t + 1 3 , - 2 = 3 t 2 + 2 6 , 4 t - 2 13.5.14 We integrate the acceleration vector to find the velocity vector
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