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Unformatted text preview: Homework 6 Chapt 8 & 9 Accelerating along a Racetrack Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity v h is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector v h at each point along the curved path and is directed toward the center of curvature of the path. The magnitude of the acceleration of an object that moves with constant speed v along a circular path of radius r is given by . 8.35. Model: Use the particle model for the ball in circular motion. Visualize: Solve: (a) The mass moves in a horizontal circle of radius 20 cm. r = The acceleration a a and the net force vector point to the center of the circle, not along the string. The only two forces are the string tension , T a which does point along the string, and the gravitational force G . F a These are shown in the freebody diagram. Newtons second law for circular motion is G cos cos 0 N z F T F T mg = = = 2 sin r r mv F T ma r = = = From the zequation, ( ) ( ) 2 0.500 kg 9.8 m/s 5.00 N cos cos11.54 mg T = = = (b) We can find the tangential speed from the requation: sin 0.63 m/s rT v m = = The angular speed is 0.63 ms 3.15 rad/s 30 rpm 0.20 m v r = = = = 8.46. Model: Masses 1 m and 2 m are considered particles. The string is assumed to be massless. Visualize:...
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This note was uploaded on 04/18/2010 for the course PHYS ? taught by Professor Zhou during the Spring '10 term at Georgia State University, Atlanta.
 Spring '10
 ZHOU

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