# Ch4 HW4 - Ch4 HW4

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Ch4 HW4 (Homework) BRANDON JARROTT FELDKAMP PHYS 2211, section M, Fall 2008 Instructor: Jennifer Curtis Web Assign Current Score: 76 out of 77 Due: Wednesday, September 24, 2008 09:00 AM EDT Description More on curving motion Instructions Reading: Sec. 4.13. Additional examples of curving motion are found in Sec. 4.18. The due date for this assignment is past. Your work can be viewed below, but no changes can be made. 1. [MI2 04.P.110] 6/7 points A child of mass 25 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 9 m/s. At this instant the cord is 4.20 m long. (a) At this instant, what is the parallel component of the rate of change of the child's momentum? = < 0 0 , 0 0 , 0 0 > (kg·m/s)/s (b) At this instant, what is the perpendicular component of the rate of change of the child's momentum? = < 0 0 , 482.143 482 , 0 0 > (kg·m/s)/s (c) At this instant, what is the net force acting on the child? net = < 0 0 , 482.143 482 , 0 0 > N (d) What is the magnitude of the force that the elastic cord exerts on the child? (It helps to draw a diagram of the forces.) | due to cord | = 2.37e2 727 N (e) The relaxed length of the elastic cord is 4.15 m. What is the stiffness of the cord? (You will be given credit for this part if it is consistent with your answer to part d, even if the answer to part d is incorrect.) k s = 4.74e3 4740 N/m Solution or Explanation (a) At this instant the magnitude of the momentum isn't changing, so the rate of change of the parallel component of momentum is zero. (b) There is a rate of change of the perpendicular component of momentum, upward. Ch4 HW4 http://www.webassign.net/v4cgibfeldkamp3@gatech/student.pl?j=2008. .. 1 of 11 11/18/2008 6:21 PM

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(c) The momentum principle says that the net force is equal in magnitude and direction to the rate of change of momentum, and the direction is upward (+y), directed toward the center of the kissing circle. The x and z components of the net force are zero. (d) If you draw a diagram of the forces, you'll see that the y component of the net force has two contributions, the upward force of magnitude F of the elastic cord and the downward force of the Earth of magnitude mg. So the y component of the net force found in part (c) is equal to F-mg; solve for F. (e) Now that you know from part (c) the force exerted by the elastic cord, you can determine the stiffness from |F| = k s |s|, where s is the stretch (change in length of the cord). 2. [MI2 04.P.94.alt01] 7/7 points A Ferris wheel is a vertical, circular amusement ride with radius 6 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9 s. Consider a rider whose mass is 50 kg. At the bottom of the ride, what is the parallel component of the rate of change of the rider's momentum? = < 0 0 , 0 0 , 0 > kg·m/s/s At the bottom of the ride, what is the perpendicular component of the rate of change of the rider's momentum?
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## This note was uploaded on 02/21/2010 for the course PHYS 2211 taught by Professor Prog during the Spring '09 term at Georgia Institute of Technology.

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Ch4 HW4 - Ch4 HW4

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