hw9 - patel (kp9583) HW9 Mackie (20211) 1 This print-out...

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Unformatted text preview: patel (kp9583) HW9 Mackie (20211) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points An amusement park builds a ride in which the victim is made to spin about a pole with a rocket strapped on his seat. The rider, box and rocket have an initial total mass of 240 kg. Neglect the mass of the rod of length 4 m . 260 m / s 75 N 4 m 240 kg What is the moment of inertia of rider, box and rocket about the pole? The acceleration of gravity is 9 . 8 m / s 2 . Treat the rider, box and rocket as a point mass. Correct answer: 3840 kg m 2 . Explanation: Let : F = 75 N , v t = 260 m / s , M = 240 kg , and = 4 m . For a point mass, the inertia is I = M 2 = (240 kg) (4 m) 2 = 3840 kg m 2 . 002 (part 2 of 4) 10.0 points The rocket develops a thrust of 75 N perpen- dicular to the path of the rider. What is the initial angular acceleration of the rider? Correct answer: 0 . 078125 rad / s 2 . Explanation: The torque of this force is = F = I = F I = (4 m) (75 N) 3840 kg m 2 = . 078125 rad / s 2 . 003 (part 3 of 4) 10.0 points After what time t is the riders velocity equal to 7 m / s? Neglect the change in mass of the rocket. Correct answer: 22 . 4 s. Explanation: Let : v = 7 m / s . = t for constant angular acceleration, so v = = t t = v = 7 m / s (4 m) (0 . 078125 rad / s 2 ) = 22 . 4 s . 004 (part 4 of 4) 10.0 points Gas exits the rocket at v t = 260 m / s. What mass per second must leave to de- velop the thrust F given above? Correct answer: 0 . 288462 kg / s. Explanation: A flow of mass d m dt with a velocity v gener- ates a thrust of F = d m dt v , so d m dt = F v t = 75 N 260 m / s = . 288462 kg / s . 005 10.0 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2 kg mass of the pulley is concentrated on its rim, which is a distance 24 . 9 cm from the axle. patel (kp9583) HW9 Mackie (20211) 2 The mass on the right is 1 . 03 kg and on the left is 1 . 51 kg. 3 . 8 m 2 kg 24 . 9 cm 1 . 51 kg 1 . 03 kg What is the magnitude of the linear acceler- ation of the hanging masses? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 03612 m / s 2 . Explanation: Let : M = 2 kg , R = 24 . 9 cm , m 1 = 1 . 03 kg , m 2 = 1 . 51 kg , h = 3 . 8 m , and v = R . Consider the free body diagrams 1 . 51 kg 1 . 03 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r is in the direc- tion of the heavier mass m 2 . For the mass m 1 , F net = m 1 a = m 1 g- T 1 T 1 = m 1 g- m 1 a and for the mass m 2 , F net = m 2 a = T 2- m 2 g T 2 = m 2 a + m 2 g ....
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hw9 - patel (kp9583) HW9 Mackie (20211) 1 This print-out...

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