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**Unformatted text preview: **billing (cab4763) – hw 10 – opyrchal – (11109) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 5 . 2 kg mass is connected by a light cord to a 1 . 7 kg mass on a smooth surface as shown. The pulley rotates about a frictionless axle and has a moment of inertia of 0 . 67 kg · m 2 and a radius of 0 . 37 m. b 5 . 2kg 1 . 7kg F 1 F 2 R What is the acceleration of the two masses? The acceleration of gravity is 9 . 81 m / s 2 . As- sume that the cord does not slip on the pulley. Correct answer: 4 . 32522 m / s 2 . Explanation: F 2 a F 1 m 1 g a Let : I = 0 . 67 kg · m 2 , R = 0 . 37 m , m 1 = 5 . 2 kg , m 2 = 1 . 7 kg , and g = 9 . 81 m / s 2 . Applying Newton’s second law to m 1 and m 2 , m 1 a = m 1 g- F 1 m 2 a = F 2 F net = F 1- F 2 = ( m 1 g- m 1 a )- m 2 a = m 1 g- ( m 1 + m 2 ) a a = αR , so applying Newton’s second law to the pulley, F net R = I α = I parenleftBig a R parenrightBig I a = F net R 2 I a = [ m 1 g- ( m 1 + m 2 ) a ] R 2 a = m 1 g R 2 I + ( m 1 + m 2 ) R 2 = (5 . 2kg)(9 . 81m / s 2 )(0 . 37m) 2 . 67kg · m 2 +(5 . 2kg+1 . 7kg)(0 . 37m) 2 = 4 . 32522 m / s 2 . 002 (part 2 of 3) 10.0 points What is the magnitude of the force F 1 ? Correct answer: 28 . 5209 N. Explanation: F 1 = m 1 g- m 1 a = (5 . 2kg)(9 . 81m / s 2- 4 . 32522 m / s 2 ) = 28 . 5209 N . 003 (part 3 of 3) 10.0 points What is the magnitude of the force F 2 ? Correct answer: 7 . 35287 N. Explanation: F 2 = m 2 a = (1 . 7kg)(4 . 32522 m / s 2 ) = 7 . 35287 N . 004 10.0 points A uniform 9 kg rod with length 42 m has a frictionless pivot at one end. The rod is released from rest at an angle of 30 ◦ beneath the horizontal. billing (cab4763) – hw 10 – opyrchal – (11109) 2 2 1 m 4 2 m 9 k g 30 ◦ What is the angular acceleration of the rod immediately after it is released? The momentimmediately after it is released?...

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