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solution21_pdf - arellano(aa39398 Practice Homework Chapter...

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arellano (aa39398) – Practice Homework Chapter 6 Solutions – weathers – (17101) 1 This print-out should have 43 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A fighter plane flying at constant speed 540 m / s and constant altitude 12000 m makes a turn of curvature radius 5700 m. On the ground, the plane’s pilot weighs (79 kg) (9 . 8 m / s 2 ) = 774 . 2 N. What is his/her apparent weight during the plane’s turn? Correct answer: 4114 . 96 N. Explanation: The plane’s altitude is not important per se , but the fact that it is constant tells us that the plane moves in a horizontal plane and its normal acceleration a N = v 2 R = 51 . 1579 m / s 2 has a horizontal direction. Furthermore, con- stant speed implies zero tangential accelera- tion, hence vectora = vectora N . The apparent weight of the pilot combines his true weight mvectorg and the inertial force mvectora due to plane’s acceleration, vector W app = m ( vectorg vectora ) . In magnitude, | vectorg vectora | = radicalbig g 2 + a 2 = 52 . 0881 m / s 2 because vectora is horizontal while vectorg is vertical. Consequently, the pilot’s apparent weight is vector W app = (52 . 0881 m / s 2 ) (79 kg) = 4114 . 96 N , which is quite heavier than his/her true weight of 774 . 2 N. 002 (part 1 of 2) 10.0 points An ant of mass m clings to the rim of a flywheel of radius r , as shown. The flywheel rotates clockwise on a horizontal shaft S with constant angular velocity ω . As the wheel rotates, the ant revolves past the stationary points I , II , III , and IV . The ant can adhere to the wheel with a force much greater than its own weight. r S I II III IV Ant ω It will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points? 1. IV 2. II 3. III correct 4. It will be equally difficult for the ant to adhere to the wheel at all points. 5. I Explanation: The sum of the ant’s gravity and its ad- hesion force is the centripetal force vector F c with magnitude m ω 2 r . Thus vector F c = mvectorg + vector F ad vector F ad = vector F c mvectorg . The maximum vector F ad is when vector F c and mvectorg are in opposite directions; i.e. , at III . 003 (part 2 of 2) 10.0 points What is the magnitude of the minimum adhe- sion force necessary for the ant to stay on the flywheel at point III ? 1. F = m ω 2 r + m g correct 2. F = m ω 2 r 2 3. F = m ω 2 r m g
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arellano (aa39398) – Practice Homework Chapter 6 Solutions – weathers – (17101) 2 4. F = m ω 2 r 2 + m g 5. F = m g Explanation: According to the explanation in the previ- ous part, the minimum adhesion force is bardbl vector G bardbl + bardbl vector F c bardbl = m ω 2 r + m g . 004 (part 1 of 2) 10.0 points A spring has a force constant of 761 N / m and an unstretched length of 8 cm. One end is attached to a post that is free to rotate in the center of a smooth table, as shown in the top view below. The other end is attached to a 4 kg disk moving in uniform circular motion on the table, which stretches the spring by 1 cm.
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