PH101 HW Rotational Kinematics 2

PH101 HW Rotational Kinematics 2 - This print-out should...

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Unformatted text preview: This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the magnitude of the torque produced by a 4.8 N force applied to a door at a per- pendicular distance of 0.21 m from the hinge. Correct answer: 1 . 008 N m. Explanation: Let : F = 4 . 8 N , d = 0 . 21 m , and = 90 . . = F d sin = (4 . 8 N) (0 . 21 m) sin 90 . = 1 . 008 N m . 002 10.0 points An Atwood machine is constructed using a disk of mass 2 . 3 kg and radius 23 . 7 cm. 3 . 7 m 23 . 7 cm 2 . 3 kg 1 . 78 kg 1 . 12 kg What is the acceleration of the system? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 59704 m / s 2 . Explanation: Let : m = 2 . 3 kg , R = 23 . 7 cm = 0 . 237 m , m 1 = 1 . 12 kg , m 2 = 1 . 78 kg , and h = 3 . 7 m . Consider the free body diagrams 1 . 78 kg 1 . 12 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r is in the di- rection of the heavier mass m 1 . 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 . The pulley is a solid disk, so I = 1 2 mr 2 and net = ccw- cw = I T 1 r- T 2 r = 1 2 mr 2 a r = 1 2 mr a Multiplying by 2 r gives ma = 2 T 1- 2 T 2 = 2 ( m 1- m 2 ) g- 2 ( m 1 + m 2 ) a a = 2 ( m 1- m 2 ) g m + 2 m 1 + 2 m 2 = 2 (1 . 12 kg- 1 . 78 kg) (9 . 8 m / s 2 ) 2 . 3 kg + 2 (1 . 12 kg) + 2 (1 . 78 kg) = 1 . 59704 m / s 2 . 003 10.0 points A 308 kg merry-go-round in the shape of a horizontal disk with a radius of 1 . 3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope....
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This note was uploaded on 10/03/2010 for the course PH 101 taught by Professor Jones,s during the Spring '08 term at Alabama.

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PH101 HW Rotational Kinematics 2 - This print-out should...

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