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Unformatted text preview: Version 076/ABADA midterm 03 Turner (58220) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 1.5 kg bicycle tire of radius 0.31 m starts from rest and rolls down from the top of a hill that is 20.0 m high. What is the translational speed of the tire when it reaches the bottom of the hill? As sume that the tire is a hoop with I = mr 2 . The acceleration of gravity is 9 . 81 m / s 2 . 1. 10.388 2. 11.2929 3. 17.1552 4. 16.2748 5. 14.0071 6. 14.6908 7. 15.3441 8. 13.6525 9. 16.8668 10. 15.6605 Correct answer: 14 . 0071 m / s. Explanation: Let : m = 1 . 5 kg , r = 0 . 31 m , h = 20 . 0 m , and g = 9 . 81 m / s 2 . The moment of inertia of the solid cylinder is I = 1 2 mr 2 v i = 0 m/s and h f = 0 m, so by conservation of energy, U i = K f U i = K trans,f + K rot,f mg h = 1 2 mv 2 f + 1 2 I 2 f mg h = 1 2 mv 2 f + 1 2 mr 2 parenleftBig v f r parenrightBig 2 g h = v 2 f v f = radicalbig g h = radicalBig (9 . 81 m / s 2 )(20 m) = 14 . 0071 m / s . keywords: 002 10.0 points Masses M 1 and M 2 are supported by wires that have equal lengths when unstretched. The wire supporting M 1 is an aluminum wire . 85 mm in diameter, and the one supporting M 2 is a steel wire 0 . 3 mm in diameter. What is the ratio M 1 M 2 if the two wires stretch by the same amount? Youngs mod ulus for aluminum is 7 10 10 N / m 2 and for steel 2 10 11 N / m 2 . 1. 1.23047 2. 1.8515 3. 7.62222 4. 2.80972 5. 1.80785 6. 2.85714 7. 1.4 8. 2.06429 9. 4.2875 10. 0.83595 Correct answer: 2 . 80972. Explanation: Let : r 1 = 0 . 425 mm , r 2 = 0 . 15 mm , Y 1 = 7 10 10 N / m 2 , and Y 2 = 2 10 11 N / m 2 . Youngs Modulus is Y = F L A . L 1 = L 2 = L , so for the aluminum wire, L 1 = M 1 g L A 1 Y Al = M 1 g L r 2 1 Y 1 and for the steel wire L 2 = M 2 g L A 2 Y steel = M 2 g L r 2 2 Y 2 . Version 076/ABADA midterm 03 Turner (58220) 2 The wires stretch by the same amount (and L 1 = L 2 ), so M 1 g L r 2 1 Y 1 = M 2 g L r 2 2 Y 2 M 1 M 2 = r 2 1 Y 1 r 2 2 Y 2 = (0 . 425 mm) 2 (7 10 10 N / m 2 ) (0 . 15 mm) 2 (2 10 11 N / m 2 ) = 2 . 80972 . 003 10.0 points The speed of a moving bullet can be deter mined by allowing the bullet to pass through two rotating paper disks mounted a distance 97 cm apart on the same axle. From the angular displacement 42 . 5 of the two bul let holes in the disks and the rotational speed 1365 rev / min of the disks, we can determine the speed of the bullet. 42 . 5 v 1365 rev / min 97 cm What is the speed of the bullet? 1. 108.732 2. 80.2058 3. 210.759 4. 122.194 5. 113.84 6. 186.925 7. 180.303 8. 248.583 9. 111.685 10. 62.5203 Correct answer: 186 . 925 m / s....
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This note was uploaded on 05/11/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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