Chapter 08 - CHAPTER 8 Quick Quizzes 1. 2. 3. 4. 5. (d)....

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C HAPTER 8 Quick Quizzes 1. (d). 2. (b). 3. (b). The hollow cylinder has the higher moment of inertial, so it will be given the smaller acceleration and take longer to stop. 4. (a). The hollow sphere has the higher moment of inertia. 5. (a). Nathan is correct. When one of the children curls up inside the tire, the rolling system closely resembles a solid cylinder or disk. In a race between a solid disk and a ring (or hollow cylinder), the disk always wins, independent of masses or radii. See the solution to problem 8.63. 6. (c). The box. All objects have the same potential energy associated with them before they are released. As the objects move down the inclines, this potential energy is transformed to kinetic energy. For the ball and cylinder, the transformation is into both rotational and translational kinetic energy. The box has only translational kinetic energy. Because the kinetic energies of the ball and cylinder are split into two types, their translational kinetic energy is necessarily less than that of the box. Consequently, their translational speeds are less than that of the box, so the ball and cylinder will lag behind. 7. (c). Apply conservation of angular momentum to the system (the two disks) before and after the second disk is added to get the result: ( ) 10 1 2 f II I ω =+ . 8. (a). The Earth already bulges slightly at the Equator, and is slightly flat at the poles. If more mass moved towards the Equator, it would essentially move the mass to a greater distance from the axis of rotation, and increase the moment of inertia. Because conservation of angular momentum requires that const zz I = , an increase in the moment of inertia would decrease the angular velocity, and slow down the spinning of the Earth. Thus, the length of each day would increase. 239
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CHAPTER 8 Problem Solutions 8.1 To exert a given torque using minimum force, the lever arm should be as large as possible. In this case, the maximum lever arm is used when the force is applied at the end of the wrench and perpendicular to the handle. Then, 40.0 N m 0.300 m min max F d τ == = 133 N 0.300 m F Pivot 8.2 The lever arm is ( ) 23 1.20 10 m cos 48.0 8.03 10 m d ° , and the torque is () ( ) 3 80.0 N 8.03 10 m Fd × = 0.642 N m counterclockwise . 8.3 ( ) ( ) ( ) ( ) 100 N sin 20.0 0.600 m 900 N sin 15.0 0.800 m A =− ° ° = ⎡⎤ ⎣⎦ 207 N m ( )( ) ( ) ( ) 800 N 0.600 m 900 N sin 15.0 0.800 m B ° ( ) ( ) 900 N cos15.0 0.600 m = 145 N m () ( ) ( ) ( ) 100 N sin 20.0 0.600 m 100 N cos 20.0 0.800 m C ° ° = 95.7 N m 8.4 In the 0° position, ( )( ) 0 mg 0 At 30°, ( ) ( ) 2 10 kg 9.8 m s 0.400 m sin 30 ° = 20 N m At 60°, ( ) ( ) 2 10 kg 9.8 m s 0.400 m sin 60 34 N m At 90°, ( ) ( ) 2 10 kg 9.8 m s 0.400 m = 39 N m 8.5 ( ) ( ) [] sin F lever arm mg L θ =⋅ = ( ) ( ) 2 3.0 kg 9.8 m s 2.0 m sin 5.0 ° = 5.1 N m 240
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CHAPTER 8 8.6 Resolve the 100-N force into components parallel to and perpendicular to the rod, as ( ) ( ) 100 N cos 20.0 37.0 54.5 N parallel F + ° = and () ( ) .
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This note was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 08 - CHAPTER 8 Quick Quizzes 1. 2. 3. 4. 5. (d)....

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