# ch08 - Chapter 8 Problems 1 CHAPTER 8 ROTATIONAL KINEMATICS...

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Chapter 8 Problems 1 CHAPTER 8 ROTATIONAL KINEMATICS

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a T a c a φ 3.0 s 8.0 s +15 rad/ s –9.0 rad/ s 0 Time (s) Angular velocity  +3.0 rad/s 35.0 ° a c a T a Race car ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) Using Equation 8.1 ( θ = Arc length / Radius) to calculate the angle (in radians) that each object subtends at your eye shows that θ Moon = 9.0 × 10 −3 rad, θ Pea = 7.0 × 10 −3 rad, and θ Dime = 25 × 10 −3 rad. Since θ Pea is less than θ Moon , the pea does not completely cover your view of the moon. However, since θ Dime is greater than θ Moon , the dime does completely cover your view of the moon. 2. 2.20 cm 3. 38.2 s 4. (a) An angular acceleration of zero means that the angular velocity has the same value at all times, as in statements A or B. However, statement C is also consistent with a zero angular acceleration, because if the angular displacement does not change as time passes, then the angular velocity remains constant at a value of 0 rad/s. 5. (c) A non-zero angular acceleration means that the angular velocity is either increasing or decreasing. The angular velocity is not constant. 6. (b) Since values are given for the initial angular velocity ϖ 0 , the final angular velocity ϖ , and the time t , Equation 8.6 [] can be used to calculate the angular displacement θ . 7. 32 rad/s 8. 88 rad 9. (c) According to Equation 8.9 ( v T = r ϖ ), the tangential speed is proportional to the radius r when the angular speed ϖ is constant, as it is for the earth. As the elevator rises, the radius, which is your distance from the center of the earth, increases, and so does your tangential speed. 10. (b) According to Equation 8.9 ( v T = r ϖ ), the tangential speed is proportional to the radius r when the angular speed ϖ is constant, as it is for the merry-go-round. Thus, the angular speed of the second child is . 11. 367 rad/s 2
Chapter 8 Problems 3 12. (e) According to Newton’s second law, the centripetal force is given by , where m is the mass of the ball and a c is the centripetal acceleration. The centripetal acceleration is given by Equation 8.11 as a c = r ϖ 2 , where r is the radius and ϖ is the angular speed. Therefore, , and the centripetal force is proportional to the radius when the mass and the angular speed are fixed, as they are in this problem. As a result, . 13. (d) Since the angular speed ϖ is constant, the angular acceleration α is zero, according to Equation 8.4. Since α = 0 rad/s 2 , the tangential acceleration a T is zero, according to Equation 8.10. The centripetal acceleration a c , however, is not zero, since it is proportional to the square of the angular speed, according to Equation 8.11, and the angular speed is not zero. 14. 17.8 m/s 2 15. (a) The number N of revolutions is the distance s traveled divided by the circumference 2 r π of a wheel: N = s /(2 r π ). 16. 27.0 m/s CHAPTER 8 ROTATIONAL KINEMATICS PROBLEMS 1. REASONING AND SOLUTION Since there are radians per revolution and it is stated in the problem that there are 100 grads in one-quarter of a circle, we find that the number of grads in one radian is

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p ROTATIONAL KINEMATICS 2. REASONING The average angular velocity
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