Chapter 7Rotational Motion and the Law of GravityQuick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsRotational Motion and the Law of Gravity 235Chapter 7Rotational Motion and the Law of GravityQuick Quizzes 1. (c). For a rotation of more than 180°, the angular displacement must be larger than3.14 radp =. The angular displacements in the three choices are (a) 6 rad 3 rad 3 rad- =, (b) ( )1 rad 1 rad 2 rad- - =, (c) 5 rad 1 rad 4 rad- =. 2. (b). Because all angular displacements occurred in the same time interval, the displacement with the lowest value will be associated with the lowest average angular speed. 3. (b). From 2 22 2002 2 2w ww waq q q--= = =D D D, it is seen that the case with the smallest angular displacement involves the highest angular acceleration. 4. (b). All points in a rotating rigid body have the same angular speed. 5. (a). Andrea and Chuck have the same angular speed, but Andrea moves in a circle with twice the radius of the circle followed by Chuck. Thus, from tv rw=, it is seen that Andrea’s tangential speed is twice Chuck’s. 6. 1. (e). Since the tangential speed is constant, the tangential acceleration is zero.2. (a). The centripetal acceleration, 2c ta v r=, is inversely proportional to the radius when the tangential speed is constant.3. (b). The angular speed, tv rw =, is inversely proportional to the radius when the tangential speed is constant. 7. (c). Both the velocity and acceleration are changing in direction, so neither of these vector quantities is constant.236 CHAPTER 7 8. (b) and (c). According to Newton’s law of universal gravitation, the force between the ball and the Earth depends on the product of their masses, so both forces, that of the ball on the Earth, and that of the Earth on the ball, are equal in magnitude. This follows also, of course, from Newton’s third law. The ball has large motion compared to the Earth because according to Newton’s second law, the force gives a much greater acceleration to the small mass of the ball. 9. (e). From 2F G Mm r=, the gravitational force is inversely proportional to the square of the radius of the orbit.10. (d). The semi-major axis of the asteroid’s orbit is 4 times the size of Earth’s orbit. Thus, Kepler’s third law (2 3T r constant=) indicates that its orbital period is 8 times that of Earth.Rotational Motion and the Law of Gravity 237Answers to Even Numbered Conceptual Questions 2. If we assume they are separated by about 10 m and their masses are estimated to be 70 kg and 40 kg, then, using the law of universal gravitation, we estimate a gravitational force on the order of 910 N-. 4. The need for a large force toward the center of the circular path on objects near the equator will cause the Earth to bulge at the equator. A force toward the center of the circular path is not needed at the poles, so the radius in this direction will be smaller than at the equator. 6. To a good first approximation, your bathroom scale reading is unaffected because you, Earth, and the scale are all in free fall in the Sun’s gravitational field, in orbit around the Sun. To a precise second approximation, you weight slightly less at noon and at midnight than you do at sunrise or sunset. The Sun’s gravitational fields is a little weaker at the center of the Earth than at the surface sub-solar point, and a little weaker still on the far side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a little less strongly than it does on the Earth below you. So you can have another doughnutwith lunch, and your bedsprings will still last a little longer. 8. The astronaut is accelerating toward the Earth at the same rate as is the spaceship. Thus, ifthe astronaut drops a wrench, it will float in space next to him. Likewise, he will float in space next to a desk or with reference to the spaceship. Thus, he believes himself to be weightless.10. Consider one end of a string connected to a spring scale and the other end connected to an object, of true weight w. The tension T in the string will be measured by the scale and construed as the apparent weight. We have - =cw T m a. This gives, = -cT w m a. Thus, the apparent weight is less than the actual weight by the term cm a. At the poles the centripetal acceleration is zero. Thus, =T w. However, at the equator the term containing the centripetal acceleration is nonzero, and the apparent weight is less than the true weight.12. If the acceleration is constant in magnitude and perpendicular to the velocity, the object is moving in a circular path at constant speed. If the acceleration is parallel to the velocity, the object is either speeding up, v and a in same direction, or slowing down, v and a in opposite directions.238 CHAPTER 714.