6 - Circular Motion and Other Applications of Newton's Laws

Obtain accurate answers without solving a quadratic

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Unformatted text preview: traight portion of the curve. Use a least-squares fit to determine this slope. t (s) 20.0° T mg Figure P6.69 70. A 9.00-kg object starting from rest falls through a viscous medium and experiences a resistive force R b v, where v is the velocity of the object. If the object’s speed reaches one-half its terminal speed in 5.54 s, (a) determine the terminal speed. (b) At what time is the speed of the object three-fourths the terminal speed? (c) How far has the object traveled in the first 5.54 s of motion? 71. Members of a skydiving club were given the following data to use in planning their jumps. In the table, d is the distance fallen from rest by a sky diver in a “free-fall d (ft) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 16 62 138 242 366 504 652 808 971 1 138 1 309 1 483 1 657 1 831 2 005 2 179 2 353 2 527 2 701 2 875 ANSWERS TO QUICK QUIZZES 6.1 No. The tangential acceleration changes just the speed part of the velocity vector. For the car to move in a circle, the direction of its velocity vector must change, and the only way this can happen is for there to be a centripetal acceleration. 6.2 (a) The ball travels in a circular path that has a larger radius than the original circular path, and so there must be some external force causing the change in the velocity vector’s direction. The external force must not be as strong as the original tension in the string because if it were, the ball would follow the original path. (b) The ball again travels in an arc, implying some kind of external force. As in part (a), the external force is directed toward the center of the new arc and not toward the center of the original circular path. (c) The ball undergoes an abrupt change in velocity — from tangent to the circle to perpendicular to it — and so must have experienced a large force that had one component opposite the ball’s velocity (tangent to the circle) and another component radially outward. (d) The ball travels in a straight line tangent to the original path. If there is an external force, it cannot have a component perpendicular to this line because if it did, the path would curve. In fact, if the string breaks and there is no other force acting on the ball, Newton’s first law says the ball will travel along such a tangent line at constant speed. 6.3 At the path is along the circumference of the larger circle. Therefore, the wire must be exerting a force on the bead directed toward the center of the circle. Because the speed is constant, there is no tangential force component. At the path is not curved, and so the wire exerts no force on the bead. At the path is again curved, and so the wire is again exerting a force on the bead. This time the force is directed toward the center of the smaller circle. Because the radius of this circle is smaller, the magnitude of the force exerted on the bead is larger here than at ....
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