PH201 Chapter 6 Solutions 6.2. Set Up: Since the stone travels in a circular path, its acceleration is directed toward the center of the circle. The only horizontal force on the stone is the tension of the string. Set the tension in the string equal to its maximum value. The free-body diagram for the stone is given in the figure below. In the diagram the stone is at a point to the right of the center of the path. (b) *6.3. Set Up: Each h travels in a circle of radius 0.750 m has mass (0.0125)(52 kg) = 0.65 kg weight 6.4 N. The period for each h is Let be toward the center of the circular path. (a) The free-body diagram for one h is given in the figure below. wrist. This force has both horizontal vertical components. is the force exerted on the h by the (b) (c) The horizontal force from the wrist is 12 times the weight of the h. Reflect: The wrist must al exert a vertical force on the h equal to its weight. *6.5. Set Up: The pern moves in a circle of radius The acceleration of the pern is directed horizontally to the left in the figure in the problem. The time for one revolution is the period
The free-body diagram is given in the figure below. is the force applied to the seat by the rod. (b) Combining these two equations Then the period is (c) The net force is proportional to m in the mass divides out the angle for a given rate of rotation is independent of the mass of the passengers. Reflect: The pern moves in a horizontal circle the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod. 6.6. Set Up: Apply to the button. The button moves in a circle, it has acceleration The situation is equivalent to that of Example 6.3. Expressing v in terms of the period T, A platform speed of 40.0 rev/min corresponds to a period of 1.50 s, (b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period (longer periods mean slower speeds, the button may be moved farther out) is inversely proportional to the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m) Reflect: The maximum radial acceleration that friction can give is At the faster rotation rate T is smaller R must be smaller to keep the same. 6.10. Set Up: The ball has acceleration directed toward the center of the circular path. When the ball is at the bottom of the swing, its acceleration is upward. Take upward, in the direction of the acceleration. The bowling ball has mass upward. (b) The free-body diagram is given in the figure below.
6.13. Set Up: A angle is of a full rotation, in a h travels through a distance of In (c) use coordinates where is upward, in the direction of at the bottom of the swing. (b) The free-body diagram is shown in the figure below. F is the force exerted by the blood vessel. (c) (d) When the arm hangs vertically at rest, upward Reflect: The acceleration of the h is only about 20% of g, the increase in the force on the blood drop when the arm swings is about 20%. 6.18. Set Up: The net force on each astronaut is the gravity force exerted by the other astronaut. Call the astronauts A B, where The free-body diagram for astronaut A is given in Figure (a) for astronaut B in Figure (b) below. for A And for B,
(b) (c) Their accelerations would increase as they moved closer the gravitational attraction between them increased. *6.19. Set Up: The gravitational force between two objects is The mass of the earth is the mass of the sun is the distance from the earth to the moon is The distance from the earth to the sun is Solve: It is more accurate to say the moon orbits the sun. Reflect: The sun is farther away but has a much greater mass than the earth. 6.22. Set Up: Use coordinates where is to the right. Each gravitational force is attractive, is toward the mass exerting it. Treat the masses as uniform spheres, the gravitational force is the same as for point masses with the same center-to-center distances. The free-body diagrams for (a) (b) are given in Figures (a) (b) below.
The net force on A is to the right. (b) The net force on A is to the left. 6.26. Set Up: The earth has radius Let the earth s mass be your mass be m. Solve: At the surface of the earth your weight is At height h above the earth s surface, 6.32. Set Up: (a) Although equation 6.7 was derived for the earth it can be applied to the asteroid, which we assume to be spherically symmetric; thus, we have Al, we know that (b) The astronaut s weight on the asteroid can be calculated using (c) We note that is nearly constant for a fall of 1.0 meter we can use the equations of motion for constant acceleration: use equation 2.12 with the positive x-axis upward, (d) Assuming that is nearly constant over the height of the jump, we can use the equations of motion for constant acceleration: use equation 2.13 ( ). At the top of the astronaut s jump we have we will assume that the astronaut jumps with the same initial velocity, on both the earth the asteroid. (b) (c) We can lve equation 2.12 for time: (d) We can lve equation 2.13 for use it to compare what happens on the asteroid to what happens on earth: Thus, we have Reflect: Contrary to our assumption, the acceleration of the astronaut does not remain perfectly constant during this incredible 560 meter jump. Using the inverse-square law, we can show that the astronaut s acceleration
decreases by only about 6% we have probably underestimated the height of the astronaut s jump slightly. *6.37. Set Up: The rotational period of the earth is The radius of the satellite s path is where h is its height above the surface of the earth r are related by applying to the motion of the satellite. (b)-(c) The free-body diagram for the satellite is given in Figure (a) below. (d) The sketch of the earth orbit of the satellite is given in Figure (b) above. Reflect: The orbital period is proportional to altitude above the earth s surface. To achieve a period as large as 24 h, the satellite is at a large *6.43. Set Up: The distance a point on the rim travels in 1 revolution is Solve: In one minute a point on the rim travels a distance The number of revolutions in one minute is The station must turn at 2.55 rpm.
6.45. Set Up: Use coordinates where +y is upward. At the maximum height, Solve: At the surface of Io, Now lve for when 6.46. Set Up: Your friend slides toward you when the friction force on her isn t sufficient to make her turn as you turn. Set apply to your friend. Set her acceleration equal to that of the car, In the absence of friction your friend moves in a straight line, turn to the right, toward her. (b) The free-body diagram for your friend is given in the figure below. Reflect: The larger the coefficient of friction, the smaller must be the radius of the turn. *6.47. Set Up: The pern moves in a horizontal circle of radius Set the static friction force equal to its maximum value, The pern has an acceleration directed toward the center of the circle. The period is
the pern has speed The free-body diagram is given in the figure below. The diagram is for when the wall is to the right of her, the center of the cylinder is to the left. (b) Combining these two equations (c) The mass m of the pern divides out of the equation for of the pern. the answer to (b) does not depend on the mass Reflect: The greater the rotation rate the larger the normal force exerted by the wall the larger the friction force. Therefore, for smaller the rotation rate must be larger. 6.54. Set Up: The block moves in a horizontal circle of radius Each string makes an angle with the vertical. The block has acceleration directed to the left in the figure in the problem. The free-body diagram for the block is given in the figure below.
(b) Reflect: The tension in the upper string must be greater than the tension in the lower string that together they produce an upward component of force that balances the weight of the block.