Physics #18 Circular Motion It All Starts with a Centripetal Force

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2 Physics #18 Circular Motion It All Starts with a Centripetal Force Adam Kirsch James H Dann, Ph.D. CK12 Editor Say Thanks to the Authors Click (No sign in required)

3 To access a customizable version of this book, as well as other interactive content, visit CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. AUTHORS Adam Kirsch James H Dann, Ph.D. CK12 Editor CONTRIBUTORS Chris Addiego Ari Holtzman Antonio De Jesus López Copyright 2012 CK-12 Foundation, The names CK-12 and CK12 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution/Non- Commercial/Share Alike 3.0 Unported (CC BY-NC-SA) License ( as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: January 29, 2013 iii

4 Contents Contents 1 An Introduction to Circular Motion 1 2 More Details About Circular Motion 6 3 A Deeper Look at the Nature of Centripetal Forces 10 4 Centripetal Force Problems 12 5 Centripetal Forces Problem Set 16 iv

5 Chapter 1. An Introduction to Circular Motion CHAPTER 1 An Introduction to Circular Motion Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion. This force changes the direction of the velocity vector of the object but not the speed. Students will also learn how to calculate that speed using the period of motion and the distance of its path (circumference of the circle it traces out). Objective The student will: Understand that objects in circular motion must experience a net acceleration (and therefore a net force) directed into the center of motion. Vocabulary Centripetal acceleration Centripetal force Introduction A ball on the end of a string is swung in a circle in Figure 1.1; a satellite orbits around the earth in Figure 1.2; a car travels around a curve in Figure 1.3. All of these objects are engaged in circular motion. Let us consider the ball and satellite first. The ball is held in the place by the tension in the string and the satellite is held in place by the earth s gravity. Both the tension and gravity hold the ball and the satellite in their respective orbits. In what direction do these forces act? If the string was to break, and the earth magically was gone, both the ball and the satellite would fly off tangent to their motion at the instant the string and gravity no longer held them. The force preventing this from happening must keep pulling them toward the center of the circle to maintain circular motion. What is the force in Figure 1.3 that prevents the car from skidding off the road? If you guessed the friction between the tires and the road you d be correct. But is it static or kinetic friction? Unless the tires skid, there can be no kinetic friction. It is static friction that prevents the tires from skidding, just as it is static friction that permits you to walk without slipping. In Figure 1.4, you can see the foot of a person who walks toward the right by pushing their foot backward with a horizontal component of force F. They move forward because the ground exerts a horizontal component force f s in the opposite direction. (Vertical forces are ignored.) The force the ground exerts on the person s foot is a static friction force. Because the foot does not slide, we know that F and f s are equal opposed forces. We can easily see which direction the static friction force must act when we walk, but what about a car performing circular motion? In what direction does the static friction act on the car in the Figure 1.3? Figure below shows the top view of a car moving around a circular track with a constant speed. Since acceleration is defined as a = v t, you may be tempted to say that since the speed remains constant, v = 0, and therefore the acceleration must also be zero. But that conclusion would be incorrect because v represents a change in velocity, not a change in speed. The velocity of the car is not constant since it is continuously changing its direction. How then do we find the acceleration of the car? Figure 1.5 shows the instantaneous velocity vectors for the car in two different positions a very small time apart. Notice that the vector V points toward the center of the circle. (Recall that V can be thought of as the sum of the vectors V 2 + ( V 1 ).) The direction of the acceleration points in the direction of V since acceleration is defined as 1

6 FIGURE 1.1 (source unknown, needs to be replaced - once that happens, this and the next two images can be made into a single composite) FIGURE 1.2 v a = ~ t. This is reasonable since if there were no force directed toward the center of the circle the car would move off tangent to the circle. We call the inward force that keeps an object in circular motion a center seeking, or centripetal force and the acceleration, centripetal acceleration. The centripetal acceleration is often denoted as ac In order to find the correct expression for the magnitude of the centripetal acceleration we ll need to use a little geometric reasoning. Figure 2

7 Chapter 1. An Introduction to Circular Motion FIGURE 1.3 FIGURE 1.4 Author: Image copyright Andre Adams, 2012; modified by CK-12 Foundation - Raymond Chou License: Used under license from Shutterstock.com Source: show two almost similar triangles. The magnitudes of V 1 and V 2 are equal and the change in location of the car occurs over a very small increment in time, t. The velocities are tangent to the circle and therefore perpendicular to the radius of the circle. As such, the radius triangle and the velocity triangle are approximately similar (see the figures above). We construct an approximate ratio between the two triangles by assuming that during the time t the car has traveled a distance s along the circle. The following ratio is constructed in order to determine the acceleration. s r. = v v, which leads to, v s. = r v, Dividing both sides of the equation by t, we have: v s t. = r v t But s t is the speed v of the car and v t is the acceleration of the car. If we allow the time to become infinitesimally small then the approximation becomes exact and we have: v 2 = ra,a = v2 r, thus the magnitude of the centripetal acceleration for an object moving with constant speed in circular motion is a c = v2 r and its direction is toward the center of the circle. Examples using Centripetal Acceleration and Force Example 1a: The 1000 kg car in Figure 1.7 moves with a constant speed 13.0 m/s around a flat circular track of radius 40.0 m. What is the magnitude and direction of the centripetal acceleration? The magnitude of the car s acceleration is a c = v2 toward the center of the track. r = 132 Example 1b: Determine the force of static friction that acts upon the car. 40 = = 4.23 m/s2 and the direction of its acceleration is 3

8 FIGURE 1.5 Author: Image copyright lana rinck, 2012; modified by CK-12 Foundation - Raymond Chou License: Used under license from Shutterstock.com Source: FIGURE 1.6 Author: CK-12 Foundation - Raymond Chou License: CC-BY-NC-SA 3.0 Using Newton s Second Law: F = f s = ma = 1000(4.225) = 4225 = 4230 N Example 1c: Determine the minimum necessary coefficient of static friction between the tires and the road. y F = F N Mg = 0,F N = Mg but f s = µ s F N = ma thus, µ s Mg = Ma,µ s g = a,µ s = a g = = = 0.43

9 Chapter 1. An Introduction to Circular Motion FIGURE 1.7 Author: CK-12 Foundation - Raymond Chou License: CC-BY-NC-SA 3.0 Check your understanding True or False? 1. Kinetic friction is responsible for the traction (friction) between the tires and the road. Answer: False. As long as the car does not skid, there is no relative motion between the instantaneous contact area of the tire and the road. 2. True or False? The force of static friction upon an object can vary. Answer: True. In attempting to move an object, a range of forces of different magnitudes can be applied until the maximum static friction between the object and the surface it rests upon is overcome and the object is set into motion. Recall that the magnitude of static friction is represented by the inequality: f s µ s F N 3. True or False? The greater the mass of the car, the greater the coefficient of friction. Answer: False. The coefficient of friction is independent of the mass of an object. Recall it is the ratio of the friction force to the normal force and as such, it is a pure number dependent only upon the nature of the materials in contact with each other; in this case rubber and asphalt. 5

10 CHAPTER 2 More Details About Circular Motion Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the objects motion. This force changes the direction of the velocity vector of the object but not the speed. Students will also learn how to calculate that speed using the period of motion and the distance of its path (circumference of the circle it traces out). Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle defined by the object s motion. This force changes the direction of the velocity vector of the object but not the speed. Students will also learn how to caclulate that speed using the period of motion and the length of its path (circumference of the circle it traces out). Key Equations v = 2πr T If a particle travels a distance 2πr in an amount of time T, then its speed is distance over time or 2πr T The Earth-Sun distance is about m The Earth-Moon distance is about m Guidance An orbital period, T, is the time it takes to make one complete rotation. If a particle travels a distance 2πr in an amount of time T, then its speed is distance over time or 2πr/T. An object moving in a circle has an instantaneous velocity vector tangential to the circle of its path. The force and acceleration vectors point to the center of the circle. Net force and acceleration always have the same direction. Centripetal acceleration is just the acceleration provided by centripetal forces. When an object moves in circular motion, it is only because real forces (such as the tension from a string, gravity, friction, etc... ) act so that the sum of all such forces is directed inward. IT IS THIS NET FORCE THAT WE CALL THE CENTRIPETAL FORCE. It is a very common misconception to believe that when there is an object moving in a circle, that "the centripetal force" just magically exists and is a force on its own. Again this cannot be overstated it is the sum of real forces acting on an object that we call "the centripetal force". 6

11 Chapter 2. More Details About Circular Motion Example 1 Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. When you make a right turn at constant speed in your car what is the force that causes you (not the car) to change the direction of yourvelocity? Choose the best possible answer. a. Friction between you and the seat b. Inertia c. Air resistance d. Tension e. All of the above f. None of the above 2. A pendulum consisting of a rope with a ball attached at the end is swinging back and forth. As it swings downward to the right the ball is released at its lowest point. Decide which way the ball attached at the end of the string will go at the moment it is released. 7

12 a. Straight upwards b. Straight downwards c. Directly right d. Directly left e. It will stop 3. A ball is spiraling outward in the tube shown to the right. Which way will the ball go after it leaves the tube? a. Towards the top of the page b. Towards the bottom of the page c. Continue spiraling outward in the clockwise direction d. Continue in a circle with the radius equal to that of the spiral as it leaves the tube e. None of the above 4. Explain using Newton s Second Law why an object moving in a circle must experience a net force towards the center of the circle. 5. Using the known distance the Earth is from the sun, calculate the speed that the Earth is moving through space in relation to the sun. 6. Using the known distance the moon is from Earth, calculate the speed that the moon is moving through space in relation to Earth. 8

13 Chapter 2. More Details About Circular Motion Answers 1. a. 2. c. 3. b. 4. The object s velocity is constantly changing, but in direction only. A force directed inward toward the center will be able to accelerate object inward, changing its direction, while not impacting the speed at which the object travels at ,000 m/s 6. 1,020 m/s 9

14 CHAPTER 3 A Deeper Look at the Nature of Centripetal Forces Summary Forces that cause objects to follow circular paths are known as centripetal, or center seeking, forces. Such forces must continuously change direction to stay perpendicular to the velocity vector. We saw in previous units on vectors and kinematics that vectors cannot impact motion in directions perpendicular to them. This is why the horizontal component of the velocity of projectiles on earth does not change (as in two dimensional motion). In this section, we look deeper into the nature of centripetal forces. Speed vs. Direction Think of a ball rolling horizontally off a cliff. At first, its velocity is perpendicular to the force of gravity. As it falls, its velocity in the x direction stays constant, but it accelerates downward due to gravity. This ball will not travel in circle, though, because gravity is only perpendicular to its velocity at the instant it leaves the cliff. Eventually, the ball s velocity components in the x and y-directions will be equal. After some time, if the cliff is tall enough, the ball s vertical velocity will dwarf its horizontal velocity. Let s now compare how gravity affects the ball s speed at different 1 second intervals during its flight. a. Let s say this ball initially had a horizontal velocity and therefore also speed of 100 m/s, and a vertical speed of 0. After the first second, its vertical velocity will about 10 m/s (assume g = 10m/s. Using the pythagorean theorem, we find that the speed is now m/s, a change of less than.5 m/s. b. Now consider the ball after 10 seconds, when its velocity components are equal. Between the 10th and 11th second, its speed goes from m/s to ( ) m/s, a much bigger increase in the same time. c. Finally, for the sake of argument, let s say the cliff is mount Everest and the ball keeps falling for 100 seconds. Now, in one second, its speed goes from m/s (1000) m/s to ( ) 2 In other words, it is much easier for a force to change the speed of an object when it points along its velocity vector. Forces are capable of changing an object s velocity speed and direction: the more parallel to motion the force, the 10

15 Chapter 3. A Deeper Look at the Nature of Centripetal Forces more it changes speed; the more perpendicular, direction (if we had analyzed the effect on the angle of the velocity vector above instead of speed, the results would have been reversed). Circular Motion Knowing this, we can understand why, when force is always perpendicular to the velocity vector, an object s speed never changes, while its direction changes continuously. If the force is constant in magnitude, the direction of the object s velocity must change at a constant rate otherwise the situation would be asymmetrical. In other words, the object will travel in a circle, with instantaneous velocity tangent to the circle and instantaneous force pointing toward the center. At any given time, the relationship between force, acceleration, and velocity is illustrated here: Circular motion is kind of a limiting case of the first second scenario above if the force had always been perpendicular to the ball s velocity, it wouldn t have accelerated downward for an entire second. No matter how weak a centripetal force, it will in principle always cause a moving object to travel in a circle. This may seem counterintuitive, but is actually a direct result of the arguments above. 11

16 CHAPTER 4 Centripetal Force Problems Students will learn how to analyze and solve Centripetal Force type problems. Students will learn how to analyze and solve Centripetal Force type problems. Key Equations Centripetal Force F C = mv2 r m v r mass (in kilograms, kg) speed (in meters per second, m/s) radius of circle Centripetal Acceleration a C = v2 r { v r speed (in meters per second, m/s) radius of circle Guidance Any force can be a centripetal force. Centripetal force is the umbrella term given to any force that is acting perpendicular to the motion of an object and thus causing it to go in a circle. Keep the following in mind when solving centripetal force type problems: The speed of the object remains constant. The centripetal force is changing the direction but not the speed of the object. Although the object feels an outward pull, this is not a true force, but merely the object s inertia, its tendancy at any given moment to keep moving in a straight line. Remember, Newton s first law maintains that the natural state of an object is to go in a straight line at constant speed. Thus, when you make a right turn in your car and the basketball in the back seat flies to the left, that is because the car is moving right and the basketball is maintaining it s position and thus from your point of view moves to the left. Your point of view in this case is different from reality because you are in a rotating reference frame. To solve problems involving circular motion, keep in mind that it is very, very similar to solving any other force problem using Newton s 2nd Law. You are summing forces (but in this case only the forces going into and/or out of the center of motion), and setting that sum equal to the object s mass multiplied by the object s acceleration. In the case of circular motion however, the acceleration is something special, namely v 2 /r. In the Context of Newton s 2nd Law: F = ma F in/outo f centero f motion = m (v 2 /r) 12

17 Chapter 4. Centripetal Force Problems Applications To find the maximum speed that a car can take a corner on a flat road without skidding out, set the force of friction equal to the centripetal force. To find the tension in the rope of a swinging pendulum, remember that it is the sum of the tension and gravity that produces a net upward centripetal force. A common mistake is just setting the centripetal force equal to the tension. To find the speed of a planet or satellite in an orbit, set the force of gravity equal to the centripetal force. Banked turns (the road is not flat, but rather at an angle) allow for a greater speed before skidding out. In this case the normal force aids the force of friction in creating a larger centripetal force than that which can be obtained by friction alone on a flat road. If you look at the picture below, can you see how both the normal force and friction would have horizontal components in the right-hand figure? Both of these contribute centripetal forces directed into the center of motion. It s not as simple as adding them together, one must consider the components of these two forces in the direction to the center of the circle. When you consider circular motion taking place in a horizontal plane (a car around a corner, a ball on the end of a string being twirled around your head, the toy airplane flying from a string hanging from the ceiling, etc... ), gravity does not factor into the summation of forces because gravity is not pointed into or out of the center of motion, but is always perpendicular to the plane of the circle. For vertical circular motion (a diving airplane, a ferris wheel, a car going over a hill, a loop-de-loop, etc... ) gravity must be one of the forces you sum when determining the total centripetal force acting on the object. Watch this Explanation MEDIA Click image to the left for more content. 13

18 Mr K s Steps When Solving Circular Motion Problems: 1. Draw a picture. 2. Establish a reference frame (generally all you will care about is in vs. out of the center of motion. Make one direction positive, and the other negative so that you sum forces properly). 3. Identify variables and check units. 4. Draw a Free Body Diagram. 5. Resolve all forces into components so that you are only summing forces (or components of forces) directed into/out of the center of motion). 6. Apply Newton s 2nd Law in the direction directed into/out of the center of motion. Also, the object s acceleration will equal the centripetal acceleration namely v 2 /r. 7. Solve for your unknowns. If this looks familar, good! Because this is almost the exact same procedure we use whenever we apply Newton s 2nd Law to problems. The only difference is this for circular motion we usually only care about (and therefore only sum in) the direction that is directed into/out of the center of motion. With the more general case, we often need to sum in both the x and the y directions. Simulation Time for Practice A 700kg car makes a turn going at 30 m/s with radius of curvature of 120m. What is the force of friction between the car s tires and the road? 2. An object of mass 10 kg is in a circular orbit of radius 10 m at a velocity of 10 m/s. a. Calculate the centripetal force (in N) required to maintain this orbit. b. What is the acceleration of this object?

19 Chapter 4. Centripetal Force Problems 3. Suppose you are spinning a child around in a circle by her arms. The radius of her orbit around you is 1 meter. Her speed is 1 m/s. Her mass is 25 kg. a. What is the magnitude and direction of tension in your arms? b. In her arms? 4. A racecar is traveling at a speed of 80.0 m/s on a circular racetrack of radius 450 m. a. What is its centripetal acceleration in m/s 2? b. What is the centripetal force on the racecar if its mass is 500 kg? c. What provides the necessary centripetal force in this case? 5. The radius of the Earth is 6380 km. Calculate the velocity of a person standing at the equator due to the Earth s 24 hour rotation. Calculate the centripetal acceleration of this person and express it as a fraction of the acceleration g due to gravity. Is there any danger of flying off? 6. Neutron stars are the corpses of stars left over after supernova explosions. They are the size of a small city, but can spin several times per second. (Try to imagine this in your head.) Consider a neutron star of radius 10 km that spins with a period of 0.8seconds. Imagine a person is standing at the equator of this neutron star. a. Calculate the centripetal acceleration of this person and express it as a multiple of the acceleration g due to gravity (on Earth). b. Now, find the minimum acceleration due to gravity that the neutron star must have in order to keep the person from flying off. Answers to Selected Problems N 2. a. 100 N b. 10 m/s 2 3. a. 25 N towards her b. 25 N towards you 4. a m/s 2 b N c. friction between the tires and the road g 6. a g m/s 2 b. The same as a. 15

20 CHAPTER 5 Centripetal Forces Problem Set 1. When you make a right turn at constant speed in your car what is the force that causes you (not the car) to change the direction of yourvelocity? Choose the best possible answer. a. Friction between your butt and the seat b. Inertia c. Air resistance d. Tension e. All of the above f. None of the above 2. You buy new tires for your car in order to take turns a little faster (uh, not advised always drive slowly). The new tires double your coefficient of friction with the road. With the old tires you could take a particular turn at a speed v o. What is the maximum speed you can now take the turn without skidding out? a. 4v o b. 2v o c. v o d. 2vo e. Not enough information given 3. A pendulum consisting of a rope with a ball attached at the end is swinging back and forth. As it swings downward to the right the ball is released at its lowest point. Decide which way the ball attached at the end of the string will go at the moment it is released. a. Straight upwards 16

21 Chapter 5. Centripetal Forces Problem Set b. Straight downwards c. Directly right d. Directly left e. It will stop 4. A ball is spiraling outward in the tube shown to the above. Which way will the ball go after it leaves the tube? a. Towards the top of the page b. Towards the bottom of the page c. Continue spiraling outward in the clockwise direction d. Continue in a circle with the radius equal to that of the spiral as it leaves the tube e. None of the above 5. An object of mass 10 kg is in a circular orbit of radius 10 m at a velocity of 10 m/s. a. Calculate the centripetal force (in N) required to maintain this orbit. b. What is the acceleration of this object? 6. Suppose you are spinning a child around in a circle by her arms. The radius of her orbit around you is 1 meter. Her speed is 1 m/s. Her mass is 25 kg. a. What is the tension in your arms? b. In her arms? 7. A racecar is traveling at a speed of 80.0 m/s on a circular racetrack of radius 450 m. a. What is its centripetal acceleration in m/s 2? b. What is the centripetal force on the racecar if its mass is 500 kg? c. What provides the necessary centripetal force in this case? 8. The radius of the Earth is 6380 km. Calculate the velocity of a person standing at the equator due to the Earth s 24 hour rotation. Calculate the centripetal acceleration of this person and express it as a fraction of the acceleration g due to gravity. Is there any danger of flying off? 9. Neutron stars are the corpses of stars left over after supernova explosions. They are the size of a small city, but can spin several times per second. (Try to imagine this in your head.) Consider a neutron star of radius 10 km that spins with a period of 0.8seconds. Imagine a person is standing at the equator of this neutron star. a. Calculate the centripetal acceleration of this person and express it as a multiple of the acceleration g due to gravity (on Earth). 17

22 b. Now, find the minimum acceleration due to gravity that the neutron star must have in order to keep the person from flying off. 10. Calculate the force of gravity between the Sun and the Earth. (The relevant data are included in Appendix B.) 11. Calculate the force of gravity between two human beings, assuming that each has a mass of 80 kg and that they are standing 1 m apart. Is this a large force? 12. Prove g is approximately 10 m/s 2 on Earth by following these steps: a. Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol m o to represent the mass of the falling object. b. Now divide that force by the object s mass to find the acceleration g of the object. 13. Our Milky Way galaxy is orbited by a few hundred globular clusters of stars, some of the most ancient objects in the universe. Globular cluster M13 is orbiting at a distance of 26,000 light-years (one light-year is m) and has an orbital period of 220 million years. The mass of the cluster is 10 6 times the mass of the Sun. a. What is the amount of centripetal force required to keep this cluster in orbit? b. What is the source of this force?# Based on this information, what is the mass of our galaxy? If you assume that the galaxy contains nothing, but Solar-mass stars (each with an approximate mass of kg), how many stars are in our galaxy? 14. Calculate the centripetal acceleration of the Earth around the Sun. 15. You are speeding around a turn of radius 30.0 m at a constant speed of 15.0 m/s. 18 a. What is the minimum coefficient of friction µ between your car tires and the road necessary for you to retain control? b. Even if the road is terribly icy, you will still move in a circle because you are slamming into the walls. What centripetal forces must the walls exert on you if you do not lose speed? Assume m = 650 kg. 16. Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances. a. If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration? b. Why, in fact, doesn t your pencil accelerate towards you? 17. A digital TV satellite is placed in geosynchronous orbit around Earth, so it is always in the same spot in the sky. a. Using the fact that the satellite will have the same period of revolution as Earth, calculate the radius of its orbit. b. What is the ratio of the radius of this orbit to the radius of the Earth? c. Draw a sketch, to scale, of the Earth and the orbit of this digital TV satellite. d. If the mass of the satellite were to double, would the radius of the satellite s orbit be larger, smaller, or the same? Why?

23 Chapter 5. Centripetal Forces Problem Set 18. A top secret spy satellite is designed to orbit the Earth twice each day (i.e., twice as fast as the Earth s rotation). What is the height of this orbit above the Earth s surface? 19. Two stars with masses kg and kg are orbiting each other under the influence of each other s gravity. We want to send a satellite in between them to study their behavior. However, the satellite needs to be at a point where the gravitational forces from the two stars are equal. The distance between the two stars is m. Find the distance from the more massive star to where the satellite should be placed. ( Hint: Distance from the satellite to one of the stars is the variable.) 20. Calculate the mass of the Earth using only: (i) Newton s Universal Law of Gravity; (ii) the Moon-Earth distance (Appendix B); and (iii) the fact that it takes the Moon 27 days to orbit the Earth. 21. A student comes up to you and says, I can visualize the force of tension, the force of friction, and the other forces, but I can t visualize centripetal force. As you know, a centripetal force must be provided by tension, friction, or some other familiar force. Write a two or three sentence explanation, in your own words, to help the confused student. 22. A space station was established far from the gravitational field of Earth. Extended stays in zero gravity are not healthy for human beings. Thus, for the comfort of the astronauts, the station is rotated so that the astronauts feel there is an internal gravity. The rotation speed is such that the apparent acceleration of gravity is 9.8 m/s 2. The direction of rotation is counter-clockwise. a. If the radius of the station is 80 m, what is its rotational speed, v? b. Draw vectors representing the astronaut s velocity and acceleration. c. Draw a free body diagram for the astronaut. d. Is the astronaut exerting a force on the space station? If so, calculate its magnitude. Her mass m = 65 kg. e. The astronaut drops a ball, which appears to accelerate to the floor, (see picture) at 9.8 m/s 2. a. Draw the velocity and acceleration vectors for the ball while it is in the air. b. What force(s) are acting on the ball while it is in the air? c. Draw the acceleration and velocity vectors after the ball hits the floor and comes to rest. d. What force(s) act on the ball after it hits the ground? 19

24 Answers to Selected Problems a. 100 N b. 10 m/s 2 6. a. 25 N towards her b. 25 N towards you 7. a m/s 2 b N c. friction between the tires and the road g 9. a m/s 2 b. The same as a N N; very small force 12. g = 9.8 m/s 2 ; you ll get close to this number but not exactly due to some other small effects 13. a N b. gravity c kg m/s a..765 b N 16. a N very small force b. Your pencil does not accelerate toward you because the frictional force on your pencil is much greater than this force. 17. a m b. 6.6 R e d. The same, the radius is independent of mass m 19. You get two answers for r, one is outside of the two stars one is between them, that s the one you want, m from the larger star a. v = 28 m/s b. v down, a right c. f right d. Yes, 640N 20