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Transcript
SECTION 1 Preview Vocabulary Latin Word Origins The word centripetal is a combination of two parts, “center” and “petal.” The second part of the word is derived from the Latin word petere, which means “seeking.” The word centripetal is used to describe an object moving in a circular path whose acceleration is directed toward the center of the path. Teach Objectives Solve problems involving centripetal acceleration. Solve problems involving centripetal force. Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the centripetal force. Circular Motion Key Term centripetal acceleration Centripetal Acceleration Consider a spinning Ferris wheel, as shown in Figure 1.1. The cars on the rotating Ferris wheel are said to be in circular motion. Any object that revolves about a single axis undergoes circular motion. The line about which the rotation occurs is called the axis of rotation. In this case, it is a line perpendicular to the side of the Ferris wheel and passing through the wheel’s center. Tangential speed depends on distance. FIGURE 1.1 Circular Motion Any point on a Ferris wheel spinning about a fixed axis undergoes circular motion. Misconception Alert! Tangential speed (vt) can be used to describe the speed of an object in circular motion. The tangential speed of a car on the Ferris wheel is the car’s speed along an imaginary line drawn tangent to the car’s circular path. This definition can be applied to any object moving in circular motion. When the tangential speed is constant, the motion is described as uniform circular motion. The tangential speed depends on the distance from the object to the center of the circular path. For example, consider a pair of horses side-byside on a carousel. Each completes one full circle in the same time period, but the horse on the outside covers more distance than the inside horse does, so the outside horse has a greater tangential speed. Some students will have difficulty with terminology at this point because of the previous familiarity with the term centrifugal. It is important to emphasize the distinction between centripetal (center-seeking) and centrifugal (center-fleeing). To avoid reinforcing this misconception, avoid using the term centrifugal. Centripetal acceleration is due to a change in direction. Suppose a car on a Ferris wheel is moving at a constant speed as the wheel turns. Even though the tangential speed is constant, the car still has an acceleration. To see why, consider the equation that defines acceleration: vf − vi a=_ tf − ti Acceleration depends on a change in the velocity. Because velocity is a vector, acceleration can be produced by a change in the magnitude of the velocity, a change in the direction of the velocity, or both. If the Ferris wheel car is moving at a constant speed, then there is no change in the magnitude of the velocity vector. However, think about the way the car is moving. It is not traveling in a straight line. The velocity vector is continuously changing direction. Demonstration Centripetal Acceleration Purpose Show an example of centripetal acceleration. Materials battery-operated toy car, 50 cm string Procedure Tie one end of the string to the midpoint of the car. Hold the other end of the string to the table or floor. Start the car. Have students observe the motion. Release the string and allow the car to move in a straight line. Point out that even though the car was traveling at constant speed, the car was accelerating because the direction of motion continued to change. Show that the change in velocity was always perpendicular to the direction of motion. 224 Chapter 7 ©Joseph Sohm/Visions of America/Corbis Plan and Prepare SECTION 1 224 Chapter 7 Differentiated Instruction Below Level Ask students to explain how a line of dancers moving in a circle demonstrates that tangential speed depends on the distance between the object and the center of the circular path. Also ask why jockeys maneuver their horses during a race to get as close to the inside rail as possible. See if students can provide other examples of how tangential speed depends on distance from the center. Untitled-27 224 5/9/2011 8:14:27 AM Untitled-27 225 The acceleration of a Ferris wheel car moving in a circular path and at constant speed is due to a change in direction. An acceleration of this nature is called a centripetal acceleration. The magnitude of a centripetal acceleration is given by the following equation: centripetal acceleration the acceleration directed toward the center of a circular path FIGURE 1.2 Centripetal Acceleration vt2 ac = _ r centripetal acceleration = Classroom Practice Centripetal Acceleration (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes. (b) For short time intervals, ∆v is directed toward the center of the circle. ___ (tangential speed)2 radius of circular path What is the direction of centripetal acceleration? To answer this question, consider Figure 1.2(a). At time ti, an object is at point A and has tangential velocity vi. At time tf , the object is at point B and has tangential velocity vf. Assume that vi and vf differ in direction but have the same magnitudes. (a) A vi B vf Centripetal Acceleration The cylindrical tub of a washing machine has a radius of 34 cm. During the spin cycle, the wall of the tub rotates with a tangential speed of 5.5 m/s. Calculate the centripetal acceleration of the clothes sitting against the tub. Answer: 8 9 m/s2 PROBLEM guide A The change in velocity (∆v = vf − vi) can be determined graphically, as shown by the vector triangle in Figure 1.2(b). Note that when ∆t is very small, vf will be almost parallel to vi. The vector ∆v will be approximately perpendicular to vf and vi and will be pointing toward the center of the circle. Because the acceleration is in the direction of ∆v, the acceleration will also be directed toward the center of the circle. Centripetal acceleration is always directed toward the center of a circle. In fact, the word centripetal means “center seeking.” This is the reason that the acceleration of an object in uniform circular motion is called centripetal acceleration. (b) Use this guide to assign problems. SE = Student Edition Textbook PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) Solving for: vf v −v i vt SE Sample, 1–2; Ch. Rvw. 8–9 PW 3–4 PB 4–6 ac SE3; Ch. Rvw. 50a PW 5–6 PB Sample, 1–3 r SE 4 PW Sample, 1–2 PB 7–10 PREMIUM CONTENT Interactive Demo Centripetal Acceleration HMDScience.com Sample Problem A A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is the car’s tangential speed? Given: ANALYZE r = 48.2 m ac = 8.05 m/s2 vt = ? *Challenging Problem Use the centripetal acceleration equation, and rearrange to solve for vt. SOLVE v 2t ac = _ r 2 �������� vt = √a �� cr = √(8.05 m/s ) (48.2 m) vt = 19.7 m/s Continued Problem Solving Circular Motion and Gravitation 225 Deconstructing Problems Use the following equations for students who are having difficulty understanding the algebraic simplifications used to derive the final equation. vt2 ac = _ r 5/9/2011 8:14:28 AM acr = vt2 cr = √ √ a v t2 cr = vt √a Circular Motion and Gravitation 225 Centripetal Acceleration Teach continued (continued) 1. A rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a centripetal acceleration of 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl’s tangential speed? Answers 2. As a young boy swings a yo-yo parallel to the ground and above his head, the yo-yo has a centripetal acceleration of 250 m/s2. If the yo-yo’s string is 0.50 m long, what is the yo-yo’s tangential speed? Practice A 1. 2.5 m/s 2. 11 m/s 3. 1.5 m/s2 4. 58.4 m 3. A dog sits 1.5 m from the center of a merry-go-round. The merry-go-round is set in motion, and the dog’s tangential speed is 1.5 m/s. What is the dog’s centripetal acceleration? 4. A racecar moving along a circular track has a centripetal acceleration of 15.4 m/s2. If the car has a tangential speed of 30.0 m/s, what is the distance between the car and the center of the track? Teaching Tip Tangential acceleration is due to a change in speed. Tangential acceleration is covered in more detail in the feature “Tangential Speed and Acceleration.” You have seen that centripetal acceleration results from a change in direction. In circular motion, an acceleration due to a change in speed is called tangential acceleration. To understand the difference between centripetal and tangential acceleration, consider a car traveling in a circular track. Because the car is moving in a circle, the car has a centripetal component of acceleration. If the car’s speed changes, the car also has a tangential component of acceleration. FIGURE 1.3 Centripetal Force Force on a Ball When a ball is whirled in a circle, it is acted on by a force directed toward the center of the ball’s circular path. Consider a ball of mass m that is tied to a string of length r and that is being whirled in a horizontal circular path, as shown in Figure 1.3. Assume that the ball moves with constant speed. Because the velocity vector, v, continuously changes direction during the motion, the ball experiences a centripetal acceleration that is directed toward the center of motion. As seen earlier, the magnitude of this acceleration is given by the following equation: vt2 ac = _ r The inertia of the ball tends to maintain the ball’s motion in a straight path. However, the string exerts a force that overcomes this tendency. The forces acting on the ball are gravitational force and the force exerted by the string, as shown in Figure 1.4(a) on the next page. The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force. This net force is directed toward the center of the circle, as shown in Figure 1.4(b). The net force that is directed toward the center of an object’s circular path is called centripetal force. Newton’s second law can be applied to find the magnitude of this force. Fc = mac 226 Chapter 7 Differentiated Instruction Inclusion Kinesthetic learners may need help understanding that Earth’s gravitational force causes a satellite in a circular orbit to change its direction but not its speed. Have students imagine that a person is running on the side of a road. Ask them to describe what would happen if the runner suddenly grabbed a lamppost or small tree trunk. Be sure students recognize that the force was perpendicular to the runner’s forward motion and caused the Untitled-27 226 226 Chapter 7 runner’s direction to change. Students can try this outdoors to feel the change in direction for themselves. 5/9/2011 8:14:29 AM ntitled-27 227 FIGURE 1.4 Centripetal Force The net force on a ball whirled in a circle (a) is directed toward the center of the circle (b). Fstring Fnet = Fc Fc Fg Demonstration v r Centripetal Force Purpose Show that the horizontal component of tension in a rope provides the centripetal force for a ball moving in a horizontal circle. m (b) (a) The equation for centripetal acceleration can be combined with Newton’s second law to obtain the following equation for centripetal force: Centripetal Force Materials 100 g mass (with hook), 30 cm string, spring scale, protractor Procedure Attach one end of the string to the hook on the mass and attach the other end to the spring scale. Use the protractor to measure the position at which the string makes an angle of 30° with the vertical. Then rotate the string with a constant speed so that the mass moves in a circular path and the string remains at the 30° angle with the vertical. While maintaining this motion, have a volunteer attempt to read the spring scale. (Maintaining a constant angle can be difficult; you may wish to practice this before demonstrating to the class.) mvt2 Fc = _ r (tangential speed)2 centripetal force = mass × ___ radius of circular path Centripetal force is simply the name given to the net force toward the center of the circular path followed by an object moving in uniform circular motion. Any type of force or combination of forces can provide this net force. For example, friction between a racecar’s tires and a circular track is a centripetal force that keeps the car in a circular path. As another example, gravitational force is a centripetal force that keeps the moon in its orbit. PREMIUM CONTENT Interactive Demo Centripetal Force HMDScience.com Sample Problem B A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5 m. The centripetal force needed to maintain the plane’s circular motion is 1.89 × 104 N. What is the plane’s mass? ANALYZE Given: vt = 56.6 m/s Unknown: m=? r = 188.5 m Fc = 1.89 × 104 N Use the equation for centripetal force. Rearrange to solve for m. SOLVE mv 2t Fc = _ r Fcr ___ (1.89 × 104 N) (188.5 m) = m=_ 2 (56.6 m/s)2 vt m = 1110 kg Continued Problem Solving Circular Motion and Gravitation 227 Deconstructing Problems Reality Check Use the following equations to help students see how the equation is rearranged to solve for mass. mv2t FC = _ r 8:14:29 AM Help students relate the equation for 5/9/2011 centripetal force to the motion of the object. A sharper turn means a greater change in direction, therefore an object moving in a small circle will require a greater centripetal force if the speed is kept constant. rFC = mv2t rFC mv2t _ 2 = _ 2 vt vt rF C _ 2 = m vt Point out that the scale reading equals the tension in the string. The vertical component balances the weight of the mass, while the horizontal component of the tension holds the mass in a circular path. Ask students to calculate the vertical tension component and to compare it with the weight (Fscalecosθ = mg) so that they can confirm whether the scale reading and angle are correct. Then have them calculate the centripetal force from the horizontal component of tension (Fscalesinθ = Fc). Repeat the demonstration at angles equal to 45° and 60°. Ask students to predict what change will occur in the force as measured on the spring scale. Circular Motion and Gravitation 227 Centripetal Force Teach continued 1. A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a tangential speed of 2.50 m/s. If the magnitude of the centripetal force is 88.0 N, what is the girl’s mass? PROBLEM guide b 2. A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N, and the combined mass of the bicycle and rider is 86.5 kg. What is the track’s radius? Use this guide to assign problems. SE = Student Edition Textbook PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) Solving for: m SE Sample, 1; Ch. Rvw. 10a*, 11* PW 3 PB 7–10 vt SE 4 Ch. Rvw. 10b, 39* PW Sample, 1–2 PB 5–7 r SE 2 PW 4 PB 6–7 Fc SE 3; Ch. Rvw. 50b, 53 PW 5 PB Sample, 1–3 (continued) 3. A dog sits 1.50 m from the center of a merry-go-round and revolves at a tangential speed of 1.80 m/s. If the dog’s mass is 18.5 kg, what is the magnitude of the centripetal force on the dog? 4. A 905 kg car travels around a circular track with a circumference of 3.25 km. If the magnitude of the centripetal force is 2140 N, what is the car’s tangential speed? Centripetal force is necessary for circular motion. FIGURE 1.5 Because centripetal force acts at right angles to an object’s circular motion, the force changes the direction of the object’s velocity. If this force vanishes, the object stops moving in a circular path. Instead, the object moves along a straight path that is tangent to the circle. Removal of Centripetal Force A ball that is on the end of a string is whirled in a vertical circular path. If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. (b) If the string breaks at the top of the ball’s path, the ball will move along a parabolic path. For example, consider a ball that is attached to a string and that is whirled in a vertical circle, as shown in Figure 1.5. If the string breaks when the ball is at the position shown in Figure 1.5(a), the centripetal force will vanish. Thus, the ball will move vertically upward, as if it has been thrown straight up in the air. If the string breaks when the ball is at the top of its circular path, as shown in Figure 1.5(b), the ball will fly off horizontally in a direction tangent to the path. The ball will then move in the parabolic path of a projectile. Describing a Rotating System *Challenging Problem (a) Answers Practice B 1. 29.6 kg 2. 40.0 m 3. 40.0 N 4. 35.0 m/s (b) To better understand the motion of a rotating system, consider a car traveling at high speed and approaching an exit ramp that curves to the left. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. At that point, the force of the door keeps the passenger from being ejected from the car. What causes the passenger to move toward the door? A popular explanation is that a force must push the passenger outward. This force is sometimes called the centrifugal force, but that term often creates confusion, so it is not used in this textbook. Inertia is often misinterpreted as a force. The phenomenon is correctly explained as follows: Before the car enters the ramp, the passenger is moving in a straight path. As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to 228 Chapter 7 Differentiated Instruction Pre-AP Have students extend the discussion of a car on an exit ramp by considering how changing some of the variables would affect the passengers. How would passengers of different masses be affected as the moving car enters the ramp? Who would experience greater force, a heavier passenger or a lighter one? How would varying the speed of the car affect the passengers? Students should explain their answers in terms of inertia and mass. Untitled-27 228 228 Chapter 7 5/9/2011 8:14:30 AM move along the original straight path. This movement is in accordance with Newton’s first law, which states that the natural tendency of a body is to continue moving in a straight line. FIGURE 1.6 Path of Car and Passenger The force of friction on the passenger is not enough to keep the passenger on the same curved path as the car. However, if a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat. If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath, as shown in Figure 1.6. Eventually, the passenger encounters the door, which provides a large enough force to enable the passenger to follow the same curved path as the car does. The passenger does not slide toward the door because of some mysterious outward force. Instead, the frictional force exerted on the passenger by the seat is not great enough to keep the passenger moving in the same circle as the car. Assess and Reteach Assess Use the Formative Assessment on this page to evaluate student mastery of the section. v Fc Reteach For students who need additional instruction, download the Section Study Guide. v Response to Intervention To reassess students’ mastery, use the Section Quiz, available to print or to take directly online at HMDScience.com. SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What are three examples of circular motion? 2. A girl on a spinning amusement park ride is 12 m from the center of the ride and has a centripetal acceleration of 17 m/s2. What is the girl’s tangential speed? 3. Use an example to describe the difference between tangential and centripetal acceleration. 4. Identify the forces that contribute to the centripetal force on the object in each of the following examples: a. a bicyclist moving around a flat, circular track b. a bicycle moving around a flat, circular track c. a racecar turning a corner on a steeply banked curve 5. A 90.0 kg person rides a spinning amusement park ride that has a radius of 11.5 m. If the person’s tangential speed is 13.2 m/s, what is the magnitude of the centripetal force acting on the person? 6. Explain what makes a passenger in a turning car slide toward the door. Critical Thinking 7. A roller coaster’s passengers are suspended upside down as it moves at a constant speed through a vertical loop. What is the direction of the force that causes the coaster and its passengers to move in a circle? What provides this force? Answers to Section Assessment 1. Answers may vary. H_CNLESE586694_C07S1.indd 2. 14229m/s 3. Specific examples may vary, but all should indicate that an object moving in a circular path undergoes centripetal acceleration because the object changes direction. Tangential acceleration occurs when the object changes its speed around the path. 4.a. the forces that the bicycle seat, handlebars, and pedals exert on the bicyclist Circular Motion and Gravitation 229 b. friction between the tires and the track c. components of (1) the normal 3/26/2013 force10:53:47 PM from the banked curve and (2) the friction between the tires and the road 5. 1.36 × 103 N 6. When the car turns, the inertia of the passenger keeps the passenger moving in the same direction that the car was initially moving. 7. toward the center of the loop; the track and gravity Circular Motion and Gravitation 229