Chapter 3 Kinematics

Transcription

1 Chapter 3 Kinematics GOALS When you have mastered the content of this chapter, you will be able to achieve the following goals: Definitions Use the following terms to describe the physical state of a system: displacement velocity uniform circular acceleration motion uniformly accelerated motion projectile motion radial acceleration tangential acceleration Equations of Motion Write the equations of motion for objects with constant velocity and for objects with constant acceleration. Motion Problems Solve problems involving freely falling and other uniformly accelerated bodies, projectile motion, and uniform circular motion. Acceleration Effects List the effects of acceleration on the human body. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 1, Human Senses, and Chapter 2, Unifying Approaches. You must also be able to use the properties of right triangles to solve problems. 24

2 OVERVIEW Chapter 3 Kinematics As you look over this chapter you will find a large number of algebraic equations. These expressions are used in describing the basic motion of objects. In this chapter, the four basic kinds of motion which are described are 1) Linear Motion (Section 3.6), 2) Uniformly Accelerated Motion (Section 3.7), 3) Projectile Motion (Section 3.8), and 4) Uniform Circular Motion (Section 3.9). SUGGESTED STUDY PROCEDURE As you begin to study this chapter, be familiar with the Chapter Goals: Definitions, Equations of Motion, Motion Problems, andacceleration Effects. (A brief explanation of each of the terms listed under Definitions is given on the next page of this.) The initial sections of this chapter require a knowledge of trigonometry. If necessary, you may want to review the properties of Right Triangles found in the Appendix, Section A.6. Next, read Chapter Sections Be sure to work through the detailed examples given in most of the sections. At the end of the chapter, read the Chapter Summary and complete Summary Exercises Check your answers against those that are given. If you don't understand a specific answer, try reviewing that section of the text. Finally, complete Algorithmic Problems 1, 2, 3, 4, 5, 6, 9, 10, 11, and 12, Exercises 3, 4, 8, 9, 10, and 11, and Problems 14, 18, and 20. Now you should be prepared to attempt a Practice Test for Kinematics. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 3.1,3.2,3.3, ,2 3.5, 3.6 Equations of Motion 3.2,3.5, ,4,5,9 3, 4 Motion Problems 3.7,3.8, ,10,11,12 8,9,10,11 14,18,20 Acceleration Effects

3 DEFINITIONS DISPLACEMENT - Change in location of an object specified by both magnitude and direction from a given origin. When you raise your coffee cup to your lips you are displacing the cup. If you raise it 30 cm and bring it toward 30 cm then the total displacement is 30 cm up and 30 cm over, a resultant displacement of about 52 cm at an angle of 45º above the horizontal from the table to your lips. You will notice that the resultant displacement is a vector that does not necessarily point in direction of actual motion but only from the starting position to the final position. VELOCITY - Ratio of change in displacement to the change in time. If you lift your coffee cup (above) straight up in 1/10 second and then pull it horizontal to your lips, the average velocity is 300 cm/sec up, followed by a horizontal average velocity of 300 cm/sec toward you. Notice again that the resultant velocity does not point in the direction of any actual motion, rather the velocity points in the direction of the change in displacement. In this case it is 520 cm/sec in a direction 45º above horizontal from the table to your lips. ACCELERATION - Ratio of the change in velocity to the change in time. Any object that you start from rest undergoes a period of acceleration while you are bringing it up to its final velocity. Notice that acceleration is a vector that points in the direction of the change in velocity and may not point in the direction of the displacement. The acceleration of an object following a curved path always has a component of the acceleration in the direction of curvature. UNIFORMLY ACCELERATED MOTION - Motion of an object whose acceleration is constant, i.e., both the magnitude and direction of the acceleration remain the same for all times. Notice the restrictions that are placed upon motion for it to be of this kind. The change in velocity must always be in the same direction and constant. Therefore, the velocity must always be in the same direction. Hence the change in displacement is always in the same direction so the displacement occurs along a straight line. There are no systems you meet in everyday life that exactly satisfy these conditions. Nevertheless, there are systems, such as freely falling objects, that are nearly described by this motion. PROJECTILE MOTION - An object has constant velocity in one direction and uniform acceleration in a direction at right angles to the constant velocity. This motion represents the idealized motion of objects near the surface of the earth when spin and wind are neglected. UNIFORM CIRCULAR MOTION - Objects moving in a circle with constant speed. Think of riding on a merry-go-round. Your speed may be constant but the direction you are going keeps changing. Perhaps you start out going East, then North, then West, then South and back toward the East again. So you go around and around. Your acceleration is a vector pointing toward the center of the circle. 26

4 RADIAL ACCELERATION - Acceleration directed toward center of curvature of motion, i.e., perpendicular to the curve of the movement of the object in space. For uniformly accelerated motion the radial acceleration is zero since the motion is in a straight line. For uniform circular motion, all of the acceleration is radial acceleration. TANGENTIAL ACCELERATION - Acceleration directed along the direction of motion, tangent to the curve of the path of movement of the object in space. For uniformly accelerated motion all of the acceleration is tangential acceleration. For uniform circular motion, none of the acceleration is tangential acceleration. ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 3.1 Introduction Your present state of motion is probably a state of rest as you move your eyes to read this page. In contrast, during a race an Olympic sprinter starts from rest, tries to increase her speed as fast as possible and then run at top speed to the finish line. On the other hand, an astronaut moving about in a spacecraft in orbit around the earth seldom uses his legs to propel himself around the craft. He seems to hover just about the floor of the space lab and move around in straight paths between objects. SECTION 3.2 Characteristics of Distance and Displacement In order to add a 3 km vector displacement to a 3 km vector displacement you must know the angle between the vectors. What angle between these vectors must you use to obtain a resultant of 3 km? The numerical answer is given at the end of Section 3.3 of the textbook. SECTION 3.5 Characteristics of Motion In many motion problems the vector nature of displacement and velocity are important aspects of the problem. To overlook the "vectorness" of certain properties of physical system is a common mistake among beginning students of physics. When reading a problem for the first time, remember that some quantities may be vectors and must be treated by the special rules of vector algebra. SECTION 3.6 Linear Motion The answers to questions 1-7 are given in the textbook. Notice that for a velocity versus time graph the area under the curve represents the displacement and the slope of the curve represents the acceleration. SECTION 3.7 Uniformly Accelerated Motion Notice that in the data shown in tables 3.2 and 3.3 the distance down the incline increases more rapidly than the time; i.e., when the time doubles the distance is more than doubled, it is about four times as large. During the time intervals from 0-1 sec to 0-2 seconds, or from 0-2 seconds to 0-4 seconds, the distance ratios are 12.4/49.2 or 49.2/198.8 for the unweighted car and 12.0/48.3 or 48.3/198.2 for the weighted car. In both cases, if the time is doubled, the distance is quadrupled. Likewise, if the time is tripled, the distance is nine times as much. These data, then, indicate a quadratic relationship between distance and time as shown in Equations 27

5 3.10. We can conclude that the car is undergoing uniformly accelerated motion down the incline. In fact, the weighted and unweighted cars appear to have almost the same acceleration. An explanation of the seeming puzzling behavior of more massive objects is discussed in Chapter 4. If you plot the data given in the Example on page 51 you will obtain a straight line graph for the velocity versus time with a positive slop of 2m/s2 and starting at (0, 0). The equation of the line is v = 2t. SECTION 3.10 Effects of Acceleration Within the body it seems likely that food is accelerated down the throat and esophagus to the stomach. It decelerates in the stomach. The blood is accelerated as it leaves the heart and decelerates on its return trip to the heart. EXAMPLES MOTION PROBLEMS 1. A physics student accidentally rolled out of bed while dreaming about uniform circular motion. What was his speed just before he hit the floor? What Data Are Given? An object (a student) of unknown mass falls from rest an unknown distance and hits a floor. What Data Are Implied? If we assume the student falls under idealized conditions, then we can treat his motion as uniformly accelerated motion down with an acceleration of 9.8 m/s2. A typical bed is about 50 cm above the floor. What Physics Principles Are Involved? If we assumed idealized falling motion, then we can use the principles of uniformly accelerated motion. The Equations 3.10 are for the special condition of starting from rest with an initial displacement of zero. Both special conditions can be met by this example. What Equations Are to be Used? In this case we know the acceleration, a = 9.8 m/s 2, and the distance of fall, s = 50 cm or 0.50 meters, and we are seeking the velocity, so we can use the following equation from 3.10: 2 2 a s = v f (3.10) where the acceleration a is given by the usual free fall value of 9.8 m/s 2, or g. Algebraic Solution 2 2gs = v f so: v f = (SQR RT) (2gs) Numerical Solution v f = (SQR ROOT) [2(9.8 m/s 2 )(0.50 m)] = (SQR RT) (9.8 m 2 /s 2 ) v f = 2.2 m/s Thinking About the Answer Notice that the units obtained for the answer, namely meters divided by seconds, are the proper units for a velocity. 28

6 2. A student is skiing down a hill and wishes to take twice as long to reach the bottom, so she starts up higher on the side of the hill. How much farther up must she start? What Data Are Given? A person of unknown mass is skiing down a slope of unknown incline. What Data Are Implied? This problem implies idealized uniformly accelerated motion of unknown acceleration a. What Physics Principles Are Involved? This problem requires the use of the concepts of uniformly accelerated motion for the special case of starting from rest, doubling the time of motion, and finding the distance travelled. Equations 3.10 may be used. What Equations Are to be Used? In this case, we infer a starting rest position at a location on the side of the hill. After a time the skier reaches the bottom. The problem is to find the distance if the time is to be doubled. The following equation combines all the known quantities s = (1/2) at 2 (3.10) Algebraic Solution Let t 1 be the first time, s 1 is the first distance, and "a" is the acceleration; then s 1 = (1/2) at 1 2 Let t 2 be the second time, twice the first time, so t 2 = 2t 1, then the second distance s 2 is given by s 2 = (1/2) at 2 2 = (1/2) a(2t 1 ) 2 s 2 = (1/2 )(at 12 )4 = 4s 1. So the skier must start up the hill a distance 4s1, four times the first starting distance. 3. A member of the university girls' basketball team dribbled the length of the court and at a distance of 15 meters from the basket, she reached up to a height of 2.5 m and launched a shot with an initial speed of 10 m/s for maximum range. Is it possible she made a basket? What Data Are Given? The distance of the origin of the launch is 15 m from the basket. The height of the origin of the shot is 2.5 m from the floor and the initial speed is 12 m/s. What Data Are Implied? The height of the basket is 3.0 m (10 ft.) from the floor. The angle of launch for maximum range is 45º (see page 56). 29

7 Idealized projectile motion of a point particle is assumed so we can draw the following sketch of the problem: 6 What Physics Principles Are Involved? The basic equations of projectile motion must be used. The problem is to determine if the ball can be 3.0 m above the floor when it is 15 m in a horizontal direction from the launch location that is 2.5 m above the floor. A combination of the equations (3.12) and (3.13) are required to solve this problem. What Equations Are to be Used? The proper horizontal and vertical displacements must be obtained by the ball to go through the loop, so we need to use the horizontal and vertical displacement equations: horizontal motion: s h = v h t (3.12) vertical motion: s v = v v t + 1/2 gt 2 (3.13) Algebraic Solution Let us proceed by assuming we know the horizontal displacement s h (which we will set equal to 15 m) and we can calculate the initial values for the vertical and horizontal components of the velocity. In the above equations the two unknown quantities are the time and the vertical distance. We can proceed by substituting a known expression, namely s h /v h for the time t and solving for the vertical displacement; s v = v v (s h /v h ) + (½) g (s h /v h ) 2 where v v = v sin θ; v n = v cos θ where θ is the angle of launch from horizontal, 45º for this problem, s v = (v sin θ) (s h /(v cos θ)) + (½)g (s h2 /v 2 cos 2 θ) but sin θ /cos θ = tan θ; so s v = s h tan θ + (½) g (s 2 h /v 2 cos 2 θ) Numerical Solution s h = 15 m tan θ = tan 45º = 1.0, cos θ = cos 45º = 1/(SQR RT( 2)) g = -9.8 m/s 2 ; v = 12 m/s s v = (15 m) (1.0) - 1/2 (9.8)m/s 2 [(15)2m 2 /((12)2m 2 /sec 2 (1/2))] s v = 15 m - (9.8)(225/144)m = -0.3 m So the shot drops below the basket by 0.8 m, since it should have been at +0.5 meters instead of -0.3 m. Thinking About the Answer This is a fairly complex problem which involves the use of several equations. It is a good idea on a problem like this one to check your procedures out carefully. 30

8 4. A skylab in orbit 435 km above the surface of the earth completes a trip around the earth in 93.0 minutes. It is 7.0 minutes after lift off when the lab finally achieves this orbit. (a) What is the average tangential acceleration necessary to obtain its orbital position? (b) What is the radial acceleration when it is in orbit? What Data Are Given? The skylab is in an orbit 435 km above the earth, since the earth's radius is 6.38 x 103 km; the radius of the orbit is ( ) x 103 km or 6.82 x 103 km. The time for one orbit is 93.0 minutes or 5580 seconds. The time required to accelerate the skylab into its orbit is 7.0 minutes or 420 seconds. What Data Are Implied? The problem implies that the skylab is in a circular orbit. The average tangential acceleration is the value obtained by dividing the final tangential velocity by the time required for acceleration. What Physics Principles Are Involved? This problem requires you to use the equations for the radial and tangential acceleration for objects moving in a circular path. The magnitude of the average tangential acceleration is given by the following equation: a t = (v t final - v t initial ) / time and the direction is toward the larger value of the tangential velocity. The radial acceleration has a magnitude v 2 /r as derived in the textbook and points towards the center of the path. What Equations Are to be Used? The two equations given above will be needed plus a knowledge of the basic properties of circles and velocity. radial acceleration = (velocity) 2 /(radius) (3.21) tangential acceleration = (v f - v i ) / time tangential velocity = circumference / time for one cycle = 2π(radius) / time Algebraic Solution (a) a t = (v f - v i ) / time of acceleration; but v i = 0 since the skylab starts from rest. Let t = time of acceleration. v t = (2π(r)) / time for one cycle; let T = time for one cycle for this problem, tangential acceleration = at = (2πr)/(T/t) = (2πr)/tT (b) radial acceleration = a v = v 2 f /r = (2πr/T) 2 /r = (4π 2 r 2 )/rt 2 = (4π 2 r)/t 2 Numerical Solution t = 420 seconds; T = 5580 seconds r = 6.82 x 10 3 km = 6.82 x 10 6 m (a) tangential acceleration = a t = (2π (6.82 x 10 6 m)) / (4.2 x 10 2 x 5.58 x 10 3 s) a t = 1.8 x 10 1 m/s 2 = 18 m/s 2 towards increasing velocity. (b) radial acceleration a r = a r = [(4π 2 )(6.82 x 10 6 m)] / (5.58 x 10 3 s) 2 a r = 8.6 m/s 2, towards the center of the earth. Thinking About the Answer Notice how this problem combines many different concepts, your knowledge of the properties of circles from geometry, as well as the derived quantities from Section 3.9 in the book. This is typical of problems in physics. Any knowledge you gained from your previous experiences with mathematics or nature can be useful in solving physics problems. Notice that both answers came out with the correct units for acceleration, meters per second squared, m/s 2. 31

9 4. Below we have given a step-by-step problem which we think will help you compare and contrast two different types of motion near the surface of the earth. Work the problem out in the spaces provided and then compare your answers with ours. At the same instant in time, two slingshots are shot from the same location on the ground. The first slingshot shoots its projective vertically up into the air at a velocity of 98 m/s. The second slingshot shoots its projectile at an angle of 30º above the horizontal with a velocity of 196 m/s. (a) How long does it take the first projectile to reach the top of its trajectory? (b) How long does it take the second projectile to reach the top if its trajectory? (c) Give the position of the first projectile when it reaches the top of its path. (Hint: Assume the start position to be 0, 0 of a coordinate system and give the location in terms of its x, y position.) (d) Give the position of the second projectile at the top of its path. (Hint: Same as in the previous problem.) (e) What is the velocity of the first projectile at the top of its path? (f) What is the velocity of the second projectile at the top of its path? (g) How long does it take the first projectile to return to the ground? (h) How long does it take the second projectile to return to the ground? (i) Give the location of the point of impact when the first projectile hits the ground. (See hint in part c.) (j) Give the location of the second projectile when it strikes the ground. (See hint in part c.) (k) Tabulate the results of your calculations below and sketch the actual path of each projectile. Discuss the similarity and differences between the motion of the two projectiles. (l) What is the acceleration of the first projectile at its maximum point? (m) What is the acceleration of the second projectile at its maximum point? 32

10 ANSWERS: (a) 10 seconds; (b) 10 seconds; (c) 0,490 m; (d) (1670 m, 490 m); (e) 0 m/s; (f) 167 m/s horizontal; (g) 20 seconds; (h) 20 seconds; (i) 0, 0; (j) 3340 m, 0 m; (k) See fig. (l) 9.8 m/s2, downward; (m) 9.8 m/s2, downward. PRACTICE TEST 1. A car traveling 54 meters/sec skids uniformly to rest in 6.0 seconds after the brakes are applied. a. What is the car's acceleration in meters/sec2? b. How many meters did the car skid in coming to rest? 2. An American space team in their command module is attempting to dock with the orbiting Skylab Spacecraft. They approach the craft along a straight course. The graph below gives the separation distance as a function of time. See Fig, a. Find the initial relative velocity of the two spacecrafts at zero seconds. b. What occurs at 200 seconds? c. What is the average velocity of relative motion during the first 400 sec. of the interval shown? 33

11 3. A golfer finds himself confronted with the golf shot diagrammed below. He has calculated that he is 90.0 meters from the green and knows that he must clear a 12 meter tree which stands half-way (and on a line) between the ball and the green. Hitting the ball with a nine iron gives the ball a velocity of 32 meters/sec at an angle θ = 45º. a. What will be the ball's time of flight assuming that it does not strike the tree? b. Assuming that the ball is driven directly toward the green, will the ball clear the tree? Prove your answer. 4. Early test with the Rocket Sled illustrated that man was capable of withstanding large units of acceleration and deceleration. The test pilot was strapped to a padded chair and the sled was accelerated rapidly from rest and then decelerated rapidly from a high velocity to rest. If the accelerations were of the order of 5 g's, compare the physiological effects of the acceleration with those of deceleration. ANSWERS: m/sec 2, 162 meters m/sec, collision, -10 m/sec sec, yes! by meters 4. For horizontal accelerations, both acceleration and decelerations have approximately the same effect: progressive tightness in chest, loss of peripheral vision, blurred vision, difficulty in speaking. Deceleration has all these with reduced chest pressure. 34