Tutorial of Kinemtatics Essay Example

Tutorial of Kinemtatics Essay Example Tutorial of Kinemtatics Essay Tutorial of Kinemtatics Essay Applied Science Department (ASD) Centre for Foundation Studies and Extension Education (FOSEE) PPH 0095 Mechanics Foundation in Engineering ONLINE NOTES Chapter 2 Kinematics FOSEE , MULTIMEDIA UNIVERSITY (436821-T) MELAKA CAMPUS, JALAN AYER KEROH LAMA, 75450 MELAKA, MALAYSIA. Tel 606 252 3594 Fax 606 231 8799 URL: http://fosee. mmu. edu. my/~asd/ PPH0095 MECHANICS Contents 2. 0 2. 1 2. 2 2. 3 2. 4 2. 5 2. 6 2. 7 2. 8 2. 9 2. 10 2. 11 2. 12 2. 3 Introduction Definitions of Linear Motion Distance Displacement Speed and Velocity Average Velocity Instantaneous Velocity Average Acceleration Instantaneous Acceleration Equations of Linear Motions Motion Graphs Free Falling Objects under gravity Projectile Motion Uniform Circular Motion ASD 2011/12 KINEMATICS 1/23 PPH0095 MECHANICS Mind Map ASD 2011/12 KINEMATICS 2/23 PPH0095 MECHANICS OBJECTIVES Upon completion of this chapter, you should be able to: 1) 2) 3) 4) 5) define distance, displacement, velocity, acceleration. know how to apply all the equation for linear motion with constant acceleration. raw graph velocity versus time , distance versus time and explain them. understand the concept of free fall and should be able to solve the problem. understand the concept of projectile motion and uniform circular motion and should be able to solve the problem. 2. 0 INTRODUCTION Kinematics is the branch of mechanics which studies the motion of objects without considering the forces that cause the motion. Vector quantities such as displacement, velocity, and acceleration are involved. The study of the motion of objects under the action of forces is called dynamics. The study of the motion of objects, and the related concepts of force and energy, form the field called mechanics. Mechanics is customarily divided into two parts i. e. kinematics and dynamics. Kinematics : the description of how objects move. Kinematics in one dimension : describing an object that moves along a straight line path, which is one dimensional motion. Kinematics in two dimensions : the description of the motion of objects that move in paths in two (or three) dimensions. Dynamics : deals with force and why objects move as they do. In this part we will solve the following questions : What akes an object at rest begin to move ? What causes a body to accelerate or decelerate ? What is involved when an object moves in a circle ? We can answer in each case that a FORCE is required. 2. 1 DEFINITIONS of LINEAR MOTION Linear motion is motion along a straight line. Three types of motion: Translational Rotational Vibrational ASD 2011/12 KINEMATICS 3/23 PPH0095 MECHANICS Fig ure 1 We only discuss objects that move without rotating (Figure 1a) Motion in straight line; Vertical Horizontal Slanting Reference Frames Any measurement of position, distance or speed must be made with respect to a frame of reference. It is always important to specify the frame of reference when stating a speed. In everyday life, we usually mean with respect to the Earth. Position For one-dimensional motion, we often choose the x axis as the line along which the motion takes place. The position of an object at any moment is given by its x coordinate. If the motion is vertical, as for a dropped object, we usually use the y axis. 2. 2 DISTANCE The length of the actual path or total path length. It depends on the frame of reference, for example, Ipoh is 200 km away from Kuala Lumpur. A set of coordinate axes represents a frame of reference. ASD 2011/12 KINEMATICS 4/23 PPH0095 MECHANICS 2. 3 DISPLACEMENT The change in position of the object, i. e. displacement is how far the objects is from its starting point. For example : A change from an initial position xi to the final position xf, the displacement is, ? x = xf xi. The symbol ? (delta) means change in. So ? x means the change in x which is the displacement. It is a quantity that has both magnitude and direction and represented in diagrams by arrows. Example 1 : A person walking 70 m to the east and then turning around and walking back (west) a distance of 30 m. Total distance = 100 m Displacement = xf xi = 40 m 0 m = 40 m Figure 2 2. 4 SPEED and VELOCITY The most obvious aspect of the motion of a moving object is how fast it is moving, i. e. its speed or velocity. Speed is simply a positive number, (i. e. a scalar: having magnitude only) with units. Velocity, on the other hand, is used to signify both the magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. (velocity is therefore a vector). Average Speed is defined as the total distance travelled along its path divided by the time it takes to travel this distance, i. . average speed = distance travelled time elapsed ASD 2011/12 KINEMATICS 5/23 PPH0095 MECHANICS 2. 5 AVERAGE VELOCITY Average velocity is defined as the displacement divided by the elapsed time, i. e. average velocity, v ave = x f xi displacement ? x = = time elapsed ? t t f ti Average velocity would be zero if starting and ending point are the same. Unit : ms-1 Fi gure 3: Velocity of a car as a function of time at constant velocity. Figure 4: Velocity of a car as a function of time with varying velocity. 2. 6 INSTANTANEOUS VELOCITY The instantaneous velocity is the velocity at any instant of time. In general the instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval. We define instantaneous velocity as the average velocity in the limit of ? t becoming extremely small, approaching zero. v = lim ?t > 0 ?x dx = ? t dt Let ? t approach zero, ? x approaches zero as well. But the ratio ? x/? t approaches some definite value, which is the instantaneous velocity at a given instant. KINEMATICS 6/23 ASD 2011/12 PPH0095 MECHANICS 2. 7 AVERAGE ACCELERATION Acceleration specifies how rapidly the velocity of an object is changing. Average acceleration is defined as the change in velocity divided by the time taken to make this change, i. e. v f vi change of velocity ? v average acceleration, aave = = = time elapsed ? t t f ti Unit : ms-2 2. 8 INSTANTANEOUS ACCELERATION The instantaneous , a , is defined as the limiting value of the average acceleration as we let ? t approach zero. instantaneous acceleration, a = lim ? dx ? d? ? 2 dv ? dt ? = d x a = = dt dt dt 2 ?t > 0 ? v dv = ? t dt since v = dx , so dt Acceleration tells us how fast the velocity changes, whereas velocity tells us how fast the position changes. x v= dt and dv d 2 x a = = dt dt 2 2. 9 EQUATIONS of LINEAR MOTIONS Many practical situations occur in which the acceleration is constant, i. e. the acceleration doesnt change over time. We now treat this situation when the magnitude of the acceleration, a, is constant and the motion is in a straight line. In this case, the instantaneous and average acceleration are equal. To simplify our notatio n, let us take the initial time in any discussion to be zero the elapsed time, t initial velocity , vo the position at time t is s the velocity at time t is v ASD 2011/12 KINEMATICS 7/23 PPH0095 MECHANICS The acceleration, which is assumed constant in time , will be a = Multiply both sides by t and get: ? v = vo + at at = v ? vo v ? vo t †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. ( 1. 9. 1 ) [omit s] velocity vo v O time t ?v +v? s=? o ? t ? 2 ? †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦( 1. 9. 2 ) [omit a] Substitute equation (1. 9. 1) into (1. 9. 2), s =( v o + v o + at )t 2 or s = vot + ? at2 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. ( 1. 9. 3 ) [ omit v ] We now derive the fourth equation, which is useful in situations where the time, t is not known. From equation ( 1. 9. 1 ) , solve for t, obtaining t= v ? vo a .. †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. ( 1. 9. 4 ) Substituting equation ( 1. . 4 ) into equation ( 1. 9. 3 ), we have ASD 2011/12 KINEMATICS 8/23 PPH0095 MECHANICS 2 2 ? v + v o v ? v o ? v ? v o s=? ?= 2a ? 2 a ? Solve for v 2 and obtain 2 v 2 = v o + 2as †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. ( 1. 9. 5 ) [ omit t ] From equation ( 1. 9. 1 ) , solve for vo, obtaining vo = v – at†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦. (1. 9. 6) Substitute equation (1. 9. 6) into (1. 9. 2), s =( v + v ? at )t 2 or s = vt ? ? at2 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. ( 1. 9. 3 ) [ omit vo ] Example 2:: Spotting a police car, you brake a porsche from 75 km/h to 45 km/h over a displacement of 88 m. a) What is the acceleration assumed to be constant ? Given: vo = 75km / h = 75 ? 103 = 20. 83 m/s 3600 45 ? 103 v = 45km / h = = 12. 5 m/s 3600 s = 88 m, a = ? v 2 = vo + 2a s (12. 5 m/s)2 = (20. 83 m/s)2 + 2a(88 m) a = -1. 6 m/s1 2 b) What is the elapsed time ? s = ? (vo + v)t 88 m = ? (12. 5 m/s + 20. 83 m/s)t t = 5. 4 s c) If you continue to slow down with the acceleration calculated in (a), how much time will elapse in bringing the car to rest from 75 km/h ? v = vo + at 20. 83 m/s = 12. 5 m/s + (-1. 6 m/s2 )t t = 13 s ASD 2011/12 KINEMATICS 9/23 PPH0095 MECHANICS d) In ( c ) what distance will be covered ? s = vot + ? at2 = (20. 3 m/s)(13 s) + ? (-1. 6 m/s2)(13 s)2 = 137 m e) Suppose that later, using the acceleration calculate in (a) but a different initial velocity , you bring your car to rest after travelling 200 m . What is the total braking time ? s = vt ? ? at2 200 m = (0 m/s) t – ? (-1. 6 m/s2) t2 t = 16 s 2. 10 MOTION GRAPHS The motion of a body can be illustrated by o a displacement-time ( x t ) graph. o a velocit y-time ( v ) graph o an acceleration-time ( a t ) graph Displacement-time graph or s t graph of a body shows how the displacement of the body varies with time. o Instantaneous velocity, v= dx = gradient of the s – t graph. dt Figure 5 shows the x – t graphs for four types of motion. Figure 5a o Constant velocity Velocity = gradient of the graph = constant ASD 2011/12 KINEMATICS 10/23 PPH0095 MECHANICS Figure 5b o Constant acceleration with initial velocity u = 0 When t = 0, gradient = 0 Gradient increases as t increases ? Velocity increases Figure 5c o Constant acceleration with initial velocity u ? 0 When t = 0, gradient ? 0 hence initial velocity ? 0 Gradient increases as t increases ? Velocity increases ASD 2011/12 KINEMATICS 11/23 PPH0095 MECHANICS Figure 5d Non-uniform acceleration When t = 0, gradient ? 0 Hence initial velocity ? 0 When t = t, gradient = 0 Hence velocity = 0 When t = t2, gradient ; 0, Hence velocity is negative. When velocity is negative, ob ject is moving in the opposite direction. Velocity–time graph or v – t graph of a body shows how the velocity of a body varies with time. Velocity, v = dx dt Displacement, s = ? v dt = area under the v – t graph. Instantaneous acceleration, a = dv = gradient of the v t graph at that instant. dt Figure 6 shows the v – t graphs for six types of motion. ASD 2011/12 KINEMATICS 12/23 PPH0095 MECHANICS Figure 6a Uniform velocity Gradient = 0, acceleration = 0 Displacement from t = t1 to t = t2, s = shaded area , A Figure 6b o Uniform acceleration Initial velocity = 0 Gradient = constant, hence Acceleration = constant Displacement from t = t1 to t = t2, s = shaded area , A ASD 2011/12 KINEMATICS 13/23 PPH0095 MECHANICS Figure 6c o Uniform acceleration Initial velocity ? 0 Gradient = constant, hence Acceleration = constant Displacement from t = t1 to t = t2, s = shaded area , A Figure 6d o Uniform acceleration Initial velocity ? 0 Constant negative gradient imp lies constant negative acceleration (constant deceleration) ASD 2011/12 KINEMATICS 14/23 PPH0095 MECHANICS Figure 6e o Non-uniform velocity Initial velocity = 0 Gradient decreases, hence acceleration decreases, Displacement from t = t1 to t = t2, s = shaded area , A Figure 6f o Increasing acceleration Initial velocity = 0 Gradient increases, hence acceleration increases. Acceleration-time graph or a – t graph of a body shows how the acceleration of the body varies with time. dv a = dt Increase in velocity = ? a dt = area under the a – t graph. Figure 7 shows four acceleration–time graphs. ASD 2011/12 KINEMATICS 15/23 PPH0095 MECHANICS Figure 7a o Constant acceleration Gradient=constant Area, A = increase in velocity from t = t1 to t = t2 Figure 7b o Acceleration increases uniformly Gradient=constant Area, A = increase in velocity from t = t1 to t = t2 Figure 7c o Decreasing acceleration Area, A = decreasing in velocity from t = t1 to t = t2 ASD 2011/12 KINEMATICS 16/23 PPH0095 MECHANICS Figure 7d o Uniform velocity When velocity = constant, acceleration , a = 0 2. 11 FREE FALLING OBJECTS UNDER GRAVITY Any object moving freely under the influence of gravity, regardless of its initial motion. When an object is in free fall, we assume that air resistance is negligible and that the only force acting on it is gravity. Object thrown upward/downward, will both experience the same acceleration as an object released from rest. Assuming air resistance is negligible, the rate of acceleration (g) of all objects in free fall is approximately 9. 8 m/s2. The vertical motion of a freely falling object is equivalent to motion in one dimension under constant acceleration. The equations for uniformly accelerated motion can be applied to free fall. Since the motion is vertical, y replaces x and y0 replaces x0 while g replaces the symbol a. It is arbitrary whether we choose y to be positive in the upward or downward direction; but we must be consistent about it throughout a problem’s solution. Thrown downward: a = g = +9. 80 m/s2 Thrown upward: a = g = 3: A boy on a bridge throws a stone vertically downward toward the river below with an initial velocity of 14. 7 m/s . If the stone hits the water 2. 00 s later, what is the height of the bridge above the water? Solution: Take y as positive downward Given: v0 = 14. 7 m/s , ( downward) , t = 2. 00 s and g = + 9. 8 m/s/s 1 y y o = v o t + at 2 2 = (14. )(2. 00) + (1 / 2)(9. 8)(2. 00) 2 = 29. 4 + 19. 60 y = 49. 0 m 2. 12 PROJECTILE MOTION It is the motion in two dimensions under the action of gravity only (downward) We Can study the motion of a projectile by considering ASD 2011/12 KINEMATICS 18/23 PPH0095 MECHANICS o The vertical component. And o The horizontal component of the motion. The vertical component of motion is motion under uniform acceleration. The h orizontal component of motion is motion under uniform velocity If air resistance is negligible, then the horizontal component of motion does not change; thus ax = 0 and vx = vx0 = constant. The vertical component of motion is affected by gravity and is described by the equations for an object in free fall. To describe it, choose a suitable origin, O and the axes (Figure 9). Let vo denote the initial velocity and ? the angle between vo and the positive x-axis. Figure 9 From diagram, the motion can divide in two components, horizontal (x-axis) and vertical (y-axis). Two assumptions: o The free-fall acceleration, g is constant over the range of the motion and is directed downward. (-g) o The effect of air resistance is negligible that is the horizontal motion has zero acceleration. x-component 0 vxo = vocos? 0 = 0 vx x y-component -g vyo = vosin? yo = 0 vy y Acceleration Initial velocity Initial position Velocity Position ASD 2011/12 KINEMATICS 19/23 PPH0095 MECHANICS From this, one can show that: Vertical component (y-component): Vertical velocity component: v y = v yo gt = v o sin ? gt Vertical position component: 1 1 y = v yo t gt 2 = (v o sin ? )t gt 2 2 2 *from v = vo + at *from y = v o t + 1 2 at 2 At maximum Height, H, the vy = 0. 2 From, v = vo + 2a y 2 0 = (v o sin ? ) 2 ? 2 gH H= v0 sin 2 ? 2g 2 If T is the time taken from O to A, to find the T, y = 0, t = T From, y = v o t + 1 2 at 2 1 gT 2 2 0 = (v o sin ? T ? T = 2v o sin ? g Horizontal component (y-component): Horizontal velocity component : vx = vxo = vo cos ? Horizontal position component : x = vxot = (vo cos ? )t To find the horizontal Range, R, t = T, x = R ASD 2011/12 KINEMATICS 20/23 PPH0095 MECHANICS From x = vo t + 1 2 at ,a =0 2 ? 2v sin ? ? ? R = (v o cos ? )? o ? ? g ? ? R= v o sin 2? g The maximum horizontal range is achieved when ? = 45o. At any time the distance, r of the projectile from the origin is r = x 2 + y2 By eliminate, the trajectory (the relation between x and y) is a parabola, Vertical position component : y = (v o sin ? )t 1 2 gt 2 Horizontal position component : x = (vo cos ? ) t x t = v o cos ? ? x ? 1 ? x ? y = (vo sin ? )? ? v cos ? ? ? 2 g ? v cos ? ? ? ? ? ? o ? ? o ? g x2 = (tan ? )x ? 2 2 2vo cos ? = x tan ? ? gx 2 sec 2 ? 2 2v 2 gx 2 y = x tan ? ? 2 (1 + tan 2 ? ) 2v ASD 2011/12 KINEMATICS 21/23 PPH0095 MECHANICS Since sin 2? = sin (180o 2? ), there would be two angles of projection, ? and (90o- ? ), that would achieve a particular range R for a certain speed of projection vo . For the speed of projection vo , however, the maximum range is obtained only when the angle of projection is 45o as shown in Figure 10. Figure 10 2. 13 UNIFORM CIRCULAR MOTION An object that moves in a circle at constant speed is said to experience uniform circular motion where the magnitude of velocity remains constant the direction of velocity continuously change Acceleration is defined as the rate of change of velocity. The rate of change of velocity depends on the change in direction as well as the change in the magnitude . Therefore, object revolving in a circle is continuously accelerating (even when the speed remains constant). An object moving in a circle of radius r with constant speed v, has an acceleration whose direction is toward the center of the circle and whose magnitude is given by the formula 2 aR = (Please refer to text book section 5. 2, pages 120, how to get this equation) v r ASD 2011/12 KINEMATICS 22/23 PPH0095 MECHANICS v1 aR aR v1 Figure 11 The acceleration vector always points toward the center of the circle. The velocity vector always points in the direction of motion (tangent to the circle or perpendicular to the radius of the circle). Circular motion is often described in terms of the frequency f as so many revolutions per second. The period T of an object revolving in a circle is the time required for one complete revolution. T= 1 f v= 2? r T END OF CHAPTER 2. ASD 2011/12 KINEMATICS 23/23