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Kinematics and Dynamics

Kinematics and Dynamics
ME 230 Kinematics and Dynamics WeiChih Wang Department of Mechanical Engineering University of Washington Instructor: WeiChih Wang Email: abongu.washington.edu Office: MEB 113 or 260 Phone: 2065432479 Office hours: M,W, F 1:302:30PM  Teaching Assistant: David Schipf schipfuw.edu Jinyuan Zhang jinyuanuw.edu Paul Murphy pgmurphyuw.edu Kebin Gu kebinu.washington.edu (grader)  Textbook: R. C. Hibbeler, Engineering Mechanics: Dynamics, 13th Ed.  Course Website: http://courses.washington.edu/engr100/me230 W. WangGeneral Policy  Homework: Homework will be assigned in class on Wed. Homework for each week is due the following Wednesday (During Class). The homework has usually 1012 problems per week. Late homework will not be accepted (partial credit will not be given). Homework solution will be available every Wednesday on the web. Please write down your section number on your homework.  Grading of Homework: Only one or two questions (chosen by the instructor) from the homework (assigned for each week) will be graded – the resulting grade will constitute the grade for that week’s homework. Therefore, answer all the questions correctly to get full credit for the homework.  Exams: Exams will be open book and open notes. There will be no alternate exams if you miss any. Exams will include materials covered in the text, class, and homework. W. WangNotes:  Homework be assigned on a weekly basis  Homework should be handwritten  TA will go over the problems with you during Lab Section and answer any questions you have on your homework.  Solutions to all problems solved in class will be posted on Thursday each week: http://courses.washington.edu/engr100/me230 W. WangGrading Homework 20  1st Midterm 25  2nd Midterm 25  Final Project 30 GPA Formula: GPA = (Score50)/40(4.02.0)+2.0 (94=4.0 and 50=2.0.) W. WangPlease make sure… • You review some maths (i.e. trigonometric identities, derivatives and integrals, vector algebra, ) • … and some STATICS… UNITS, Vector addition, free body diagram (FDB) (Hibbeler Statics: Ch. 1,2 and 5) W. WangExamples (1) W. WangExamples (2) W. WangExamples (3) W. WangWhat is Dynamics Important contributors: Galileo Galilei, Newton, Euler, Lagrange Mechanics (The study of how bodies react to forces acting on them) Study of equilibrium of a Study of accelerated motion Statics Dynamics body that is at rest/moves of a body with constant velocity • Kinematics: geometric aspects of the motion • Kinetics: Analysis of forces which cause the motion W. WangAn Overview of Mechanics Mechanics: The study of how bodies react to forces acting on them. Statics: The study of Dynamics: The study of force bodies in equilibrium or and torque and their effect on a in constant speed. accelerated moving body 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics concerned with the forces causing the motion W. WangMechanics  Statics – effects of forces on bodies at rest  Dynamics  Theoretically, kinematics and kinetics constitute dynamics.  Kinematics – study of motion of bodies without reference to forces which cause the motion  Kinetics – relates action of forces on bodies to their resulting motion  Kinematics and kinetics almost occur together all the time in practice. W. WangHowever…  From Wikipedia, the free encyclopedia: Dynamics is a branch of physics (specifically classical mechanics) concerned with the study of forces and torques and their effect on motion, as opposed to kinematics, which studies the motion of objects without reference to its causes. Isaac Newton defined the fundamental physical laws which govern dynamics in physics, especially his second law of motion. Also why this class is called kinematics and dynamics. W. WangWhy is dynamics important  Understanding dynamics is key to predicting performance, designing systems, etc.  The ability to control a system (say, a car) depends upon understanding the dynamics  It is fundamental to advanced topics, such as fluid mechanics, structural dynamics, or vibration. W. WangApplications of Dynamics  Modern machines and structures operated with high speed (acceleration)  Analysis design of  Moving structure  Fixed structure subject to shock load  Robotic devices  Automatic control system  Rocket, missiles, spacecraft  Ground air transportation vehicles  Machinery  Human movement (Biomechanics) W. WangExample: The Coriolis Force Kinematics: coordinate reference frames matter, as in this merrygo round W. WangExample: Car Crush Test Kinetics: Impact , impulse and moment. Crash Test of a New Mercedes SLS AMG 2010 W. WangExample: Three Phase Diamagnetic Levitation Motor Studying of rotational motion of a motor, kinetics: magnetic forces, Kinematics: rotation speed and angles W. WangExample: SelfAssemble Robots Block communicate through Wireless Communication Studying of kinematics and kinetics of a moving robot Kinetics: forces on latches, kinematics: position tracking W. WangTopics to be covered Chapter 12:Introduction Kinematics of a particle Chapter 13: Kinetics of a particle: Force and Acceleration Chapter 14: Kinetics of a particle: Work and Energy Chapter 15: Kinetics of a particle: Impulse and Momentum Chapter 16: Planar kinematics of a Rigid Body Chapter 17: Planar kinetics of a Rigid Body: Force and Acceleration W. Wangcont’d Chapter 18: Planar kinetics of a Rigid Body: Work and Energy Chapter 19: Planar kinetics of a Rigid Body: Impulse and Momentum Chapter 20 and 21: ThreeDimensional Kinematics of a Rigid Body Overview of 3D Kinetics of a Rigid Body Chapter 22: Vibrations: underdamped free vibration, energy method, undamped forced vibration, viscous damped vibrations W. WangChapter 12: Kinematics of a Particle translational motion Two classes of motion W. Wang CoordinatesParticle Kinematic • Kinematics of a particle (Chapter 12) 12.112.2 W. WangObjectives Students should be able to: 1. Find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path (Continuous motion) (12.2) Next lecture; Determine position, velocity, and acceleration of a particle using graphs (Erratic motion) (12.3) W. WangQuestion 1. In dynamics, a particle is assumed to have . A) both translation and rotational motions B) only a mass C) a mass but the size and shape cannot be neglected D) no mass or size or shape, it is just a point W. WangW. WangW. WangW. WangShape of the motion F = ma W. WangF = ma W. WangKinematics of a Particle Type: • Constrained Motion: Pendulum, roller coaster, swing. • Unconstraint motion: Football trajectory, balloon in air Contents: • Rectilinear Motion: Moving along a straight line • Curvilinear Motion: 2D or 3D motion (a) rectangular coordinates (b) Normal and tangential coordinates (c) cylindrical (or Polar) coordinates • Relative motion: For complicated motion (a) Translating axes (b) rotating axes W. Wang Vector r r y  r x Recall in your high school math, a vector quantity is a quantity that is described by both magnitude and direction where W. WangRectilinear kinematics: Continuous motion A particle travels along a straightline path defined by the coordinate axis s The POSITION of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar (magnitude) s. Scalar s can be positive or negative. Typical units for r and s are meters (m or cm ) or feet (ft or inches). The displacement of the particle is defined as its change in position. Vector form: r = r’ r Scalar form:  s = s’ s The total distance traveled by the particle, s , is a positive scalar that represents the T total length of the path over which the particle travels. W. WangDepends on how you define your origin and positive and negative direction W. WangVelocity Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s. The average velocity of a particle during a time interval t is v = r/t avg The instantaneous velocity is the timederivative of position. v=dr/dt at P Speed is the magnitude of velocity: v = ds/dt Average speed is the total distance traveled divided by elapsed time: W. Wang (v ) = s /  t sp avg TAverage vs. Instantaneous Speed During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car. It shows your speed at a particular instant in time. The instantaneous speed of an object is not to be confused with the average speed. Average speed is a measure of the distance traveled in a given period of time; it is sometimes referred to as the distance per time ratio. Suppose that during your trip to school, you traveled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). The average speed of your car could be determined as On the average, your car was moving with a speed of 25 miles per hour. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour. Yet, on average, you were moving with a speed of 25 miles per hour. W. WangAverage Speed and Average Velocity Now let's consider the motion of that physics teacher again. The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity. The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that the displacement refers to the change in position and the velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average velocity of 0 m/s. W. WangAcceleration Acceleration is the rate of change in the velocity of a particle. It is a vector 2 2 quantity. Typical units are m/s or ft/s . The instantaneous acceleration is the time derivative of velocity. Vector form: a = dv/dt 2 2 Scalar form: a = dv/dt = d s/dt Acceleration can be positive (speed increasing) or negative (speed decreasing). As the book indicates, the derivative equations for velocity and W. Wang acceleration can be manipulated to get a ds = v dvDirection of Acceleration and Velocity Consider the motion of a Hot Wheels car down an incline, across a level and straight section of track, around a 180degree curve, and finally along a final straight section of track. Such a motion is depicted in the animation below. The car gains speed while moving down the incline that is, it accelerates. Along the straight sections of track, the car slows down slightly (due to air resistance forces). Again the car could be described as having an acceleration. Finally, along the 180degree curve, the car is changing its direction; once more the car is said to have an acceleration due to the change in the direction. Accelerating objects have a changing velocity either due to a speed change (speeding up or slowing down) or a direction change. W. WangThis simple animation above depicts some additional information about the car's motion. The velocity and acceleration of the car are depicted by vector arrows. The direction of these arrows are representative of the direction of the velocity and acceleration vectors. Note that the velocity vector is always directed in the same direction which the car is moving. A car moving eastward would be described as having an eastward velocity. And a car moving westward would be described as having a westward velocity. The direction of the acceleration vector is not so easily determined. As shown in the animation, an eastward heading car can have a westward directed acceleration vector. And a westward heading car can have an eastward directed acceleration vector. So how can the direction of the acceleration vector be determined A simple rule of thumb for determining the direction of the acceleration is that an object which is slowing down will have an acceleration directed in the direction opposite of its motion. Applying this rule of thumb would lead us to conclude that an eastward heading car can have a westward directed acceleration vector if the car is slowing down. Be careful when discussing the direction of the acceleration of an object; slow down, apply some thought and use the rule of thumb. W. WangW. WangSUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrate acceleration for velocity and position. Velocity: Position: v t v s s t a a  dv dt or v dv ds ds v dt  vo o vo so so o •Note that s and v represent the initial position and velocity of the particle at t o o = 0. W. WangFour types of Acceleration (I) a= constant (constant acceleration) e.g. gravitational acceleration (II) a= a(t) e.g. acceleration of a rocket with a constant thrust (III) a=a(v) e.g. deceleration from air drag (IV) a= a(s) e.g. acceleration from a spring load W. Wang(I) a= constant (Constant acceleration) The three kinematic equations can be integrated for the special case when acceleration is constant (a = a ) to obtain very useful equations. A c common example of constant acceleration is gravity; i.e., a body freely 2 2 falling toward earth. In this case, a = g = 9.81 m/s = 32.2 ft/s c downward. These equations are: v t  dv a dt v v a t yields c o c  v o o s t 2  ds v dt s s v t (1/2)a t yields  o o c o s o v s 2 2  v dv a ds yields v (v ) 2a (s s )  c c o o v o so W. WangW. WangW. WangW. WangFunction of v not t W. WangW. WangW. WangW. WangW. WangW. WangW. WangSUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrate acceleration for velocity and position. Velocity: Position: v t v s s t a a  dv dt or v dv ds ds v dt  vo o vo so so o •Note that s and v represent the initial position and velocity of the particle at t o o = 0. W. WangEXAMPLE Given: A particle travels along a straight line to the right 2 with a velocity of v = ( 4 t – 3 t ) m/s where t is in seconds. Also, s = 0 when t = 0. Find: The position and acceleration of the particle when t = 4 s. Plan: Establish the positive coordinate, s, in the direction the particle is traveling. Since the velocity is given as a function of time, take a derivative of it to calculate the acceleration. Conversely, integrate the velocity function to calculate the position. W. WangEXAMPLE (continued) Solution: 1) Take a derivative of the velocity to determine the acceleration. 2 a = dv / dt = d(4 t – 3 t ) / dt = 4 – 6 t 2  a = – 20 m/s (decelerating in direction) when t = 4 s Hard to tell 2 unless you Originally shown as a=20m/s (or in the  direction) solve “S” 2) Calculate the distance traveled in 4s by integrating the velocity using s = 0: o s t 2 (4 t – 3 t ) dt  v = ds / dt ds = v dt ds  2 3 s o o  s – s = 2 t –t o 2 3  s – 0 = 2(4) –(4) s = – 32 m ( or 32m going in  direction) Originally shown as s = – 32 m ( or ) W. WangSign Convention A simple rule of thumb for determining the direction of the acceleration is that an object which is slowing down will have an acceleration directed in the direction opposite of its motion. Applying this rule of thumb would lead us to conclude that an eastward heading car can have a westward directed acceleration vector if the car is slowing down. Be careful when discussing the direction of the acceleration of an object; slow down, apply some thought and use the rule of thumb. W. Wangright s a left v 4/6 4/3 Time (t) sec a = 0 means constant velocity a = dv / dt = 4 – 6 t v = 0 means stopping possible 2 changing direction v = 4 t – 3 t s = 0 means at original position 2 3 s – s = 2 t –t W. Wang oDo this  First define which direction is your positive direction.  Just remember slowing down (deceleration) is negative and speed up (acceleration) is positive, but hard to tell which direction it’s going unless you know position. So just reference to your defined positive reference in acceleration answer.  Velocity or position negative means going opposite direction of the direction you define as positive and Positive velocity or position means you are going in the same direction as you define as positive direction. Again indicate your reference direction. W. WangExample 1. A particle has an initial velocity of 3 ft/s to the left at s = 0 ft. Determine its position when t = 3 s if the 0 2 acceleration is 2 ft/s to the right. A) 0.0 ft B) 6.0 ft C) 18.0 ft D) 9.0 ft 2. A particle is moving with an initial velocity of v = 12 ft/s 2 and constant acceleration of 3.78 ft/s in the same direction as the velocity. Determine the distance the particle has traveled when the velocity reaches 30 ft/s. A) 50 ft B) 100 ft C) 150 ft D) 200 ft W. WangSolution for 1 v t  dv a dt v v a t yields  c o c v o o s t 2  ds v dt s s v t (1/2)a t yields  o o c o s o v s 2 2  v dv a ds v (v ) 2a (s s ) yields  c c o o vo so 2 S= 3 (m/s) x3 (sec) + (1/2) (2m/t ) x 3 (sec) = 0 Conservation of Energy equation W. WangSolution for 2 v t  dv a dt v v a t yields  c o c v o o s t 2  ds v dt s s v t (1/2)a t yields  o o c o s o v s 2 2  v dv a ds v (v ) 2a (s s ) yields  c c o o vo so Conservation of Energy equation W. WangAnalysing problems in dynamics Coordinate system • Establish a position coordinate S along the path and specify its fixed origin and positive direction • Motion is along a straight line and therefore s, v and α can be represented as algebraic scalars • Use an arrow alongside each kinematic equation in order to indicate positive sense of each scalar Kinematic equations • If any two of α, v, s and t are related, then a third variable can be obtained using one of the kinematic equations (one equation can only solve one unknown) • When performing integration, position and velocity must be known at a given instant (…so the constants or limits can be evaluated) W. Wang • Some equations must be used only when a is constantProblem solving MUSTS 1. Read the problem carefully (and read it again) 2. Physical situation and theory link 3. Draw diagrams and tabulate problem data 4. Coordinate system 5. Solve equations and be careful with units 6. Be critical. A mass of an aeroplane can not be 50 g 7. Read the problem carefully W. WangImportant points • Dynamics: Accelerated motion of bodies • Kinematics: Geometry of motion • Average speed = average velocity • Rectilinear kinematics or straightline motion • Acceleration is negative when particle is slowing down •α ds = v dv; relation of acceleration, velocity, displacement W. WangHomework Assignment Chapter 12: 10, 22, 24, 26, 28, 32, 37, 62, 71, 92, 98, 112, 120, 122, 144, 163, 175, 179 Due Next Wednesday W. WangW. WangAdditional Information  Lecture notes are online: http://courses.washington.edu/engr100/me230 W. WangLecture 1: Particle Kinematic • Kinematics of a particle (Chapter 12) 12.3 W. WangKinematics of a particle: Objectives • Concepts such as position, displacement, velocity and acceleration are introduced • Study the motion of particles along a straight line. Graphical representation • Investigation of a particle motion along a curved path. Use of different coordinate systems • Analysis of dependent motion of two particles • Principles of relative motion of two particles. Use of translating axis W. WangMaterial covered • Kinematics of a particle Rectilinear kinematics: Erratic motion Next lecture; General curvilinear motion, rectangular components and motion of a projectile W. WangObjectives Students should be able to: 1. Determine position, velocity, and acceleration of a particle using graphs (12.3) W. WangErratic (discontinuous) motion Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve W. WangW. Wangst graph construct vt Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt) Therefore, the vt graph can be constructed by finding the slope at various points along the st graph W. Wangvt graph construct at Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt) Therefore, the at graph can be constructed by finding the slope at various points along the vt graph Also, the distance moved (displacement) of the particle is the area under the vt graph during time t W. WangIntegrate acceleration for velocity and position. Velocity: Position: v t v s s t a a  dv dt or v dv ds ds v dt  vo o vo so so o •Note that s and v represent the initial position and velocity of the particle at t o o = 0. W. Wangat graph construct vt Given the at curve, the change in velocity (v) during a time period is the area under the at curve. So we can construct a vt graph from an at graph if we know the initial velocity of the particle W. Wangvt graph construct st We begin with initial position S0 and add algebraically increments Δs determined from the vt graph Equations described by v t graphs may be integrated in order to yield equations that describe segments of the st graph W. WangPlease remember the link handle complex motions graphing visual description of motion differentiation and integration slope and area under curve W. WangExplanation of Example 12.7 (A) W. WangExplanation of Example 12.7 (B) W. WangExample Given: The vt graph shown Find: The at graph, average speed, and distance traveled for the 30 s interval Hint Find slopes of the curves and draw the at graph. Find the area under the curvethat is the distance traveled. Finally, calculate average speed (using basic definitions) W. WangExample For 0 ≤ t ≤ 10 a = dv/dt = 0.8 t ft/s² For 10 ≤ t ≤ 30 a = dv/dt = 1 ft/s² a(ft/s²) 8 1 t(s) 30 10 W. WangExample 3 s =  v dt = (1/3) (0.4)(10) = 400/3 ft 010 2 2 s =  v dt = (0.5)(30) + 30(30) – 0.5(10) – 30(10) 1030 = 1000 ft s = 1000 + 400/3 = 1133.3 ft 030 v = total distance / time avg(030) = 1133.3/30 = 37.78 ft/s W. WangA couple of cases more… A couple of cases that are a bit more …COMPLEX… and therefore need more attention W. WangIntegrate acceleration for velocity and position. Velocity: Position: v t v s s t a a  dv dt or v dv ds ds v dt  vo o vo so so o •Note that s and v represent the initial position and velocity of the particle at t o o = 0. W. Wangas graph construct vs A more complex case is presented by the as graph. The area under the acceleration versus position curve represents the change in velocity (recall  a ds =  v dv ) This equation can be solved for v , allowing you to solve for the 1 velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance W. Wangvs graph construct as Another complex case is presented by the vs graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. a = v (dv/ds) Thus, we can obtain a plot of a vs. s from the vs curve W. WangDetermine position, velocity, and acceleration of a particle using graphs  Experimental data (very complicate motion)  Nonlinear motion  Find a function has the closely match curve or break it up and analyze it section by section.  Allow one to quickly analyze the changes in direction, velocity acceleration. W. WangPlease think about it If a particle in rectilinear motion has zero speed at some instant in time, is the acceleration necessarily zero at the same instant NO W. Wangright s a left v 4/6 4/3 Time (t) sec a = 0 means constant velocity a = dv / dt = 4 – 6 t v = 0 means stopping possible 2 changing direction v = 4 t – 3 t s = 0 means at original position 2 3 s – s = 2 t –t W. Wang oHomework Assignment Chapter 12: 10, 22, 24, 26, 28, 32, 37, 62, 71, 92, 98, 112, 120, 122, 144, 163, 175, 179 Due Next Wednesday W. WangW. WangWater Calculator http://www.youtube.com/watchv=dYlvr1lsmj0 W. WangMechanical System  Conveyor belt system Kinematic equations(position, velocity and acceleration) make sure it stop at the right position and not moving and stop and an appropriate speed to prevent water tipping over. Kinetic equations (gear system that provide enough torque to move the belt, enough friction on the wheels to catch the belt and rotate the belt) W. WangWater Calculator Powerpoint  Faculty.washington.edu/abong W. WangChapter 12: Kinematics of a Particle translational motion Two classes of Coordinates motion W. WangLecture 2: Particle Kinematic • Kinematics of a particle (Chapter 12) 12.412.6 W. WangKinematics of a particle: Objectives • Concepts such as position, displacement, velocity and acceleration are introduced • Study the motion of particles along a straight line. Graphical representation • Investigation of a particle motion along a curved path. Use of different coordinate systems • Analysis of dependent motion of two particles • Principles of relative motion of two particles. Use of translating axis W. WangMaterial covered • Kinematics of a particle General curvilinear motion Curvilinear motion: Rectangular components (Cartesian coordinate) Motion of a projectile Next lecture; Curvilinear motion: Normal tangential components and cylindrical components W. WangObjectives Students should be able to: 1. Describe the motion of a particle traveling along a curved path 2. Relate kinematic quantities in terms of the rectangular components of the vectors 3. Analyze the freeflight motion of a projectile W. WangRelated applications The path of motion of each plane in this formation can be tracked with radar and their x, y, and z coordinates (relative to a point on earth) recorded as a function of time How can we determine the velocity or acceleration at any instant A roller coaster car travels down a fixed, helical path at a constant speed If you are designing the track, why is it important to be able to predict the W. Wang acceleration of the carGeneral curvilinear motion A particle moving along a curved path undergoes curvilinear motion. Since the motion is often threedimensional, vectors are used to describe the motion A particle moves along a curve defined by the path function, s The position of the particle at any instant is designated by the vector r = r(t). Both the magnitude and direction of r may vary with time If the particle moves a distance s along the curve during time interval t, the displacement is determined by vector subtraction:  r = r’ r W. WangVelocity Velocity represents the rate of change in the position of a particle The average velocity of the particle during the time increment t is v = r/t avg The instantaneous velocity is the timederivative of position v = dr/dt The velocity vector, v,is always tangent to the path of motion The magnitude of v is called the speed. Since the arc length s approaches the magnitude ofr as t→0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector W. WangAcceleration Acceleration represents the rate of change in the velocity of a particle If a particle’s velocity changes from v to v’ over a time increment t, the average acceleration during that increment is: a =v/t = (v v’)/t avg The instantaneous acceleration is the timederivative of velocity: 2 2 a =dv/dt = d r/dt A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function W. WangSo pretty much the same set of equations we were describing in the rectilinear motion applies to curvilinear motion except in acceleration where due to the fact that when it is moving around a curve, in addition to the magnitude change along the direction of the path, there is a velocity direction change as well. When taking a time derivative of velocity for acceleration, it actually produce an additional acceleration that is not considered in the rectilinear motion. W. WangCurvilinear motion: Rectangular components It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference The position of the particle can be defined at any instant by the position vector r = x i + y j + z k The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t) 2 2 2 0.5 The magnitude of the position vector is: r = (x + y + z ) The direction of r is defined by the unit vector: u =(1/r)r r W. Wang Vector r r y  r x Recall in your high school math, a vector quantity is a quantity that is described by both magnitude and direction where W. WangUnit Vector Basically it is projection of the unit vector to x,y,z coordinates. Let our unit vector be: u = u i+ u j + u k r 1 2 3 P On the graph, u is the unit vector (in black) pointing in the same direction as vector OP, and i, j, and k (the unit vectors in the x, y and zdirections respectively) are marked in green. W. Wang We now zoom in on the vector u, and change orientation slightly, as follows: Now, if in the diagram in the last page, α is the angle between u and the xaxis (in dark red), β is the angle between u and the yaxis (in green) and γ is the angle between u and the zaxis (in pink), = u = 1 x 1 x cos cos = u = 1 x 1 x cos cos = u = 1 x 1 x cos cos So the direction of r is u = cos i+cos j cos k defined by the unit vector: r u =(1/r)r r 2 2 2 cos = x/sqrt(x +y +z ) 2 2 2 u = u i+ u j + u k cos = y/sqrt(x +y +z ) r 1 2 3 2 2 2 cos = z/sqrt(x +y +z ) W. WangRectangular components: Velocity The velocity vector is the time derivative of the position vector: v=dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since the unit vectors i, j, k are constant in magnitude and direction, this equation reduces to v = v i+v j+v k x y z Where; v = dx/dt, v = dy/dt, v = dz/dt x y z The magnitude of the velocity vector is 2 2 2 0.5 v = (v ) + (v ) + (v ) x y z The direction of v is tangent to the path of motion. W. Wang Rectangular components: Acceleration The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): 2 2 a =dv/dt=d r/dt =a i +a j+a k x y z where a = v x = dv /dt, a = v y = dv /dt, x x y y a =v z = dv /dt z z The magnitude of the acceleration vector is 2 2 2 0.5 a = (a ) + (a ) + (a ) x y z The direction of a is usually not tangent to the path of the particle W. WangW. WangEXAMPLE Given: The box slides down the slope described by the equation 2 y = (0.05x ) m, where x is in meters. 2 v = 3 m/s, a = 1.5 m/s at x = 5 m. x x Find: The y components of the velocity and the acceleration of the box at at x = 5 m. Plan: Note that the particle’s velocity can be related by taking the first time derivative of the path’s equation. And the acceleration can be related by taking the second time derivative of the path’s equation. Take a derivative of the position to find the component of the velocity and the acceleration. W. WangEXAMPLE (continued) Solution: Find the ycomponent of velocity by taking a time 2 derivative of the position y = (0.05x )    y = 2 (0.05) x x = 0.1 x x Find the acceleration component by taking a time derivative  of the velocity y    y = 0.1 x x + 0.1 x x Substituting the xcomponent of the acceleration, velocity at  x=5 into y and y. W. WangEXAMPLE (continued)   2 Since x = v = 3 m/s, x = a = 1.5 m/s at x = 5 m x x   y = 0.1 x x = 0.1 (5) (3) = 1.5 m/s    y = 0.1 x x + 0.1 x x 2 = 0.1 (3) + 0.1 (5) (1.5) = 0.9 – 0.75 2 = 0.15 m/s At x = 5 m v = – 1.5 m/s = 1.5 m/s  y 2 a = 0.15 m/s y W. WangProjectile motion… www.glenbrook.k12.il.us/gbssci/phys/mmedia/index.html W. WangMotion of a projectile Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., gravity) For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant W. WangHorizontal motion Vertical motion Y direction X direction V = V sin V = V cos oy o ox o 2 2 2 2 V = V +2aY V = V fy oy fx ox V = V +at V = V fy oy fx ox 2 Y= V t+1/2 at X= V t oy ox Y=1/2(V +V )t X=1/2(V +V )t fy oy fx ox W. WangKinematic equations: Horizontal Vertical motion Since a = 0, the velocity in the x horizontal direction remains constant (v = v ) and the x ox position in the x direction can be determined by: x = x + (v )(t) o ox Since the positive yaxis is directed upward, a = g. Application of the y constant acceleration equations yields: v = v – g(t) y oy 2 y = y + (v )(t) – ½g(t) o oy 2 2 v = v –2g(y –y ) W. Wang y oy oEXAMPLE I Given: v and θ A Find: Horizontal distance it travels and v . C Plan: Apply the kinematic relations in x and y directions. o o Solution: Using v = 10 cos 30 and v = 10 sin 30 Ax Ay o We can write v = 10 cos 30 x o v = 10 sin 30 – (9.81) t y o x = (10 cos 30 ) t o 2 y = (10 sin 30 ) t – ½ (9.81) t Since y = 0 at C o 2 0 = (10 sin 30 ) t – ½ (9.81) t t = 0, 1.019 s W. WangEXAMPLE I (continued) Velocity components at C are; o v = 10 cos 30 Cx = 8.66 m/s  o v = 10 sin 30 – (9.81) (1.019) Cy = 5 m/s = 5 m/s  Horizontal distance the ball travels is; o x = (10 cos 30 ) t o x = (10 cos 30 ) 1.019 = 8.83 m W. WangEXAMPLE II Given: Projectile is fired with v =150 m/s at point A. A Find: The horizontal distance it travels (R) and the time in the air. Plan: How will you proceed W. WangEXAMPLE II Given: Projectile is fired with v =150 m/s at point A. A Find: The horizontal distance it travels (R) and the time in the air. Plan:Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and ydirections. W. WangEXAMPLE II (continued) Solution: 1) Place the coordinate system at point A. Then, write the equation for horizontal motion. +  x = x + v t B A Ax AB where x = R, x = 0, v = 150 (4/5) m/s B A Ax Range, R, will be R = 120 t AB 2) Now write a vertical motion equation. Use the distance equation. 2 + y = y + v t –0.5 g t B A Ay AB AB where y = – 150, y = 0, and v = 150(3/5) m/s B A Ay 2 We get the following equation: –150 = 90 t + 0.5 (– 9.81) t AB AB Solving for t first, t = 19.89 s. AB AB Then, R = 120 t = 120 (19.89) = 2387 m AB W. WangExample Given: Snowmobile is going 15 m/s at point A. Find: The horizontal distance it travels (R) and the time in the air. Solution: First, place the coordinate system at point A. Then write the equation for horizontal motion. + x = x + v t and v = 15 cos 40˚ m/s B A Ax AB Ax Now write a vertical motion equation. Use the distance equation. 2 + y = y + v t –0.5g t v = 15 sin 40˚ m/s B A Ay AB c AB Ay Note that x = R, x = 0, y = (3/4)R, and y = 0. B A B A Solving the two equations together (two unknowns) yields W. Wang R = 19.0 m t = 2.48 s AB Example of Final Project Engineering with Circus The Human canon ball InClass Team Competition: The circus is in town A recently layoff Boeing engineer is trying out to become a member of the human canon ball team in the circus. The first test he is asked to do is to figure out how to fly over a newly constructed water tower and land safely on a trampoline without injuring himself. Before he actually does the stunt, he decides to make a scale model to test and see if he will be able to make the jump. Please help him Problem: Find the angle, the height and the distance the engineer need to travel to land safely on the trampoline. y max  tower trampolline x/2 x/2 Given: The distance between the tower and the trampoline is same as the distance between tower to canon. The height of the tower is 28cm. Initial velocity is 3.885m/s Gravitational acceleration is 9.8m/s W. WangAdditional Information: Y direction X direction V = V sin V = V cos oy o ox o 2 2 2 2 V = V +2aY V = V fy oy fx ox V = V +at V = V fy oy fx ox 2 Y= V t+1/2 at X= V t oy ox Y=1/2(V +V )t X=1/2(V +V )t fy oy fx ox Additional Problem Constraints:  Can you think of any other problems that are not being addressed in the model Judging: We will see who made the jump. The winners get 1 extra credit points. Turn in the correct calculation get another 2 extra credit. W. WangHomework Assignment Chapter 12: 10, 22, 24, 26, 28, 32, 37, 62, 71, 92, 98, 112, 120, 122, 144, 163, 175, 179 Due Next Wednesday W. WangW. Wang
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