ENT 142 ENGINEERING DYNAMICS Programme B Eng Hons

ENT 142 ENGINEERING DYNAMICS Programme: B. Eng. (Hons) (Mechanical Eng. ) School of Mechatronic Engineering Universiti Malaysia Perlis Uni. MAP

Course Description Unit Name: ENT 142 Engineering Dynamics (3 credits). Lecturer: Pn. Fauziah Mat Room No. 59, MEKA 4 D, PPK Mekatronik, Kampus Pauh Putra. Email: fauziah@unimap. edu. my Contact Hours: 3 hours or 4 hours/week. Tuition Pattern: 3 hours lecture or 2 hours lecture + 2 hours tutorial. Lecture Schedule: Monday (2. 00 pm – 3. 00 pm): BKN 5 Thurday (10. 00 am – 12. 00 am): BK C Tutorial Schedule: Monday (5. 00 pm – 7. 00 pm): MN 4 E 0

Course Description Pre-Requisites : None Assessment : Examination : 70% Mid Term Examination: 10 % Final Examination: 60 % Coursework (30 %) Assignments (2 x 5%) : 10 % Quizzes (4 x 5%): 20 %

Synopsis The objective of the course is to enable students to evaluate problems related to mechanics concepts in dynamic condition. The course covers topics of force and acceleration, work and energy, and also impulse and momentum for both kinematics of a particle and planar kinetics of a rigid body problems.

Course Outcomes CO 1: Ability to analyze problems related to rectilinear kinematics, law of motions, and also concepts mechanics and vector mechanics. CO 2: Ability to evaluate problems related to kinematics of particle, involving force and acceleration, work and energy, and also impulse and momentum. CO 3: Ability to evaluate problems related to planar kinetics or a rigid body, involving force and acceleration, work and energy, and also impulse and momentum.

Syllabus Chapter 1. Kinematics of a Particle 1. 1 Introduction. 1. 2 Rectilinear Kinematics: Continuous Motion. 1. 3 Rectilinear Kinematics: Erratic Motion 1. 4 General Curvilinear Motion 1. 5 Curvilinear Motion: Rectangular Component 1. 6 Motion of a Projectile 1. 7 Curvilinear Motion: Normal and Tangential Components Chapter 2. Kinetics of a Particle: Force and Acceleration 2. 1 Newton’s Law of Motion 2. 2 The Equation of Motion 2. 3 Equations of Motion for a System of Particles 2. 4 Equations of Motion: Rectangular Coordinates 2. 5 Equation of Motion: Normal and Tangential Coordinates

Chapter 3. Kinetics of a Particle: Work and Energy 3. 1 The Work of a Force 3. 2 Principle of Work and Energy 3. 3 Principle of Work and Energy for a System of Particles 3. 4 Power and Efficiency 3. 5 Conservative Forces and Potential Energy 3. 6 Conservation of Energy. Chapter 4. Kinetics of a Particle: Impulse and Momentum 4. 1 Principle of Linear Impulse and Momentum 4. 2 Principle of Linear and Momentum for a System of Particles 4. 3 Conservation of Linear Momentum for a System of Particles 4. 4 Impact

Chapter 5. Planar Kinematics of a Rigid Body 5. 1 Rigid-Body Motion 5. 2 Translation 5. 3 Rotation About a Fixed Axis 5. 4 Relative-Motion Analysis: Velocity 5. 5 Relative-Motion Analysis: Acceleration Chapter 6. Planer Kinetics of a Rigid Body: Force and Acceleration 6. 1 Moment of Inertia 6. 2 Planar Kinetic Equations of Motion 6. 3 Equation of Motion: Translation

Chapter 7. Planar Kinetics of a Rigid Body: Work and Energy 7. 1 Kinetic Energy 7. 2 The Work of a Force 7. 3 The Work of Couple 7. 4 Principle of Work and Energy 7. 5 Conservation of Energy Chapter 8. Planar Kinetics of a Rigid Body: Impulse and Momentum 8. 1 Linear and Angular Momentum 8. 2 Principle of Impulse and Momentum 8. 3 Conservation of Momentum

Required Textbook : • R. C. Hibbler, Engineering Mechanics: Dynamics, 13 th Ed. , Prentice Hall, 2013. Recommended Books: • J. L. Meriam and L. Glenn Kraige, Engineering Mechanics: Dynamics, 2001, John Willey & Sons, Inc. • F. P. Beer, E. R. Johnston and W. E. Clausen, Vector Mechanics for Engineers: Dynamics, 2004, Mc Graw Hill.

An Overview of Mechanics: the study of how bodies react to forces acting on them Statics: the study of bodies in equilibrium Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics - concerned with the forces causing the motion

CHAPTER 1 KINEMATICS OF A PARTICLE

1. 1 Introduction & 1. 2 Rectilinear Kinematics: Continuous Motion Today’s Objectives: Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path. In-Class Activities: • Applications • Relations between s(t), v(t), and a(t) for general rectilinear motion • Relations between s(t), v(t), and a(t) when acceleration is constant

APPLICATIONS The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles. Why? If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?

APPLICATIONS (continued) A train travels along a straight length of track. Can we treat the train as a particle? If the train accelerates at a constant rate, how can we determine its position and velocity at some instant?

POSITION AND DISPLACEMENT A particle travels along a straight-line 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 s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft). 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. T, is a positive scalar that represents the total length of the path over which the particle travels.

VELOCITY 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 vavg = s/ t The instantaneous velocity is the time-derivative of position. v = ds/dt Speed is the magnitude of velocity: v = ds/dt Average speed is the total distance traveled divided by elapsed time: (vsp)avg = s. T/ t

ACCELERATION Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s 2 or ft/s 2. The instantaneous acceleration is the time derivative of velocity. Vector form: a = dv/dt Scalar form: a = dv/dt = d 2 s/dt 2 Acceleration can be positive (speed increasing) or negative (speed decreasing). As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

SUMMARY 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. Position: Velocity: v t v s vo o vo so ò dv = ò a dt or ò v dv = ò a ds s t so o ò ds = ò v dt • Note that so and vo represent the initial position and velocity of the particle at t = 0.

CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i. e. , a body freely falling toward earth. In this case, ac = g = 9. 81 m/s 2 = 32. 2 ft/s 2 downward. These equations are: v t ò dv = ò a dt c vo o s t ò ds = ò v dt so v v = vo + act yields s = s o + v ot + (1/2)a ct 2 yields v 2 = (vo )2 + 2 ac(s - so) o s ò v dv = ò ac ds vo yields so

EXAMPLE Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6 t m/s 2. Find: The distance the motorcycle travels before it stops. Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrate it once to calculate the velocity and again to calculate the position.

EXAMPLE (continued) Solution: 1) Integrate acceleration to determine the vvelocity. t a = dv / dt => dv = a dt => ò dv = ò (-6 t )dt vo o => v – vo = -3 t 2 => v = -3 t 2 + vo 2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use vo = 27 m/s. 0 = -3 t 2 + 27 => t = 3 s 3) Now calculate the distance traveled in 3 s by integrating the velocity using so = 0: s t v = ds / dt => ds = v dt => ò ds = ò (-3 t 2 + vo)dt so o => s – so = -t 3 + vot => s – 0 = (3)3 + (27)(3) => s = 54 m

1. 3 Rectilinear Kinematics: Erratic Motion Today’s Objectives: Students will be able to determine position, velocity, and acceleration of a particle using graphs. In-Class Activities: • Applications • s-t, v-t, a-t, v-s, and a-s diagrams

APPLICATION In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the rocket sled, can we determine its acceleration at position s = 300 meters ? How?

GRAPHING Graphing provides a good way to handle complex motions (e. g. erratic or discontinuity) 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.

S-T GRAPH 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 v-t graph can be constructed by finding the slope at various points along the s-t graph.

V-T GRAPH 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 a-t graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t.

A-T GRAPH Given the a-t curve, the change in velocity ( v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.

A-S GRAPH A more complex case is presented by the a-s graph. The area under the acceleration versus position curve represents the change in velocity (recall ò a ds = ò v dv ). s 2 ½ (v 1² – vo²) = ò a ds = area under the s 1 a-s graph This equation can be solved for v 1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.

V-S GRAPH Another complex case is presented by the v-s 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 v-s curve.

EXAMPLE Given: v-t graph for a train moving between two stations Find: a-t graph and s-t graph over this time interval Think about your plan of attack for the problem!

EXAMPLE (continued) Solution: For the first 30 seconds the slope is constant and is equal to: a 0 -30 = dv/dt = 40/30 = 4/3 ft/s 2 Similarly, a 30 -90 = 0 and a 90 -120 = -4/3 ft/s 2 a(ft/s 2) 4 3 t(s) -4 3

EXAMPLE (continued) s(ft) The area under the v-t graph represents displacement. 3600 3000 s 0 -30 = ½ (40)(30) = 600 ft s 30 -90 = (60)(40) = 2400 ft s 90 -120 = ½ (40)(30) = 600 ft 600 30 90 t(s) 120

1. 4 General Curvilinear Motion & 1. 5 Curvilinear Motion: Rectangular Components Today’s Objectives: Students will be able to: a) Describe the motion of a particle traveling along a curved path. b) Relate kinematic quantities in terms of the rectangular components of the vectors. In-Class Activities: • Applications • General curvilinear motion • Rectangular components of kinematic vectors

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 of each plane at any instant? Should they be the same for each aircraft?

APPLICATIONS (continued) A roller coaster car travels down a fixed, helical path at a constant speed. How can we determine its position or acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car?

POSITION AND DISPLACEMENT A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, 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

VELOCITY Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment t is vavg = r/ t. The instantaneous velocity is the time-derivative 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!

ACCELERATION 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: aavg = v/ t = (v - v’)/ t The instantaneous acceleration is the timederivative of velocity: a = dv/dt = d 2 r/dt 2 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.

RECTANGULAR COMPONENTS: POSITION 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=xi+yj+zk. The x, y, z components may all be functions of time, i. e. , x = x(t), y = y(t), and z = z(t). The magnitude of the position vector is: r = (x 2 + y 2 + z 2)0. 5 The direction of r is defined by the unit vector: ur = (1/r)r

RECTANGULAR 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 = vxi + vyj + vzk • • • x y z where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt The magnitude of the velocity vector is v = [(vx)2 + (vy)2 + (vz)2]0. 5 The direction of v is tangent to the path of motion.

RECTANGULAR COMPONENTS: ACCELERATION The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a = dv/dt = d 2 r/dt 2 = axi + ayj + azk where • • • • ax = vx = dvx /dt, ay = vy = dvy /dt, • • az = vz = dvz /dt The magnitude of the acceleration vector is a = [(ax)2 + (ay)2 + (az)2 ]0. 5 The direction of a is usually not tangent to the path of the particle.

EXAMPLE Given: The motion of two particles (A and B) is described by the position vectors r. A = [3 t i + 9 t(2 – t) j] m r. B = [3(t 2 – 2 t +2) i + 3(t – 2) j] m Find: The point at which the particles collide and their speeds just before the collision. Plan: 1) The particles will collide when their position vectors are equal, or r. A = r. B. 2) Their speeds can be determined by differentiating the position vectors.

Solution: EXAMPLE (continued) 1) The point of collision requires that r. A = r. B, so x. A = x. B and y. A = y. B. x-components: 3 t = 3(t 2 – 2 t + 2) Simplifying: t 2 – 3 t + 2 = 0 Solving: t = {3 [32 – 4(1)(2)]0. 5}/2(1) => t = 2 or 1 s y-components: 9 t(2 – t) = 3(t – 2) Simplifying: 3 t 2 – 5 t – 2 = 0 Solving: t = {5 [52 – 4(3)(– 2)]0. 5}/2(3) => t = 2 or – 1/3 s So, the particles collide when t = 2 s. Substituting this value into r. A or r. B yields x. A = x. B = 6 m and y. A = y. B = 0

EXAMPLE (continued) 2) Differentiate r. A and r. B to get the velocity vectors. . . v. A = dr. A/dt =. x. A i + y. A j = [3 i + (18 – 18 t)j] m/s At t = 2 s: v. A = [3 i – 18 j] m/s • • v. B = dr. B/dt = x. Bi + y. Bj = [(6 t – 6)i + 3 j] m/s At t = 2 s: v. B = [6 i + 3 j] m/s Speed is the magnitude of the velocity vector. v. A = (32 + 182) 0. 5 = 18. 2 m/s v. B = (62 + 32) 0. 5 = 6. 71 m/s

1. 6 MOTION OF A PROJECTILE Today’s Objectives: In-Class Activities: Students will be able to • Check homework, if any analyze the free-flight motion of • Applications a projectile. • Kinematic equations for projectile motion

APPLICATIONS A kicker should know at what angle, q, and initial velocity, vo, he must kick the ball to make a field goal. For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

APPLICATIONS (continued) A fireman wishes to know the maximum height on the wall he can project water from the hose. At what angle, q, should he hold the hose?

CONCEPT OF PROJECTILE MOTION 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.

KINEMATIC EQUATIONS: HORIZONTAL MOTION Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by: x = xo + (vox)(t) Why is ax equal to zero (assuming movement through the air)?

KINEMATIC EQUATIONS: VERTICAL MOTION Since the positive y-axis is directed upward, ay = -g. Application of the constant acceleration equations yields: vy = voy – g(t) y = yo + (voy)(t) – ½g(t)2 vy 2 = voy 2 – 2 g(y – yo) For any given problem, only two of these three equations can be used. Why?

Example 1 Given: vo and θ Find: The equation that defines y as a function of x. Plan: Eliminate time from the kinematic equations. Solution: Using vx = vo cos θ We can write: x = (vo cos θ)t or and t = vy = vo sin θ x vo cos θ y = (vo sin θ)t – ½ g(t)2 By substituting for t: y = (vo sin θ) ( x vo cos θ ) ( )( – g 2 x vo cos θ 2 )

1. 7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS Today’s Objectives: Students will be able to determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path. In-Class Activities: • Applications • Normal and tangential components of velocity and acceleration • Special cases of motion

APPLICATIONS Cars traveling along a clover-leaf interchange experience an acceleration due to a change in speed as well as due to a change in direction of the velocity. If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration? Why would you care about the total acceleration of the car?

APPLICATIONS (continued) A motorcycle travels up a hill for which the path can be approximated by a function y = f(x). If the motorcycle starts from rest and increases its speed at a constant rate, how can we determine its velocity and acceleration at the top of the hill? How would you analyze the motorcycle's “flight” at the top of the hill?

NORMAL AND TANGENTIAL COMPONENTS When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle). The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued) The positive n and t directions are defined by the unit vectors un and ut, respectively. The center of curvature, O’, always lies on the concave side of the curve. The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.

VELOCITY IN THE n-t COORDINATE SYSTEM The velocity vector is always tangent to the path of motion (t-direction). The magnitude is determined by taking the time derivative of the path function, s(t). . v = vut where v = s = ds/dt Here v defines the magnitude of the velocity (speed) and ut defines the direction of the velocity vector.

ACCELERATION IN THE n-t COORDINATE SYSTEM Acceleration is the time rate of change of velocity: . . a = dv/dt = d(vut)/dt = vut + vut. Here v represents the change in. the magnitude of velocity and ut represents the rate of change in the direction of ut. After mathematical manipulation, the acceleration vector can be expressed as: . a = vut + (v 2/r)un = atut + anun.

ACCELERATION IN THE n-t COORDINATE SYSTEM (continued) There are two components to the acceleration vector: a = a t ut + a n un • The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. . at = v or at ds = v dv • The normal or centripetal component is always directed toward the center of curvature of the curve. an = v 2/r • The magnitude of the acceleration vector is a = [(at)2 + (an)2]0. 5

SPECIAL CASES OF MOTION There are some special cases of motion to consider. 1) The particle moves along a straight line. . 2 r => an = v /r = 0 => a = at = v The tangential component represents the time rate of change in the magnitude of the velocity. 2) The particle moves along a curve at constant speed. . at = v = 0 => a = an = v 2/r The normal component represents the time rate of change in the direction of the velocity.

SPECIAL CASES OF MOTION (continued) 3) The tangential component of acceleration is constant, at = (at)c. In this case, s = so + vot + (1/2)(at)ct 2 v = vo + (at)ct v 2 = (vo)2 + 2(at)c(s – so) As before, so and vo are the initial position and velocity of the particle at t = 0. How are these equations related to projectile motion equations? Why? 4) The particle moves along a path expressed as y = f(x). The radius of curvature, r, at any point on the path can be calculated from 2 ]3/2 [ 1 (dy/dx) + r = ________ d 2 y/dx 2

THREE-DIMENSIONAL MOTION If a particle moves along a space curve, the n and t axes are defined as before. At any point, the t-axis is tangent to the path and the n-axis points toward the center of curvature. The plane containing the n and t axes is called the osculating plane. A third axis can be defined, called the binomial axis, b. The binomial unit vector, ub, is directed perpendicular to the osculating plane, and its sense is defined by the cross product ub = ut x un. There is no motion, thus no velocity or acceleration, in the binomial direction.

EXAMPLE PROBLEM Given: Starting from rest, a motorboat travels around a circular path of r = 50 m at a speed that increases with time, v = (0. 2 t 2) m/s. Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 3 s. Plan: The boat starts from rest (v = 0 when t = 0). 1) Calculate the velocity at t = 3 s using v(t). 2) Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector.