MAE 242 Dynamics – Section I Dr. Kostas Sierros

1. MAE 242 Dynamics – Section I Dr. Kostas Sierros

2. Design project 1 …because of the make – up quiz…

3. Make-up quiz • Make – up quiz will take place next Tuesday (18th September) • People who solved the Quiz 1 problem will get bonus points • People that did not do well on Quiz 1, will start from scratch • Next Monday (17th September) I can offer a help session. What time (after 5:00 PM) do you prefer??

4. Cylindrical coordinates (13.6) This approach to solving problems has someexternal similarityto the normal & tangential method just studied. However, the path may be more complex or the problem may have other attributes that make it desirable to use cylindrical coordinates Equilibrium equations or “Equations of Motion” in cylindrical coordinates (using r, q , and z coordinates) may be expressed in scalar form as:  Fr = mar  Fq = maq  Fz = maz

5. Cylindrical coordinates (13.6) continued… If the particle is constrained to move only in the r – q plane (i.e., the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system. In this case, the path is only a function of q.  Fr = mar  Fq = maq Note that a fixed coordinate system is used, not a “body-centered” system as used in the n – t approach

6. Tangential and normal forces If a forcePcauses the particle to move along a path defined by r = f (q ), the normal forceNexerted by the path on the particle is always perpendicular to the path’s tangent. The frictional forceF always acts along the tangent in the opposite direction of motion. The directions ofNandFcan be specified relative to the radial coordinate by using the angle y

7. r d r q y tan = = dr dr q d Determination of angle ψ The angle y, defined as the angle between the extended radial line and the tangent to the curve, can be required to solve some problems. It can be determined from the following relationship; If y is positive, it is measured counterclockwise from the radial line to the tangent. If it is negative, it is measured clockwise

8. Problem 1

9. Problem 2

10. Problem 3

11. Kinetics of a particle: Work & Energy Chapter 14 Chapter objectives • Develop the principle of work and energy and apply it in order to solve problems that involve force, velocity and displacement • Problems that involve power and efficiency will be studied • Concept of conservative force will be introduced and application of theorem of conservation of energy, in order to solve kinetic problems, will be described

12. Lecture 8 • Kinetics of a particle: Work and Energy (Chapter 14) • -14.1-14.3

13. Material covered • Kinetics of a particle: Work & Energy • The work of a force • Principle of Work and Energy • Principle of Work and Energy for a • system of particles • …Next lecture…Power and efficiency, conservative forces and potential energy, conservation of energy…and MAKE – UP QUIZ

14. Today’s Objectives • Students should be able to: • Calculate the work of a force • Apply the principle of work and energy to a particle or system of particles

15. Applications I A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the “valleys” of the track How can we design the track (e.g., the height, h, and the radius of curvature, r to control the forces experienced by the passengers?

16. Applications II Crash barrels are often used along roadways for crash protection. The barrels absorb the car’s kinetic energy by deforming If we know the typical velocity of an oncoming car and the amount of energy that can be absorbed by each barrel, how can we design a crash cushion?

17. Work and Energy Another equation for working kinetics problems involving particles can be derived byintegratingtheequation of motion (F = ma)with respect to displacement By substituting at = v (dv/ds) into Ft = mat, the result is integrated to yield an equation known as theprinciple of work and energy This principle is useful for solving problems that involveforce,velocity, anddisplacement. It can also be used to explore the concept ofpower To use this principle, we must first understand how to calculate thework of a force

18. r2 ò r1 U1-2 = F•dr Work of a force (14.1) A force doeswork on a particle when the particle undergoes adisplacement along the line of action of the force Work is defined as theproductofforceanddisplacement componentsacting in thesame direction. So, if the angle between the force and displacement vector is q, the increment of work dU done by the force is; dU = F ds cos q By using the definition of thedot productand integrating, the total work can be written as;

19. If F is a function of position (a common case) this becomes s2 ò F cos q ds U1-2 = s1 Work of a force (14.1) continued… If both F and q are constant (F = Fc), this equation further simplifies to U1-2 = Fc cos q (s2 - s1) Work ispositive if the force and the movement are in thesame direction. If they areopposing, then the work isnegative. If the force and the displacement directions areperpendicular, the work iszero

20. U1-2 = - W dy = - W (y2 - y1) = - W Dy y2 ò y1 Work of a weight The work done by the gravitational force acting on a particle (orweight of an object) can be calculated by using; The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Dy isupward, the work isnegativesince the weight force always acts downward

21. The work of the spring force moving from position s1 to position s2 is; s2 s2 =ò = ò U1-2 Fs ds k s ds = 0.5k(s2)2 - 0.5k(s1)2 s1 s1 Work of a spring force When stretched, alinear elastic springdevelops a force of magnitude Fs = ks, where k is thespring stiffnessand s is thedisplacement from the unstretched position If a particle is attached to the spring, the force Fs exertedon the particle is oppositeto that exerted on the spring. Thus, the work done on the particle by the spring force will benegativeor U1-2 = – [ 0.5k (s2)2 – 0.5k (s1)2 ]

22. http://www.mech.uwa.edu.au/DANotes/springs/intro/springAnimation.gifhttp://www.mech.uwa.edu.au/DANotes/springs/intro/springAnimation.gif

23. Spring forces It is important to note the following about spring forces: 1.The equations just shown are forlinearsprings only! Recall that a linear spring develops a force according to F = ks (essentially the equation of a line) 2.The work of a spring isnotjust spring force times distance at some point, i.e., (ksi)(si).Beware, this is a trap that students often fall into! 3.Alwaysdouble checkthe sign of the spring work after calculating it. It is positive work if the force put on the object by the spring and the movement are in the same direction

24. Principle of work and energy (14.2 & 14.3) By integrating the equation of motion,  Ft = mat = mv(dv/ds), theprinciple of work and energycan be written as  U1-2 = 0.5m(v2)2 – 0.5m(v1)2 orT1 +  U1-2 = T2 U1-2 is thework done by all the forcesacting on the particle as it moves from point 1 to point 2. Work can be either apositive or negativescalar T1 and T2 are thekinetic energiesof the particle at the initial and final position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2. The kinetic energy is always apositive scalar(velocity is squared!) So, the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle’s final kinetic energy

25. Principle of work and energy (continued…) Note that the principle of work and energy (T1 +  U1-2 = T2) isnot a vector equation! Each term results in a scalar value Both kinetic energy and work have the same units, that of energy! In the SI system, the unit for energy is called ajoule(J), where 1 J = 1 N·m. In the FPS system, units are ft·lb The principle of work and energycannotbe used, in general, to determine forces directednormalto the path, since these forces do no work The principle of work and energy can also be applied to asystem of particlesby summing the kinetic energies of all particles in the system and the work due to all forces acting on the system

26. Example Given:A 0.5 kg ball of negligible size is fired up a vertical track of radius 1.5 m using a spring plunger with k = 500 N/m. The plunger keeps the spring compressed 0.08 m when s = 0 Find:The distance s the plunger must be pulled back and released so the ball will begin to leave the track when q = 135° Plan:1) Draw the FBD of the ball at q = 135°. 2) Apply the equation of motion in the n-direction to determine the speed of the ball when it leaves the track. 3) Apply the principle of work and energy to determines

27. t N n 45° W => + Fn = man = m (v2/r) => W cos45° = m (v2/r) => (0.5)(9.81) cos 45° = (0.5/1.5)v2 => v = 3.2257 m/s Example (continued) Solution: 1)Draw the FBD of the ball atq = 135° The weight (W)acts downward through the center of the ball. The normal force exerted by the track is perpendicular to the surface. The friction force between the ball and the track has no component in the n-direction 2)Apply the equation of motion in the n-direction.Since the ball leaves the track at q = 135°, set N = 0

28. Example (continued) 3)Apply the principle of work and energybetween position 1 (q = 0) and position 2 (q = 135°). Note that the normal force(N)does no work since it is always perpendicular to the displacement direction. (Students: Draw a FBD to confirm the work forces) T1 + U1-2 = T2 0.5m (v1)2 – W Dy – (0.5k(s2)2 – 0.5k (s1)2) = 0.5m (v2)2 and v1 = 0, v2 = 3.2257 m/s s1 = s + 0.08 m, s2 = 0.08 m Dy = 1.5 + 1.5 sin 45° = 2.5607 m => 0 – (0.5)(9.81)(2.5607) – [0.5(500)(0.08)2 – 0.5(500)(5 + 0.08)2] = 0.5(0.5)(3.2257)2 => s = 0.179 m = 179 mm

29. Homework Hibbeler 13.2 13.18 13.22 13.25 13.43 13.50 To be handed in Tuesday 18th September