Engineering science lesson 2


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Engineering science lesson 2

  1. 1. Chapter 2- Dynamic Engineering Systems 2.1 Uniform acceleration • linear and angular acceleration • Newton’s laws of motion • mass, moment of inertia and radius of gyration of rotating components • combined linear and angular motion • effects of friction 2.2 Energy transfer • gravitational potential energy • linear and angular kinetic energy • strain energy • principle of conservation of energy • work-energy transfer in systems with combine linear and angular motion • effects of impact loading 2.3 Oscillating mechanical systems • simple harmonic motion • linear and transverse systems; • qualitative description of the effects of forcing and damping
  2. 2. Potential energy• Stored energy that an object has due to its position or condition• When work gets done on an object, its potential and/or kinetic energy increases• There are different types of potential energy: 1. Gravitational energy 2. Elastic potential energy (energy in an stretched spring) 3. Others (magnetic, electric, chemical, …)
  3. 3. Conservative and non-conservative forces• A force for which W12= - W21 is called a conservative forces.• The net work done by a conservative force around any closed path (return back to the initial configuration) is zero.• A force that is not conservative is called a nonconservative force.• We cannot define potential energy associated with a nonconservative forces.– Examples of conservative forces • gravitational force • spring force– Examples of non-conservative forces • fluid drag force
  4. 4. Gravitational potential energy In general : x y z ∆U = − W = − ∫ Fx dx − ∫ Fy dy − ∫ Fz dz f f f xi y i z i Gravitational potential energy, ∆Ug y y r ∆U g = − Wg = − ∫ Fg dy = − ∫ (−mg ) dy Q Fg has only y-component: (-mg ) ˆ f f j y i y i = mg  y    yf yi ⇒ ∆U g = mg ( y − y ) = mg ∆yf i Rewrite Uf as U and yf as y and take U i to be the reference level and assign the values U i =0 and yi =0. ⇒ U ( y ) = mg y 4/8
  5. 5. Elastic potential energy xi xf 1∆U s = −W s = k (x f − x i ) 2 2 2 Elastic potential energy stored in a spring: 1 2 The spring is stretched or U s = kx compresses from its equilibrium position by x 2
  6. 6. Exercise• What P.E. is gained when a 100. kg object is raised 4.00 m straight up?• What PE would be gained if an object were moved 4 m to the right?
  7. 7. Kinetic energy• Energy due to motion of the object• The unit of kinetic energy is Joules (J).• Kinetic energy is a scalar (magnitude only)• Kinetic energy is non-negative (zero or positive)• Basically divided into two – Linear Kinetic energy linear kinetic energy is given by: m = mass K = 1 mv 2 2 v = speed (magnitude of velocity) v2 = vx2+ vy2 – Angular Kinetic energy angular kinetic energy is given by: K=1/2 Iω2 ω = angular speed I = inertia
  8. 8. Remember work-kinetic energy theorem for linear motion: 1 1 Wnet = mv f − mvi 2 2 2 2 Net work done on an object changes its kinetic energy There is an equivalent work-rotational kinetic energy theorem: 1 1 Wnet = I ω f − I ωi 2 2 2 2Net rotational work done on an object changes its rotational kinetic energy
  9. 9. Conservation of energy• Energy cannot be created or destroyed (Energy conservation law)• Energy can be transformed from one form to another form• The total energy of an isolated system cannot change.• When we stated the conservation of energy for a system two conditions are applied: • Isolated system (no external forces) • Only conservative forces in the system
  10. 10. Work• How do we give an object kinetic energy? By doing work! How do we do work? By means of a force acting on the object!!• Work W is energy transferred to or from an object by means of a force acting on the object. – Energy transferred to an object is defined as positive work – Energy extracted from an object is defined as negative work – ‘Work ’ ‘transferred energy’ – ‘Doing work’ ‘the act of transferring energy’• Work = (force along displacement)X(displacement) or mathematically W = F . d = Fd cos(φ )• The work done by the force F is zero if: – d = 0: displacement equal to zero φ = 90deg. : force perpendicular to displacement
  11. 11. IllustrationDefinition:The work done on an object by an external CONSTANTforce is- the product of the component of the force in thedirection of the displacement and the magnitude of thedisplacement. r r W = F ×d ×cos θ = F .d = F|| ×d
  12. 12. • Work is a scalar – Work has only magnitude, no direction. – The value of W is independent of how we draw the coordinate system (unlike the components of r or F)• The SI unit of work is the Joule (J) 1 Joule ≡ 1 Newton·meter 1 J = 1 Nm = 1 kg m2/s2• Work can be positive, negative or zero depending on the angle between the force and the displacement
  13. 13. Exercise F = 10 NWhat is the work done by- the gravitational force θ = 60° d = 10m- the normal force- the force Fwhen the block is displacedalong the horizontal.
  14. 14. Exercises• What is the potential energy of a 10kg mass: 1. 100m above the surface of the earth 2. at the bottom of a vertical mine shaft 1000m deep• Find the work done in raising 100 kg of water through a vertical distance of 3m.• A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a constant breaking force over a distance of 75m. Find: 1. The cars initial kinetic energy 2. The final kinetic energy 3. The breaking force
  15. 15. Combined linear and angular motion• Two bicycles roll down a hill that is 20m high. Both bicycles have a total mass of 12kg and 700 mm diameter wheels (r=.350 m). The first bicycle has wheels that weigh 0.6kg each, and the second bicycle has wheels that weigh 0.3kg each. Neglecting air resistance, which bicycle has the faster speed at the bottom of the hill? (Consider the wheels to be thin hoops).
  16. 16. Effects of impact loadingImpulse• Accumulated effect of force exerted on an object for a period of time • Impulse = (force)(time) • Increase in F or t ⇒ increase in I • Vector • Equal to the change in momentum of the systemImpact• Collision characterized by the exchange of a large force during a short time period• Impact Types – Perfectly elastic – Perfectly plastic – Elastic• Characterized by Coefficient of Restitution [COR] (e=1, 0<e<1, e=0)
  17. 17. Strategies in SportsGiving impulse: – Maximize (batting, throwing, etc.)Receiving impulse: – Minimize (landing, receiving, etc.) – Reduce impulse force by increasing time
  18. 18. Applications of impact• Ball drop and COR• COR vs surfaces
  19. 19. Exercises• An object weighing 15. N is lifted from the ground to a height of 0.22 m. Find objects change in PE • In the figure, a 2.0 kg package of tamale slides along a floor with speed v1 = 4.0 m/s. It then runs into and compresses a spring, until the package momentarily stops. Its path to the initially relaxed spring is frictionless, but as it compresses the spring, a kinetic frictional force from the floor, of magnitude 15 N, acts on it. The spring constant is 10,000 N/m. By what distance d is the spring compressed when the package stops?
  20. 20. Exercises• A roller coaster cart of mass m = 300 kg starts at rest at point A, whose height off the ground is h1 = 25 m, and a little while later reaches point B, whose height off the ground is h2 = 7 m. What is the potential energy of the cart relative to the ground at point A? What is the speed of the cart at point B, neglecting the effect of friction?
  21. 21. Exercises• Consider the diagram. The mass of block A is 75 kg and the mass of block B is 15 kg. The coefficient of static friction between the two blocks is µ= 0.45. The horizontal surface is frictionless. What minimum force F must be exerted on block A in order to prevent block B from falling?
  22. 22. Exercises• A pirate drags a 50 kg treasure chest over the rough surface of a dock by exerting a constant force of 95N acting at an angle of 15 above the horizontal. The chest moves 6m in a straight line, and the coefficient of kinetic friction between the chest and the dock is 0.15. How much work does the pirate perform? How much energy is dissipated as heat via friction? What is the final velocity of the chest?