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Outline. Kinetics Forces in human motion Impulse- momentum Mechanical work, power, & energy Locomotion Energetics. mv. Linear momentum (G). Product of the mass and linear velocity of an object G = mv Units: kg * m / s G: vector quantity direction of velocity vector. Impulse.

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  1. Outline • Kinetics • Forces in human motion • Impulse-momentum • Mechanical work, power, & energy • Locomotion Energetics

  2. mv Linear momentum (G) Product of the mass and linear velocity of an object G = mv Units: kg * m / s G: vector quantity direction of velocity vector

  3. Impulse Impulse = ∫F dt F = force applied to object area under a force-time curve product of the average force and time of application if constant force (F): Impulse = F * t Units: N * s

  4. An impulse imparted on an object causes a change in momentum ∆G= Impulse ∆G= ∫F dt if constant force (F): ∆G = F * t Gfinal=Ginitial+ F * t if average force (F): ∆G = F * t Gfinal = Ginitial + F * t Gfinal = Ginitial + Impulse

  5. A person (90 kg) on a bike (10 kg) increases v from 0 to 10 m/s. What impulse was required? Gfinal = Ginitial+ Impulse mvf= mvi + Impulse Impulse = mvf - mvi vi = 0; vf = 10 m/s; m = 90kg +10kg = 100 kg

  6. Spiking a volleyball What is the impulse applied to the ball? vinitial=3.6m/s (towards spiker) vfinal=25.2m/s (away from spiker) m=0.27 kg tcontact=18ms 5.83 Ns 324 N 7.776 Ns 432 N I don’t understand

  7. Vertical jump: impulse-momentum analysis Stance: Vertical impulse increases momentum Fy = Fg,y– mg mvtakeoff= mvi + ∫(Fg,y - mg)dt vi = 0 vtakeoff = ∫(Fg,y - mg)dt To maximize jump height: maximize stance impulse increase time of force application, increase Fg,y mg Fg,y

  8. mvtakeoff = ∫(Fg,y - mg)dt Stance time = 0.52 s Fgy (ave) = 750 N mg = 570 N Fg,y(N)

  9. A jumping person (m = 57 kg; vi = 0) has an average Fg,y of 750 N for 0.5 seconds. What is the person’s vertical takeoff velocity? mvy,takeoff = mvi + ∫(Fg,y - mg)dt 57 * vy,takeoff = 0 + (Fg,y - mg) * t 57 * vy,takeoff = (750 – 559.17) * 0.5 vy,takeoff = 1.67 m/s

  10. A jumping person (m = 57 kg; vi = 0) has an average Fg,y of 750 N for 0.5 seconds. What is the person’s vertical takeoff velocity? mvy,takeoff = mvi + ∫(Fg,y - mg)dt 57 * vy,takeoff = 0 + (Fg,y - mg) * t 57 * vy,takeoff = (750 – 559.17) * 0.5 vy,takeoff = 1.67 m/s How high did they jump? (1.67)2/2g=0.14m

  11. Walking or running at a constant average speed On average, forward velocity of body does not change during stance ∆ vx = 0 ∫ Fg,xdt = 0

  12. Walk: 1.25 m/s (constant avg. v) ∫Fg,x dt = 0 ---> A1 = A2 Fgx (body weights) A2 A1

  13. Run: 3.83 m/s (constant avg, v) ∫Fg,x dt = 0 ---> A1 = A2 A2 Fgx (body weights) A1

  14. Run: 3.83 m/s (constant avg. v)

  15. Accelerating: ∫Fg,x dt > 0 Accelerating Fg,x A1 < A2 A2 0 A1 Time

  16. Decelerating: ∫Fg,x dt < 0 Decelerating A1 > A2 Fg,x A2 0 A1 Time

  17. A person (100 kg) on a bicycle (10 kg) can apply a decelerating force of 200N by maximally squeezing the brake levers. How long will it take for the bicyclist to stop if he is traveling at 13.4 m/sec (30 miles per hour) and the braking force is the only force acting to slow him down?

  18. A soccer ball (4.17N) was travelling at 7.62 m/s until it contacted the head of a player and sent travelling in the opposite direction at 12.8 m/sec. If the ball was in contact with the player’s head for 22.7 milliseconds, what was the average force applied to the ball?

  19. Outline • Kinetics • Forces in human motion • Impulse-momentum • Mechanical work, power, & energy • Locomotion Energetics

  20. Mechanical Work & Energy Principle of Work and Energy Work to overcome fluid and friction forces gravitational and elastic forces Mechanical Energy Kinetic energy Potential energy Gravitational Elastic Conservation of Energy Units for Work and Mechanical energy Joule = Nm

  21. Work (U) F U = force * distance U =|F| *|r| * cos (θ) F: force applied r: distance moved θ: angle between force vector and line of displacement Scalar 1 N * m = 1 Joule r U = Fr F r θ = 0

  22. Work (U) F U = force * distance U =|F| *|r| * cos (θ) F: force applied r: distance moved θ: angle between force vector and line of displacement Scalar 1 N * m = 1 Joule θ = 30 r U = Frcos (θ) F r θ = 30

  23. Work (U) F U = force * distance U =|F| *|r| * cos (θ) F: force applied r: distance moved θ: angle between force vector and line of displacement Can be positive or negative Positive work: Force and displacement in same direction Negative Work: Force and displacement in opposite directions Scalar 1 N * m = 1 Joule θ = 30 r U = Frcos (θ)

  24. Work against Resistive (Non-Conservative) Forces Work to overcome resistances (friction, aero/hydro) 1 N * m = 1 Joule Dissipative (lost as heat)

  25. Which of the following is NOT and example of a non-conservative force? • Friction • Air Resistance C) Water Resistance D) Gravity E) None of the above

  26. Work against Conservative Forces Work to overcome gravity or spring forces Work leads to energy conservation

  27. Potential energy • When work on an object by a force can be expressed as the change in an object’s position. • Work done by gravitational forces • Gravitational potential energy • Work done by elastic forces • Elastic (strain) potential energy Potential energy arises from position of an object

  28. Gravitational potential energy (Ep,g) U = F*r = mg*ry mg = weight of object ry= vertical position of object mg Ep,g = mgry ry

  29. Elasticpotential energy: energy stored when a spring is stretched or compressed Ep,s= 0.5kx2 Spring Rest length (no energy stored) Stretched (Energy stored) Compressed (Energy stored)

  30. Kinetic energy (Ek,t) m = mass v = velocity k = kinetic, t = translational v m Ek,t = 0.5 mv2 Kinetic energy is based on velocity of an object

  31. Work-Energy TheoremMechanical work = ∆ Mechanical energy U= DE = DEk+DEp When positive mechanical work is done on an object, its mechanical energy increases. U=F*r=DE When negative mechanical work is done (e.g. braking) on an object, its mechanical energy decreases. U=-F*r=DE

  32. Ep,g = 100 J Ep,g = 0 ∆ Mechanical Energy = Mechanical work A 200 Newton object is lifted up 0.5 meter. ∆Ep,g = mg∆ry ∆Ep,g = 200 • 0.5 = 100 J U = 100 J

  33. ∆ Mechanical Energy = Mechanical work A 200 Newton object is lowered 0.5 meter. Negative work Ep,g = 100 J U = -100 J Ep,g = 0

  34. Mechanical work in uphill walking A person (mg = 1000 N) walks 1000m on a 45°uphill slope. How much mechanical work is required to lift the c.o.m. up the hill? A) -1,000 kJ B) 1,000 kJ C) 707 kJ D) – 707kJ E) I am lost 1000 m ∆ry 45°

  35. Law of Conservation of Energy DE=U DEk+DEp=Uext If only conservative forces are acting on the system: DEk+DEp=0 Ek+Ep=Constant Ek1+Ep1=Ek2+Ep2

  36. A woman with a mass of 60kg dives from a 10m platform, what is her potential and kinetic energy 3m into the dive? A) PE = 0 J, KE = 1765.8 J B) PE = 4120.2 J, KE = 0 J C) PE = 4120.2 J, KE = 1765.8 J D) PE= 1765.8 J, KE = 4120.2 J

  37. Mechanical Power (Pmech): Rate of performing mechanical work Pmech = U / ∆t Pmech= (F * r * cosθ ) / ∆t Pmech = F * vcosθ Units J / s = Watts (W)

  38. A sprinter (80 kg) increases forward velocity from 2 to 10 m/s in 5 seconds. U & Pmech ? A) U = 2560 J, P = 12.8 kW B) U = 3840 J, P = 19.2 kW C) U = 2560 J, P = 512 W D) U= 3840 J, P = 768 W

  39. A sprinter (80 kg) increases forward velocity from 2 to 10 m/s in 5 seconds. U & Pmech ? U = Ek,t(final) - Ek,t(initial) = 0.5m(vx,f2 - vx,i2) vx,i = 2 m/s vx,f = 10 m/s U = (0.5)(80)(100 - 4) = 3840 J Pmech = U / ∆t = 3840 J / 5 s = 768 W

  40. Swimming: work & power to overcome drag forces A person swims 100 meters at 1 m/s against a drag force of 150N. Work: Power:

  41. Swimming: work & power to overcome drag forces A person swims 100 meters at 1 m/s against a drag force of 150N. Work: U = F * d = 150 N * 100 meters = 15, 000 J Power: Pmech = F * v = 150 N * 1 m/s = 150W or you could calculate time (100 seconds) and use work/time

  42. Mechanical power to overcome drag Pmech = Fdrag* v Pmech = -0.5CDAv2* v = (-0.5CDA) * v3 Swimming, bicycling: most of the muscular power output is used to overcome drag

  43. Summary Work: result of force applied over distance Energy: capacity to do work Kinetic Energy: energy based on velocity of an object Potential Energy: energy arising from position of an object Power: rate of Work production

  44. Outline • Kinetics (external) • Forces in human motion • Impulse-momentum • Mechanical work, power, & energy • Locomotion Energetics

  45. Kinetic energy (Ek,t) m = mass v = velocity k = kinetic, t = translational v m Ek,t = 0.5 mv2

  46. Gravitational potential energy (Ep,g) mg = weight of object ry= vertical position of object mg Ep,g = mgry ry

  47. Elastic energy: energy stored when a spring is stretched or compressed Spring Rest length (no energy stored) Stretched (Energy stored) Compressed (Energy stored)

  48. Mechanical energy in level walking Some kinetic energy Some gravitational potential energy Little work done against aerodynamic drag Unless slipping, no work done against friction Not much bouncing (elastic energy)

  49. Mechanical energy fluctuations in level walking Average Ek,t constant (average vx constant) Average Ep,g constant (average ry constant) HOWEVER Ek,t and Ep,g fluctuate within each stance

  50. Walk vx decreases ry increases vx increases ry decreases

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