Power, propulsion
Power, propulsion ? NOTE: Many engines convert chemical energy into work at a constant rate, i.e. constant power P (=Fv), NOT constant force ! ? Exceptions: — jets and rockets rated by thrustforce (Newtons) — Gravity provides constant force mg, independent of speed — Electric motors -> larger force (“ kick in the pants” ) at lower rpm. — Human power: Exercise bicycle/stair climber (Calories/min?)
? Constant power constant acceleration! — acceleration a = F/m= (P/v) /m decreases with speed — If no friction, work done at constant power is: W = Pt = 1/2mv2 - 1/2mv02
Power, Propulsion with friction ? If there is a friction force f, vehicle can accelerate as long as applied force F (=P/v) overcomes friction, i.e. (F - f)> 0. Applying Newton’s laws: dv P i.e. Acceleration = (applied force - friction)/mass F ma m f dt v dv Can rewrite as : mv P fv, dt d i.e. (1 2 mv2 ) P fv dt (rate of increase of kinetic energy) = dK or P fv (power input) - (rate of work done against friction) dt
f
F=P/v
Power, Friction, terminal speed dv P Equation of motion : m f dt v
OR
dK P fv dt
? (Try to!) solve either form of these differential equations to get v(t) ? For solid friction: fk=constant. For fluids: f = f(v) - tough to solve!!! ? BUT: Can see that speed increases until F=f and power P=fv. THEN: —a=0, vehicle reaches (constant) terminal speed vmax= P / f — all engine work used against friction, no work (& no net force) left to accelerate ? ! ? E.g. A 4.5kW (6hp) golf cart loaded with CEOs has a weight of 4000N. If =0.2 (rolling friction), what is its maximum speed: (a) on a level surface? (b) on a 3? uphill slope?
AutoTrader: 1972 Apollo 17 Lunar Rover Low mi, 1 ownr, nds new battery. Buyer collects.
? m = 720 kg loaded ? P =1 hp = 746W ? roll = 0.2 ? g = 1.6 m/s2 What is its top speed? a. on level ground ? b. on a 3? uphill slope ?
Mid-term 2 Review ? Newton’s 3 laws of Motion: — Inertia. If v = constant (including 0) then net force F =0 —a = F/m. First find net external force, then divide by m —F AB= -F BA. Forces arise in “ interaction pairs” ? i.e. “ (force of A on B) = - (force of B on A) ” ? object cannot exert a net force on itself ? Equal+opposite forces are of same type (gravitational, surface/surface) ? In free-body diagram for “ A” , onlyF BA should appear
? Forces, motion in 2D (vectors): — Displacement, velocity, acceleration & so force are vectors. — Remember that F (and a) may not be parallel to motion v — Apply Newton’s laws independently alongx,y): ( ax = F x /m etc.
Force types: ? Weight: force of gravity on object F g=mg ? Tension: acts away from ends of rope, transmits force from supplier to attachment point ? If object is pushed against a surface: — Normal force n pushes back on object surface ? n NOT always equal to mg (slopes, lifting suitcase? ) — Surface friction depends only onn|,| not speed, contact area, ... ? Static friction opposes impending motion: f sn ? Kinetic friction opposes actual motion: f = kn
Solving problems with “ free-body diagrams” ? Isolate each object. — Consider only forces ON that object (not forces exerted by it on other ones).
? Is the object accelerating? — YES: add forces along motion direction, e.g.ax = Fx/m — NO: we have equilibrium : Fx= 0, Fy= 0 (e.g. Force table lab)
? If objects’ motions are coupled (pushing/pulling each other), can solve for 2 equations of motion using: — Tension has same magnitude (but opposite direction, away from ends) for connecting rope — Distance, speed, acceleration have same magnitude
Example: Atwood Machine (lab)
Work, Energy, Power Expanded Work-EnergyTheorem: For any system: {Change in mechanical energy} = {External work input} - {Work done by system}
i.e. K + U = Win - Wout (so if no other changes to system, K + U= 0) For constant friction f, Wout = f d: system has to do work, so loses mechanical energy (K+U) decreases (“ lost” enegy transferred into heat). Win =F r = F r cos, from external source or conversion of internal energy into work K= {1/2mv2 - 1/2mv02 } - sum over all objects which change motion U= {mg h} : positive for height gain, negative for height loss Power P dW/dt = Fv cos (Watt)
Force & Energy method equivalence (by example) ? How high can you jump? No friction so K + U= 0 ? Inclined planes: first check for FRICTION (usually, f= mgcos and grav. force down slope = mgsin) ? Chapter 7 Problem 21:
Review materials ? Lecture notes (NOT text book? ) and in-class examples ? Reading Quizzes ? Homework Problems ? Online self-test questions (website) ? Lab manuals and reports
? Constant power constant acceleration! — acceleration a = F/m= (P/v) /m decreases with speed — If no friction, work done at constant power is: W = Pt = 1/2mv2 - 1/2mv02
Power, Propulsion with friction ? If there is a friction force f, vehicle can accelerate as long as applied force F (=P/v) overcomes friction, i.e. (F - f)> 0. Applying Newton’s laws: dv P i.e. Acceleration = (applied force - friction)/mass F ma m f dt v dv Can rewrite as : mv P fv, dt d i.e. (1 2 mv2 ) P fv dt (rate of increase of kinetic energy) = dK or P fv (power input) - (rate of work done against friction) dt
f
F=P/v
Power, Friction, terminal speed dv P Equation of motion : m f dt v
OR
dK P fv dt
? (Try to!) solve either form of these differential equations to get v(t) ? For solid friction: fk=constant. For fluids: f = f(v) - tough to solve!!! ? BUT: Can see that speed increases until F=f and power P=fv. THEN: —a=0, vehicle reaches (constant) terminal speed vmax= P / f — all engine work used against friction, no work (& no net force) left to accelerate ? ! ? E.g. A 4.5kW (6hp) golf cart loaded with CEOs has a weight of 4000N. If =0.2 (rolling friction), what is its maximum speed: (a) on a level surface? (b) on a 3? uphill slope?
AutoTrader: 1972 Apollo 17 Lunar Rover Low mi, 1 ownr, nds new battery. Buyer collects.
? m = 720 kg loaded ? P =1 hp = 746W ? roll = 0.2 ? g = 1.6 m/s2 What is its top speed? a. on level ground ? b. on a 3? uphill slope ?
Mid-term 2 Review ? Newton’s 3 laws of Motion: — Inertia. If v = constant (including 0) then net force F =0 —a = F/m. First find net external force, then divide by m —F AB= -F BA. Forces arise in “ interaction pairs” ? i.e. “ (force of A on B) = - (force of B on A) ” ? object cannot exert a net force on itself ? Equal+opposite forces are of same type (gravitational, surface/surface) ? In free-body diagram for “ A” , onlyF BA should appear
? Forces, motion in 2D (vectors): — Displacement, velocity, acceleration & so force are vectors. — Remember that F (and a) may not be parallel to motion v — Apply Newton’s laws independently alongx,y): ( ax = F x /m etc.
Force types: ? Weight: force of gravity on object F g=mg ? Tension: acts away from ends of rope, transmits force from supplier to attachment point ? If object is pushed against a surface: — Normal force n pushes back on object surface ? n NOT always equal to mg (slopes, lifting suitcase? ) — Surface friction depends only onn|,| not speed, contact area, ... ? Static friction opposes impending motion: f sn ? Kinetic friction opposes actual motion: f = kn
Solving problems with “ free-body diagrams” ? Isolate each object. — Consider only forces ON that object (not forces exerted by it on other ones).
? Is the object accelerating? — YES: add forces along motion direction, e.g.ax = Fx/m — NO: we have equilibrium : Fx= 0, Fy= 0 (e.g. Force table lab)
? If objects’ motions are coupled (pushing/pulling each other), can solve for 2 equations of motion using: — Tension has same magnitude (but opposite direction, away from ends) for connecting rope — Distance, speed, acceleration have same magnitude
Example: Atwood Machine (lab)
Work, Energy, Power Expanded Work-EnergyTheorem: For any system: {Change in mechanical energy} = {External work input} - {Work done by system}
i.e. K + U = Win - Wout (so if no other changes to system, K + U= 0) For constant friction f, Wout = f d: system has to do work, so loses mechanical energy (K+U) decreases (“ lost” enegy transferred into heat). Win =F r = F r cos, from external source or conversion of internal energy into work K= {1/2mv2 - 1/2mv02 } - sum over all objects which change motion U= {mg h} : positive for height gain, negative for height loss Power P dW/dt = Fv cos (Watt)
Force & Energy method equivalence (by example) ? How high can you jump? No friction so K + U= 0 ? Inclined planes: first check for FRICTION (usually, f= mgcos and grav. force down slope = mgsin) ? Chapter 7 Problem 21:
Review materials ? Lecture notes (NOT text book? ) and in-class examples ? Reading Quizzes ? Homework Problems ? Online self-test questions (website) ? Lab manuals and reports
