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UW-Madison PHYSICS 207 - Lecture 11

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Page 1Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 1Lecture 11Goals:Goals:Assignment:Assignment: Read through Chapter 10, 1stfour section MP HW6, due Wednesday 3/3Chapter 8: Employ rotational motion models with friction or in free fallChapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse Understand what momentum is and how it relates to forces Employ momentum conservation principles  In problems with 1D and 2D Collisions  In problems having an impulse (Force vs. time)Physics 207: Lecture 11, Pg 2 One last reprisal of theFree Body Diagram Remember1 Normal Force is ⊥⊥⊥⊥ to the surface2 Friction is parallel to the contact surface3 Radial (aka, centripetal) acceleration requires a net forceZero Gravity RidePage 2Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 3Zero Gravity RideA rider in a horizontal “0 gravity ride” finds herself stuck with her back to the wall.Which diagram correctly shows the forces acting on her?Physics 207: Lecture 11, Pg 4Banked CurvesIn the previous car scenario, we drew the following free body diagram for a race car going around a curve at constant speed on a flat track.Because the acceleration is radial (i.e., velocity changes in direction only) we need to modify our view of friction. nmgFfSo, what differs on a banked curve?Page 3Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 5Banked Curves (high speed)1 Draw a Free Body Diagram for a banked curve.2 Use a rotated x-y coordinates3 Resolve into components parallel and perpendicular to bankmarxxy y θNmgFfPhysics 207: Lecture 11, Pg 6Banked Curves (constant high speed)1 Draw a Free Body Diagram for a banked curve.2 Use a rotated x-y coordinates3 Resolve into components parallel and perpendicular to bank( Note: For very small banking angles, one can approximate that Ffis parallel to mar. This is equivalent to the small angle approximation sin θ = tan θ, but very effective at pushing the car toward the center of the curve!!)marxxy y θNmgFfPage 4Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 7Banked Curves, high speed4 Apply Newton’s 1stand 2ndLawsmarxxy y θNmg cos θFfmg sin θθmarcos θmarsin θΣ Fx= -marcos θ = - Ff- mg sin θΣ Fy= marsin θ = 0 - mg cos θ + NFriction model Ff≤ µ N (maximum speed when equal)Physics 207: Lecture 11, Pg 8Banked Curves, low speed4 Apply Newton’s 1stand 2ndLawsmarxxy y θNmg cos θFfmg sin θθmarcos θmarsin θΣ Fx= -marcos θ = + Ff- mg sin θΣ Fy= marsin θ = 0 - mg cos θ + NFriction model Ff≤ µ N (minimum speed when equal butnot less than zero!)Page 5Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 9Banked Curves, constant speed vmax= (gr)½[ (µ+ tan θ) / (1 -µtan θ)] ½vmin= (gr)½[ (tan θ -µ) / (1 + µtan θ)] ½Dry pavement“Typical” values of r = 30 m, g = 9.8 m/s2, µ= 0.8, θ = 20°vmax= 20 m/s (45 mph)vmin= 0 m/s (as long as µ> 0.36 )Wet Ice“Typical values” of r = 30 m, g = 9.8 m/s2, µ= 0.1, θ =20°vmax= 12 m/s (25 mph)vmin= 9 m/s(Ideal speed is when frictional force goes to zero)Physics 207: Lecture 11, Pg 10Banked Curves, Testing your understandingFree Body Diagram for a banked curve.Use rotated x-y coordinatesResolve into components parallel and perpendicular to bankNmgFfAt this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you begin to accelerate in that direction.How does the radial acceleration change? maryyx x θPage 6Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 11Navigating a hillKnight concept exercise: A car is rolling over the top of a hillat speed v. At this instant,A. n > w.B. n = w.C. n < w.D.We can’t tell about n without knowing v.This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion.Fc= mg = m v2/r v = (gr)½(just like an object in orbit)Note this approach can also be used to estimate the maximum walking speed.At what speed does the car lose contact?Physics 207: Lecture 11, Pg 12Orbiting satellites vT= (gr)½Page 7Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 13Locomotion: how fast can a biped walk?Physics 207: Lecture 11, Pg 14How fast can a biped walk?What about weight?(a) A heavier person of equal height and proportions can walk faster than a lighter person(b) A lighter person of equal height and proportions can walk faster than a heavier person(c) To first order, size doesn’t matterPage 8Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 15How fast can a biped walk?What about height?(a) A taller person of equal weight and proportions can walk faster than a shorter person(b) A shorter person of equal weight and proportions can walk faster than a taller person(c) To first order, height doesn’t matterPhysics 207: Lecture 11, Pg 16How fast can a biped walk?What can we say about the walker’s acceleration if there is UCM (a smooth walker) ?Acceleration is radial !So where does it, ar, come from?(i.e., what external forces act on the walker?)1. Weight of walker, downwards2. Friction with the ground, sidewaysPage 9Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 17Impulse & Linear Momentum Transition from forces to conservation lawsNewton’s Laws Conservation LawsConservation Laws Newton’s LawsThey are different faces of the same physicsNOTE: We have studied “impulse” and “momentum”but we have not explicitly named them as suchConservation of momentum is far more general thanconservation of mechanical energyPhysics 207: Lecture 11, Pg 18Forces vs time (and space, Ch. 10) Underlying any “new” concept in Chapter 9 is (1) A net force changes velocity (either magnitude or direction) (2) For any action there is an equal and opposite reaction If we emphasize Newton’s 3rdLaw and emphasize changes with time then this leads to the Conservation of Momentum PrinciplePage 10Physics 207 – Lecture 11Physics 207: Lecture 11, Pg 19Example 1A 2 kg block, initially at rest on frictionless horizontal surface, isacted on by a 10 N horizontal force for 2 seconds (in 1D).What is the final velocity? F is to the positive & F = ma thus a = F/m = 5 m/s2 v = v0+ a ∆t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction)Notice: v - v0= a ∆t m (v - v0) = ma ∆t m ∆v = F ∆tIf the mass had been 4 kg …what is the final velocity?Fr-+F (N)1002time (s)Physics 207: Lecture 11, Pg 20Twice the mass Same force Same time Half the


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UW-Madison PHYSICS 207 - Lecture 11

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