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UW-Madison PHYSICS 207 - Lecture 11 Notes

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Lecture 11Zero Gravity RideSlide 3Banked CurvesBanked Curves (high speed)Banked Curves, high speedBanked Curves, low speedBanked Curves, constant speedNavigating a hillOrbiting satellites vT = (gr)½Locomotion: how fast can a biped walk?How fast can a biped walk?Slide 15Slide 16Impulse & Linear MomentumForces vs time (and space, Ch. 10)ExampleForce curves are usually a bit different in the real worldExample with Action-ReactionSlide 25Slide 26Applications of Momentum ConservationSlide 28Momentum ConservationSlide 30Exercise 2 Momentum ConservationSlide 32Physics 207: Lecture 11, Pg 1Lecture 11Goals:Goals:Assignment:Assignment:Read through Chapter 10, 1st four sectionMP HW6, due Wednesday 3/3Chapter 8: Employ rotational motion models with friction or in free fall Chapter 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 the Free 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 RidePhysics 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?Physics 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 bankmarxx y y NmgFfPhysics 207: Lecture 11, Pg 7Banked Curves, high speed4 Apply Newton’s 1st and 2nd Lawsmarxx y y Nmg cos Ffmg sin mar cos mar sin  Fx = -mar cos  = - Ff - mg sin  Fy = mar sin  = 0 - mg cos + NFriction model  Ff ≤  N (maximum speed when equal)Physics 207: Lecture 11, Pg 8Banked Curves, low speed4 Apply Newton’s 1st and 2nd Lawsmarxx y y Nmg cos Ffmg sin mar cos mar sin  Fx = -mar cos  = + Ff - mg sin  Fy = mar sin  = 0 - mg cos + NFriction model  Ff ≤  N (minimum speed when equal but not less than zero!)Physics 207: Lecture 11, Pg 9Banked Curves, constant speed vmax = (gr)½ [ ( + tan ) / (1 -  tan )] ½ vmin = (gr)½ [(tan  - ) / (1 +  tan )] ½ Dry pavement“Typical” values of r = 30 m, g = 9.8 m/s2,  = 0.8, = 20°vmax = 20 m/s (45 mph)vmin = 0 m/s (as long as  > 0.36 )Wet Ice“Typical values” of r = 30 m, g = 9.8 m/s2,  = 0.1, =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 11Navigating a hillKnight concept exercise: A car is rolling over the top of a hill at 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)½Physics 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 matterPhysics 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, sidewaysPhysics 207: Lecture 11, Pg 17Impulse & Linear MomentumTransition from forces to conservation lawsNewton’s Laws  Conservation LawsConservation Laws  Newton’s Laws They are different faces of the same physics NOTE: 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 3rd Law and emphasize changes with time then this leads to the Conservation of Momentum PrinciplePhysics 207: Lecture 11, Pg 22ExampleThere many situations in which the sum of the product “mass times velocity” is constant over time To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s)A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” F (N)1002Time (sec)Area under curve : “impulse”With: m v = Favg tPhysics 207: Lecture 11, Pg 23Force curves are usually a bit different in the real worldPhysics 207: Lecture 11, Pg 24Example with Action-ReactionNow the 10 N force from before is applied by person A on person B while standing on a frictionless surfaceFor the force of A on B there is an equal and opposite force of B on AF (N)1002Time (sec)0-10A on BB on A MA x VA = Area of top curve MB x VB = Area of bottom curve Area (top) + Area (bottom) = 0Physics 207: Lecture 11, Pg 25Example with Action-ReactionMA VA + MB VB = 0MA [VA(final) - VA(initial)] + MB


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UW-Madison PHYSICS 207 - Lecture 11 Notes

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