PHYS 1443 – Section 501Lecture #9AnnouncementsForces of FrictionExample w/ FrictionNewton’s Second Law & Uniform Circular MotionExample of Uniform Circular MotionExample of Banked HighwayPHYS 1443 – Section 501Lecture #9Wednesday, February 18, 2004Dr. Andrew Brandt•Forces of Friction•Uniform and Non-uniform Circular MotionsWednesday, Feb. 18, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt1Announcements• HW #3 on Ch. 4 due tonight at midnight• Beware negative points• HW#4 on Ch. 5 due Mon Feb. 23• If you are struggling, read text before+after class, start HW early, ask questions, physics clinic--RM 10• Remember the 1stmidterm, Weds., Feb. 25,– Covers chapters 1-5• Short Review on Monday, start ch. 6 unless questionsWednesday, Feb. 18, 2004 2PHYS 1443-501, Spring 2004Dr. Andrew BrandtForces of FrictionResistive force exerted on a moving object due to viscosity or other types of frictional properties of the medium in or surface on which the object moves.These forces are either proportional to velocity or normal forceThe resistive force exerted on the object until just before the beginning of its movementForce of static friction, fs:Empirical Formula What does this formula tell you? Frictional force is variable—as you push/pull harder it increases until it reaches the static limit!!nfssµ≤Force of kinetic friction, fkBeyond the limit, there is no more static frictional force but kinetic frictional force takes it over.The resistive force exerted on the object during its movementnfkkµ=Wednesday, Feb. 18, 2004 3PHYS 1443-501, Spring 2004Dr. Andrew BrandtexampleExample w/ FrictionSuppose a block is placed on a rough surface inclined relative to the horizontal. The inclination angle is increased till the block starts to move. Show that by measuring this critical angle, θc, one can determine coefficient of static friction, µs.Free-bodyDiagramθxyMaFgnnF= -Mgθfs=µknxyNet force =aMsgfnF ++=Fx comp.=−sgxfF=−sfMgθsin0=sf=ns cMgθsin=xFµy comp.=yMa=−gyFn=−cMgnθcos0=n=gyFcMgθcos=yF=nMgcθsin=ccMgMgθθcossincθtan=sµquiz&exampleWednesday, Feb. 18, 2004 4PHYS 1443-501, Spring 2004Dr. Andrew Brandtwhat if 0cθ=Newton’s Second Law & Uniform Circular MotionFrFrrmrvar2=The centripetal acceleration is always perpendicular to the velocity vector, v, for uniform circular motion.Are there forces in this motion? If so, what do they do?The force that causes centripetal acceleration acts toward the center of the circle and causes a change in the direction of the velocity vector. This force is called centripetal force.=rmarvm2∑=rFWhat do you think will happen to the ball if the string that holds the ball breaks? Why?Based on Newton’s 1stlaw, since the external force no longer exist, the ball will continue its motion without change and will fly away following a tangential direction to the circle.Wednesday, Feb. 18, 2004 5PHYS 1443-501, Spring 2004Dr. Andrew BrandtExample of Uniform Circular MotionA ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? mFrrvar2=Centripetal acceleration:∑=rFWhen does the string break?When the centripetal force is greater than the sustainable tension.=rmarvm2T>Trvm =2()smmTrv /2.12500.05.10.50=×==Calculate the tension of the cord when speed of the ball is 5.00m/s.()()NrvmT 33.85.100.5500.022=×==Wednesday, Feb. 18, 2004 6PHYS 1443-501, Spring 2004Dr. Andrew BrandtExample of Banked HighwayWednesday, Feb. 18, 2004 7PHYS 1443-501, Spring 2004Dr. Andrew Brandt(a) For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required to keep the car from skidding. rmvn2sin =θ=∑yF=∑xFx comp.y comp.xy=−rmanθsincosmgnθ=0=−mgnθcosrmvmgmgn2tancossinsin ===θθθθgrv2tan =θ=−rmvn2sinθ0mgn=θcos(b) What is this angle for an expressway off-ramp curve of radius 50m at a design speed of 50km/h? ()4.08.95014tan2=×=θ()o224.0tan1==−θsmhrkmv /14/50
View Full Document