Slide 1AnnouncementsReminder: Special Project for Extra CreditSpecial Project for Extra CreditComponents and Unit VectorsUnit VectorsForces of Friction SummarySlide 8Example 4.16 w/ FrictionDefinition of the Uniform Circular MotionSpeed of a uniform circular motion?Ex. : A Tire-Balancing MachineCentripetal AccelerationCentripetal AccelerationNewton’s Second Law & Uniform Circular MotionEx. 5.11 of Uniform Circular MotionExample 5.15: Banked HighwayForces in Non-uniform Circular MotionEx. 5.12 for Non-Uniform Circular MotionWednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #7Wednesday, June 15, 2011Dr. Jaehoon Yu•Force of friction•Uniform Circular Motion•Motion Under Resistive ForcesWednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu2Announcements•Mid-term exam–In the class on Tuesday, June 21, 2011–Covers: CH 1.1 – what we finish Monday, June 20 plus Appendices A and B–Mixture of free response problems and multiple choice problems•Reading assignments–CH5.5 and 5.6Wednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu3Reminder: Special Project for Extra CreditA large man and a small boy stand facing each other on frictionless ice. They put their hands together and push against each other so that they move apart. a) Who moves away with the higher speed, by how much and why? b) Who moves farther in the same elapsed time, by how much and why?•Derive the formulae for the two problems above in much more detail and explain your logic in a greater detail than what is in this lecture note.•Be sure to clearly define each variables used in your derivation.•Each problem is 10 points.•Due is Monday, June 20.Wednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu4Special Project for Extra CreditA 92kg astronaut tied to an 11000kg space craft with a 100m bungee cord pushes the space craft with a force P=36N in space. Assuming there is no loss of energy at the end of the cord, and the cord does not stretch beyond its original length, the astronaut and the space craft get pulled back to each other by the cord toward a head-on collision. Answer the following questions.•What are the speeds of the astronaut and the space craft just before they collide? (10 points)•What are the magnitudes of the accelerations of the astronaut and the space craft if they come to a full stop in 0.5m from the point of initial contact? (10 points)•What are the magnitudes of the forces exerting on the astronaut and the space craft when they come to a full stop? 6 points)•Due is Wednesday, June 22.Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu5Components and Unit VectorsCoordinate systems are useful in expressing vectors in their components Aur= Ax2+Ay2(Ax,Ay)AθAyAxxyxA = Aur=}Components(+,+)(-,+)(-,-) (+,-)yA =} Magnitude Aurcosθ( )2+ Aursinθ( )2 = Aur2cos2θ +sin2θ( ) = Aur Aurcosθ AursinθWednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu6Unit Vectors•Unit vectors are the ones that tells us the directions of the components•Dimensionless •Magnitudes these vectors are exactly 1•Unit vectors are usually expressed in i, j, k or ir, jr, kr Aur=So a vector A can be expressed as yA = Aurcosθ + Aursinθ xA +irirjrjrWednesday, June 15, 2011Forces of Friction Summary furs≤μsFurNResistive force exerted on a moving object due to viscosity or other types frictional property of the medium in or surface on which the object moves.Force of static friction, fs:Force of kinetic friction, fkThe resistive force exerted on the object until just before the beginning of its movementThe resistive force exerted on the object during its movement furk=μkFurNThese forces are either proportional to the velocity or the normal force.Empirical Formula What does this formula tell you? Frictional force increases till it reaches the limit!!Beyond the limit, the object moves, and there is NO MORE static friction but the kinetic friction takes it over.Which direction does kinetic friction apply?Opposite to the motion!PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu7Wednesday, June 15, 20118Look at this problem again…MSuppose you are pulling a box on a rough surfice, using a rope.TWhat are the forces being exerted on the box?Gravitational force: FgNormal force: nTension force: Tn= -FgTFree-body diagramFg=MgNet force: F=Fg+n+T+FfxFIf T is a constant force, ax, is constantxfvyFax=T −FfM0yaxn= -FgFg=MgTT −Ff=Max−Fg+ n =yMa0vxi+axt =vxi+T −FfM⎛⎝⎜⎞⎠⎟txf−xi=vxit +12T −FfM⎛⎝⎜⎞⎠⎟t2PHYS 1443-001, Spring 2011 Dr. Jaehoon YuFriction force: FfFfFfn =FgWednesday, June 15, 20119Example 4.16 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. rF =Free-bodyDiagramθxyMaFgnnF= -Mgθfs=μsnyFxFμs=Net forcex comp.y comp.fs=nxy Mra = rFg+rn+rfsFgx−fs=Mgsinθ −fs=0μsn =cMgsinyMagyFncMgncos0gyFcMgcosnMgcsinccMgMgcossinctanPHYS 1443-001, Spring 2011 Dr. Jaehoon Yufs=μsnWednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon YuUniform circular motion is the motion of an object traveling at a constant speed on a circular path.Definition of the Uniform Circular Motion10Is there an acceleration in this motion?Yes, you are absolutely right! There is an acceleration!!Wednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon YuLet T be the period of this motion, the time it takes for the object to travel once around the complete circle whose radius is r isv =rSpeed of a uniform circular motion? 2π r T distancetime=11Wednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon YuThe wheel of a car has a radius of 0.29m and is being rotated at 830 revolutions per minute on a tire-balancing machine. Determine the speed at which the outer edge of the wheel is moving. 1830revolutions min=T =v =Ex. : A Tire-Balancing Machine 1.2 ×10−3min revolution 1.2 ×10−3 min= 0.072 s 2πrT= 2π 0.29 m( )0.072 s= 25m s12Wednesday, June 15, 2011PHYS 1443-001, Spring 2011 Dr. Jaehoon YuIn uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant.aab q=Centripetal Accelerationb+ =90oq+ =90o0b q- =13The
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