PHYS 1441 – Section 002 Lecture #11AnnouncementsSlide 3Forces of Friction SummaryExample w/o FrictionExample w/ FrictionDefinition of the Uniform Circular MotionSpeed of a uniform circular motion?Ex. : A Tire-Balancing MachineCentripetal AccelerationSlide 11Newton’s Second Law & Uniform Circular MotionEx. Effect of Radius on Centripetal AccelerationExample of Uniform Circular MotionWednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #11Wednesday, Oct. 13, 2010Dr. Jaehoon Yu•Force of Friction Review•Example for Motion with friction•Uniform Circular Motion•Centripetal Acceleration•Newton’s Law & Uniform Circular Motion •Unbanked and Banked highways•Newton’s Law of Universal GravitationAnnouncements•Reading Assignments–CH5.4, 5.5, 5.9 and 6.2•Quiz next Monday, Oct. 18–Beginning of the class–Covers CH4.5 to CH5.3•Colloquium today–Our own Dr. Q. ZhangWednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2Wednesday, Oct. 13, 2010 3PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Oct. 13, 2010Forces 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 movementNkkf Fm=ur urThese 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 kinetic friction takes it over.Which direction does kinetic friction apply?Opposite to the motion!PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu4Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5Example w/o FrictionA crate of mass M is placed on a frictionless inclined plane of angle θ. a) Determine the acceleration of the crate after it is released. Fur=Free-bodyDiagramθxyMdaFgnnF= -MgθSupposed the crate was released at the top of the incline, and the length of the incline is d. How long does it take for the crate to reach the bottom and what is its speed at the bottom?d =yFxFax=gsinθxy∴ t =2dgsinθxfvxy Furg+nr=amxMagxFθsinMgMay=gyFnn −mgcosθ =0vixt +12axt2=12gsinθ t2vix+axt =gsinθ2dgsinθ=2dgsinθ∴ vxf= 2dgsinθWednesday, Oct. 13, 2010Example 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.F =urFree-bodyDiagramθxyMaFgnnF= -MgθFs=μsnyFxFsNet forcex comp.y comp.sfnxyM a =r Furg+nr+ fursFgx−fs=sfMgθsin0nscMgθsinyMagyFncMgnθcos0gyFcMgθcosnMgcθsinccMgMgθθcossincθtanPHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6FsWednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 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 Motion7Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 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.v =rSpeed of a uniform circular motion? 2π r T distancetime=8Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 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. 1830 revolutions 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 s9Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 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- =10The change of direction of the velocity is the same as the change of the angle in the circular motion!Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2vvD=vtD=Dca =Centripetal Acceleration2v trD2vr2vrFrom the geometryCentripetal AccelerationWhat is the direction of ac?Always toward the center of circle!carcar sinθ 2=11Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuNewton’s Second Law & Uniform Circular Motionca =The centripetal * acceleration is always perpendicular to the velocity vector, v, and points to the center of the rotational axis (radial direction) in a uniform circular motion. The force that causes the centripetal acceleration acts toward the center of the circular path and causes the change in the direction of the velocity vector. This force is called the centripetal force.Are there forces in this motion? If so, what do they do?cF =�What do you think will happen to the ball if the string that holds the ball breaks? The external force no longer exist. Therefore, based on Newton’s 1st law, the ball will continue its motion without changing its velocity and will fly away following the tangential direction to the circle.cma =2vmrr2v*Mirriam Webster: Proceeding or acting in the direction toward the center or rotational axis 12Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuEx. Effect of Radius on Centripetal AccelerationThe bobsled track at the 1994 Olympics in Lillehammer, Norway, contained turns with radii of 33m and 23m. Find the centripetal acceleration at each turn for a speed of 34m/s, a speed that was achieved in the two –man event. Express answers as multiples of g=9.8m/s2.Centripetal acceleration:R=33m2vmr=ra =2vr( )2342333335r ma m s== = =( )2342242448r ma m s== = =R=24m 3.6g4.9g13Wednesday, Oct. 13, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuExample 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? ra =Centripetal acceleration:rFWhen does the string break?when the required centripetal force is greater than the sustainable tension. mv2r=Calculate the tension
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