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UT Arlington PHYS 1441 - Lecture Notes

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Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #3Wednesday, June 9, 2004Dr. Jaehoon YuToday’s homework is HW #2, due 6pm, next Wednesday, June 15!!• Chapter two: Motion in one dimension– Free Fall • Chapter three: Motion in two dimension– Coordinate systems– Vector and Scalar– Two dimensional equation of motion– Projectile motion• Chapter four: Newton’s Laws of MotionWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu2Announcements• E-mail distribution list: 25 of you have registered– Remember 3 extra credit points if done by midnight tonight– Next Wednesday is the last day of e-mail registration– -5 extra points if you don’t register by next Wednesday– A test message will be sent out this Friday• Homework: You are supposed to download the homework assignment, solve it offline and input the answers back online.– 44 registered– 40 submittedWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu3Kinematic Equations of Motion on a Straight Line Under Constant Acceleration()tavtv xxixf +=221tatvxxxxiif += +()11 22fxixfxixx vt vvt+−= =()ifxf xxavv xxi−+= 222Velocity as a function of timeDisplacement as a function of velocities and timeDisplacement as a function of time, velocity, and accelerationVelocity as a function of Displacement and accelerationYou may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu4Falling Motion• Falling motion is a motion under the influence of gravitational pull (gravity) only; Which direction is a freely falling object moving?– A motion under constant acceleration – All kinematic formula we learned can be used to solve for falling motions. • Gravitational acceleration is inversely proportional to the distance between the object and the center of the earth• The gravitational acceleration is g=9.80m/s2on the surface of the earth, most of the time.•The direction of gravitational acceleration is ALWAYS toward the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”• Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu5Example for Using 1D KinematicEquations on a Falling object Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building,g=-9.80m/s2fv =fy =(a) Find the time the stone reaches at the maximum height.What is the acceleration in this motion?What happens at the maximum height? The stone stops; V=0t=(b) Find the maximum height.yi yvat+ =20.0 9.80t+− =0.00 /msSolve for t20.09.80=2.04s2150.0 20 2.04 ( 9.80) (2.04)2+× +×− ×50.0 20.4 70.4( )m=+=212iyi yyvt at++ =Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu6Example of a Falling Object cnt’dt =yfv =fy =Positionyfv =Velocity(c) Find the time the stone reaches back to its original height.(d) Find the velocity of the stone when it reaches its original height.(e) Find the velocity and position of the stone at t=5.00s.2.04 2×= 4.08syi yvat+=20.0 ( 9.80) 4.08+−×=20.0( / )ms−yi yvat+=20.0 ( 9.80) 5.00 29.0( / )ms+−×=−2150.0 20.0 5.00 ( 9.80) (5.00)2=+×+×−×212iyiyyvt at++27.5( )m=+Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu7Coordinate Systems• Makes it easy to express locations or positions• Two commonly used systems, depending on convenience– Cartesian (Rectangular) Coordinate System• Coordinates are expressed in (x,y)– Polar Coordinate System • Coordinates are expressed in (r,θ)• Vectors become a lot easier to express and computeO (0,0)(x1,y1)=(r,θ)rθ1x=How are Cartesian and Polar coordinates related?r=y1x1+x+y1y=cosrθsinrθtanθ=()2211xy+11yxWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu8Example Cartesian Coordinate of a point in the x-y plane are (x,y)= (-3.50,-2.50)m. Find the polar coordinates of this point.yx(-3.50,-2.50)mrθθsr=sθθ+=180tan sθ=sθ=θ∴=()()()223.50 2.50=− +−18.5 4.30( )m==()22xy+2.50 53.50 7−=−15tan7−⎛⎞=⎜⎟⎝⎠35.5D180sθ+=180 35.5 216+=DDDWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu9Vector and ScalarVector quantities have both magnitude (size) and directionScalar quantities have magnitude onlyCan be completely specified with a value and its unitForce, gravitational acceleration, momentumNormally denoted in BOLD BOLD letters, FF, or a letter with arrow on topFTheir sizes or magnitudes are denoted with normal letters, F, or absolute values:For FEnergy, heat, mass, weightNormally denoted in normal letters, EBoth have units!!!Wednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu10Properties of Vectors• Two vectors are the same if their and the are the same, no matter where they are on a coordinate system.xyAABBEEDDCCFFWhich ones are the same vectors?A=B=E=DA=B=E=DWhy aren’t the others?C:C: The same magnitude but opposite direction: C=C=--A:A:A negative vectorF:F: The same direction but different magnitude sizes directionsWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu11Vector Operations• Addition: – Triangular Method: One can add vectors by connecting the head of one vector to the tail of the other (head-to-tail)– Parallelogram method: Connect the tails of the two vectors and extend– Addition is commutative: Changing order of operation does not affect the results A+B=B+AA+B=B+A, A+B+C+D+E=E+C+A+B+DA+B+C+D+E=E+C+A+B+DAABBAABB=AABBA+BA+B• Subtraction: – The same as adding a negative vector:A A -B = A B = A + (-BB)AA--BBSince subtraction is the equivalent to adding a negative vector, subtraction is also commutative!!!• Multiplication by a scalar is increasing the magnitude A, BA, B=2A A AAB=2AB=2AAB 2=A+BA+BA+BA+BAA--BBORWednesday, June 9, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu12Example of Vector AdditionA car travels 20.0km due north followed by 35.0km in a direction 60.0owest of north. Find the magnitude and direction of resultant displacement.NE60οθ1r()()22sincosθθBBAr ++=20AABB1sin 601tancos60BABθ−=+JGJG JGFind other ways to solve this problem…()θθθcos2sincos2222ABBA +++=θcos222ABBA ++=()()60cos0.350.2020.350.2022××++=)(2.482325 km==60cos0.350.2060sin0.35tan1+=−N W wrt


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UT Arlington PHYS 1441 - Lecture Notes

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