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# UT Arlington PHYS 1441 - Lecture Notes

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Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #6 Monday, Sept. 27, 2010 Dr. Jaehoon Yu • Components of the 2D Vector • Understanding the 2 Dimensional Motion • 2D Kinematic Equations of Motion • Projectile Motion • Maximum Range and Height Today’s homework is homework #4, due 10pm, Tuesday, Oc. 5!!Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 3 2D Coordinate Systems • They make it easy and consistent to express locations or positions • Two commonly used systems, depending on convenience, are – Cartesian (Rectangular) Coordinate System • Coordinates are expressed in (x,y) – Polar Coordinate System • Coordinates are expressed in distance from the origin ® and the angle measured from the x-axis, θ (r,θ) • Vectors become a lot easier to express and compute O (0,0) (x1,y1) r1 θ1"How are Cartesian and Polar coordinates related? y1 x1 +x +y tanθ1= x12+ y12( ) r1 r1= (r1,θ1) θ1= tan−1y1x1⎛⎝⎜⎞⎠⎟ cosθ1 sinθ1Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 4 Components and Unit Vectors Coordinate systems are useful in expressing vectors in their components (Ax,Ay) θ"Ay Ax x y A=}Components (+,+) (-,+) (-,-) (+,-) } Magnitude Acosθ( )2+ Asinθ( )2 = A2cos2θ+ sin2θ( ) = A Acosθ Asinθ A= A= Ax2+ Ay2Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 Unit 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 i, j, k A= = Acosθ + AsinθMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 Examples of Vector Operations C= D= C= D=θ= 2.0i+ 2.0 j( )+ 2.0i− 4.0 j( ) = 2.0 + 2.0( )i tan−1CyCx= tan−1−2.04.0= −27 = 15 + 23 − 13( )i d1+ d2+ d3= 15i+ 30 j+ 12k( )+ 23i+ 14 j− 5.0k( )+ −13i+ 15 j( ) 25( )2+ 59( )2+ 7.0( )2= 65(cm) 4.0( )2+ −2.0( )2 = 16 + 4.0 = 20 = 4.5(m) + 2.0 − 4.0( )j = 4.0i −2.0 jm( ) A+ B= + 30 + 14 + 15( )j + 12 − 5.0( )k +59 j +7.0k(cm)MagnitudeMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 7 Displacement: 2D Displacement Δr = r −roMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 8 v =Average velocity is the displacement divided by the elapsed time. r −ro = t − to ΔrΔt2D Average VelocityMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 9 The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time. v = limΔt →0ΔrΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 10 a = vo v Δv2D Average Acceleration v −vo = t − to ΔvΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 11 Displacement, Velocity, and Acceleration in 2-dim • Displacement: Δr=• Average Velocity: v≡• Instantaneous Velocity: v≡• Average Acceleration a≡• Instantaneous Acceleration: a≡How is each of these quantities defined in 1-D? rf− ri ΔrΔt= rf− ritf− ti limΔt →0ΔrΔt ΔvΔt= vf− vitf− ti limΔt →0ΔvΔtMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 12 Kinematic Quantities in 1D and 2D Quantities 1 Dimension 2 Dimension Displacement Average Velocity Inst. Velocity Average Acc. Inst. Acc. Δx = Δr=vx=ΔxΔt=xf− xitf− ti v≡ΔrΔt=rf− ritf− ti v≡ limΔt →0ΔrΔt vx≡ limΔt →0ΔxΔt a≡ΔvΔt=vf− vitf− tiax≡ΔvxΔt=vxf− vxitf− ti a≡ limΔt →0ΔvΔt ax≡ limΔt →0ΔvxΔtWhat is the difference between 1D and 2D quantities? xf− xi rf− riMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 13 This is a motion that could be viewed as two motions combined into one. (superposition…) A Motion in 2 DimensionMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 14 Motion in horizontal direction (x) x =12vxo+ vx( )t x = vxot +12axt2 vx= vxo+ axt vx2= vxo2+ 2axxMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 15 Motion in vertical direction (y) y =12vyo+ vy( )t y = vyot +12ayt2 vy= vyo+ ayt vy2= vyo2+ 2ayyMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 16 Imagine you add the two 1 dimensional motions on the left. It would make up a one 2 dimensional motion on the right. A Motion in 2 DimensionMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 17 x-component x =12vxo+ vx( )t Δx = vxot +12axt2 vx= vxo+ axt vx2= vxo2+ 2axx y =12vyo+ vy( )t Δy = vyot +12ayt2 vy= vyo+ ayt vy2= vyo2+ 2ayyKinematic Equations in 2-Dim y-componentMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 18 In the x direction, the spacecraft in zero-gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s. Ex. A Moving SpacecraftMonday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 19 How do we solve this problem? 1. Visualize the problem  Draw a picture! 2. Decide which directions are to be called positive (+) and negative (-). 3. Write down the values that are given for any of the five kinematic variables associated with each direction. 4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y separately. Select the appropriate equation. 5. When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem.Monday, Sept. 27, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 20 In the x direction, the spacecraft in a zero gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the

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