19 September 2008

Kinematics With Vector Analysis

Modul Fisika Bab: Kinematics With Vector Analysis, yang dibuat oleh Supatmi, S.Pd kemudian diedit kembali oleh MMA Arientatmi, S.Pd dibantu oleh Kautsar Saimima.

MODULE PHYSICS XI CHAPTER 1 Competence Standard Basic Competence : : 1. Analyze the natural indication and regularity in scope of dot object mechanic. 1.1. Analyze the rectilinear motion, rotation motion and parabolic motion with vector Subjects : analysis. Characteristic rectilinear motion with vector analysis. Parabolic motion. Rotation motion with constant angle acceleration. KINEMATICS WITH VECTOR ANALYSIS A particle called motion if the particle position to reference point always change. Example of motion equation : x = xo + vt y = yo ± vot ± ½ gt vector or unit vector. A. RECTALINEAR MOTION WITH VECTOR ANALYSIS 1. The Vector on a plane ( 2 dimension ) The vector A = OA The vector A can be written A = Axi + Ayj I and j are unit vector on x axis and y axis with magnitude = 1. 2 motion axis x with constant velocity. motion axis y in gravitation field. Motion equation of 1 dimension While for motion in 2 or 3 dimension, the motion equation can be written in form The magnitude of A vector can be written: A= Ax + Ay 2 2 KINEMATICS MMA. ARIENTATMI, S.PD -1- MODULE PHYSICS XI The direction of A vector can be written: Tan Θ = Ay Ax Ay = A sin Θ Ax = A cos Θ Ay A sin θ = = tan θ Az A cos θ 2. The Vector on 3 dimension The A vector has component vector Axi, Ayj and Azk. The A vector can be written: A = Axi + Ayj + Azk i, j and k are unit vector on x axis, y axis and z axis, with magnitude = 1. The magnitude of A vector can be written: A= Ax + Ay + Az 2 2 2 3. Differential method and Integral method. Differential method: f=(x) = axn f'(x) = anxn-1 Integral method: ∫x n dx = 1 n +1 x + c; n ≠ −1 n +1 4. Displacement. A particle that move as far as a curve in room. Position of particle as position vector from reference point O and called position vector r. If a particle is at point ( x, y ), so position vector of the point can be represented as: r =xi + yj Study the following figure! KINEMATICS MMA. ARIENTATMI, S.PD -2- MODULE PHYSICS XI At the time t = t1, the particle is at point P1. r1 is position vector that drawed for refence point O to P1. At the time t = t2, particle is at point P2 and has position vector r2. If the coordinate P1( x1, y1) and P2 (x2, y2), so position vector r1 and r2 can be written as follow: r1 = x1i + y1j r2 = x2i + y2j So displacement vector is change of position vector. Δr = = = Δr = r 2 – r1 (x2i + y2j) – (x1i + y1j) (x2 – x1)i + (y2 - y1)j Δxi + Δyj And then displacement magnitude can be written Δr = ∆x 2 + ∆y 2 Direction of displacement can be written Tan Θ = Sample problem 1.1 Suppose the position vector of a particle is given by r = (t3 + t)i + (2t2) j; t is in second and r in meter. Determine the magnitude and direction of the particle displacement from t1 = 1 s to t2 = 2 s. Solution r = (t3 + t)i + ( 2t2 )j for t1 r1 = = = For t2 r2 KINEMATICS Δy / Δx 1 s, then ( 13 + 1 )i + ( 2 x 12 )j 2i + 2j 2 s, then ( 23 + 2 ) i + ( 2 x 22 ) j = = MMA. ARIENTATMI, S.PD -3- MODULE PHYSICS XI = 10i + 8j So the displacement vector Δr Δr = = = = r 2 – r1 (10i + 8j) – (2i + 2j) (10 – 2)i + (8 – 2)j 8i + 6j The magnitude of displacement Δr = = = ∆x 2 + ∆y 2 82 + 6 2 m 10 m Direction of displacement Tan Θ = = = Θ = = 5. Velocity a. Average Velocity Average velocity ( v average Δy / Δx 6/8 3/4 tan-1 (3/4) 36,87˚ ) at the time interval Δt have defined as product of displacement at each time interval. Figure 1.2 direction determining of average velocity Average velocity on a plane is mathematically can be determined as follows vaverage = = Explanation: KINEMATICS ∆r ∆t r2 − r1 t 2 − t1 MMA. ARIENTATMI, S.PD -4- MODULE PHYSICS XI v Δr r1 r2 = = = = average velocity position vector of particle position vector at t1 position vector at t2 By substituting equation Δr = Δxi + Δyj to equation vaverage = vaverage = vaverage = if vx ∆r so we obtain the following relation: ∆t ∆xi + ∆yj ∆t ∆x ∆y i+ j ∆t ∆t = ∆x and vy ∆t = ∆y , so ∆t vaverage = vxi + vyj The magnitude of average velocity: vaverage = vx + v y vy vx 2 2 Direction of average velocity: Tan Θ = b. Instantaneous velocity Instantaneous velocity is a vector quantity that expresses the velocity of an object at certain time. Mathematically, Instantaneous velocity is expressed by the following equation: v = ∆t → 0 lim vaverage = = ∆t → 0 lim ∆r ∆t  ∆x ∆y  lim  i + j ∆t →0 ∆t ∆t   v = dx dy i+ j dt dt dx dt dy , dt if vx = and vy = so the equation of Instantaneous velocity can be written: vins = vxi + vyj The magnitude of Instantaneous velocity: KINEMATICS MMA. ARIENTATMI, S.PD -5- MODULE PHYSICS XI vins = vx + v y 2 2 Direction of Instantaneous velocity: Tan Θ = vy vx 6. Determining Position from velocity function In x direction : vx = dx dt xo ∫ dx x = ∫ vx dt 0 1 x - xo = ∫ vx dt 0 1 x = xo + ∫ vx dt 0 1 In y direction : vy = dy dt yo ∫ dy y = ∫ vy dt 0 1 y - yo = ∫ vy dt 0 1 y = yo + ∫ vy dt 0 1 Explanation: (xo, yo): the particle initial position coordinate Thus, the particle position vector on a plane can be determined by substituting the value of x and y in the equation r = xi + yj, so we obtain the following equation r = ( xo + ∫ vx 0 1 dt )i + ( yo + ∫ vy 0 1 dt )j Sample problem 1.2 A particle moves on an x-y plane. The particle initial position is at coordinate ( 2, 4 ) m. The components of the particle velocity complay the equation vx = 5t and vy = 4 + 3t2, vx and vy are in m/s and t is in s. KINEMATICS MMA. ARIENTATMI, S.PD -6- MODULE PHYSICS XI Determine: a. The equation of particle position vector! b. The particle position at t = 3 second! Solution a. The particle initial position is ( 2, 4 ) m, so xo = 2m and yo = 4m Therefore we obtain x = xo + 1 ∫ vx 0 1 dt = 2+ ∫ 5t 0 dt = = y = 2 2 + 5  t m 1 2   ( 2 + 2,5 t2 ) m yo + 1 ∫ vy dt 0 1 = 4+ ∫ 4 + 3t 0 2 dt = = r r = = 4 + 4t + 3 13 t 3 ( 4 + 4t + t3 ) m xi + yj { ( 2 + 2,5 t2 )i + (4 + 4t + t3 )j } meter The particle position vector is b. The particle position at t = 3 seconds is x = = = y = = = follows: r = ( 24,5 i + 43 j ) meter. ( 2 + 2,5 t2 ) m 2 + 2,3 (3)2 m 24,5 m ( 4 + 4t + t3 ) m 4 + 4 (3) + (3)3 m 43 m Thus at t = 3 seconds the particle position vector can be written as 7. Acceleration a. Average acceleration KINEMATICS MMA. ARIENTATMI, S.PD -7- MODULE PHYSICS XI ∆v ∆t aaverage = if v = vxi + vyj then the equation of average acceleration can be written as follow: aaverage = so aaverage ∆vx ∆vy i+ j ∆t ∆t = axi + ayj The magnitude of average acceleration is: aaverage = ax + ay 2 2 The direction of average acceleration is Tan θ = ay ax b. Instantaneous acceleration The instantaneous acceleration is the average acceleration for time interval Δt approaching lim aaverage lim ∆v ∆t zero. Mathematically, instantaneous acceleration can be written: ains = = = = = ains = ∆t→0 ∆t→0 dv dt d (v = vxi+ vyj ) dt dvy dvx i+ j dt dt axi + ayj The magnitude of instantaneous acceleration is: ains = ax + ay 2 2 The direction of instantaneous acceleration is Tan θ = ay ax 8. Determining velocity from acceleration If acceleration a= dv then velocity is determined as follows dt KINEMATICS MMA. ARIENTATMI, S.PD -8- MODULE PHYSICS XI dv → dv = a dt dt a= Then integrate both segments, so we obtain: vo ∫ v dv = ∫ a dt 0 1 1 v - vo = ∫ a dt 0 v = vo + ∫ a dt 0 1 In mathematics, the value of integral ∫ a dt is the same as the region 0 1 area below at curve ( acceleration against time ) where lower limit t = 0 and upper limit t = t. Sample problem 1.3 An object moves from a rest where the acceleration vector is expressed by a = ( 6t – 4 )i + 6j, a is m/s2 and t is in s. Determine the object velocity at t = 4 second. Solution To calculate the object velocity, firstly we integrate the equation a = ( 6t – 4 )i + 6j v = vo + 1 ∫ a dt 0 v=0+ ∫ [ ( 6t − 4) i + 6j] dt 0 1 6 v = ( t2 − t)i + 6tj 2 The object velocity vector at t = 4s is 6 v = ( (4)2 − 4(4)i + 6(4))i + 6 (4)j 2 KINEMATICS MMA. ARIENTATMI, S.PD -9- MODULE PHYSICS XI = 32 i + 24 j the object velocity at t = 4s is v= 322 + 242 = 40 m/s B. PARABOLIC MOTION Figure 1.4 parabolic motion of an object which is thrown upward at initial velocity vo and forms an angle of θo to earth surface. The parabolic motion on x-axis is uniform rectilinear motion while motion on yaxis is accelerated uniform rectilinear motion. Therefore from the figure above we can obtain the following equation. 1. Position Equation and Velocity on Parabolic Motion Initial velocity component on x-axis and y-axis vox voy vx = = = vo cos θo vo sin θo vox → x= = motion) v vy = = vo + at → vo - gt → x y y The object velocity at t The object at P ( look at the picture 1.4 ), the object velocity is v (v x and vy ) = = = = vot + ½ at2 ( a = g ) voy . t – ½ gt2 vo cos θo . t vo sin θo . t – ½ gt2 vox . t vo cos θo . t Position and velocity on x-axis (glb = uniform rectilinear motion) Position and velocity on y-axis (glbb = accelerated uniform rectilinear KINEMATICS MMA. ARIENTATMI, S.PD - 10 - MODULE PHYSICS XI The magnitude of velocity v = vx + v y vy vx 2 2 The direction of velocity Tan θ = 2. Determine of Maximum Height The component of object velocity at the highest point ( point H ) can be written as follows vy = 0 vx = vox = vo cos Θo since at the highest point H, vy = 0, so the volocuty at the highest point v H is vH 0 g tH tH explanation tH = time to reach the highest point ( s ) = voy – gt = vo sin Θo – d tH = = vo sin Θo v o sin θ g To reach the farthest point ( point A ) look at the picture 1.4, requires twice the time to reach the highest point. tA tA = = 2tH 2 vo sin θ o g Determine of the highest point coordinate H For horizontal coordinate ( x ) xH xH = vox tH = (vo cos Θo)   2  vo sin θ o g      xH = vo ( 2 sin θ o cosθ o ) 2g 2 xH v = o sin 2θ o 2g For vertical coordinate ( y ) yH yH = voy tH – ½ gtH2 2  v sin θ o = ( vo sin θ o )  o  g   1  vo sin θ o  − g 2 g       KINEMATICS MMA. ARIENTATMI, S.PD - 11 - MODULE PHYSICS XI 2vo sin 2 θ o vo sin 2 θ o = − 2g 2g 2 2 yH yH v = o sin 2 θ o 2g 2 Thus, the highest poiny coordinate H is 2  vo 2  vo 2  H ( xH, yH ) → H =   2 g sin 2θ o , 2 g sin θ o    3. Determine of Fartest Point ( Point A ) x xA xA = vox t and y = voy t – ½ gt2 = vox tA = ( vo cos Θo )    2vo sin θ o g      xA  vo 2  1  =  g 2 sin θ o cosθ o → sin θ o cosθ o = 2 sin 2θ o    vo 2   =  g  sin 2θ o   = voy tA – ½ gtA2 xA yA yA  2v sin θ o = ( vo sin Θo )  o  g   2vo 2 sin 2 θ o =  g  =0  1  2vo sin θ o − g  2 g           2 yA   2vo 2 sin 2 θ o −  g  yA Therefore, the farthest point coordinate is  vo 2 A ( xA, yA ) → A (  g  Sample Problem 1.4   sin 2θ o , 0)   A bullet is shot at initial velocity 40 m/s. And elevation angle 60˚ from a flat ground, air friction is neglected and gravitational acceleration g = 10 m/s2. Determine: a. The time needed by the bullet to reach the highest point H. b. The maximum height reached by the bullet. KINEMATICS MMA. ARIENTATMI, S.PD - 12 - MODULE PHYSICS XI c. The farthest distance reached by the bullet. Solution: vo = 40 m/s Θo = 60˚ g = 10 m/s2 a. tH = 1 v o sin θ 40 × 3 o = 40 sin 60 s. 2 = =2 3 g 10 10 b. yH 3 2  40 2  2 o (1600 )   vo  2 = sin θ o =   4  = 60 m  2 × 10  sin 60 =  2g   2  vo 2   2   sin 2θ o =  40  sin 2 60o = 160 x sin 120o = 160 x ½ =  10     g = 80 c. x 3 3m C. CIRCULAR MOTION 1. The Particle Position in Polar Coordinate The position of P point with polar coordinate ( r, Θ ) Explanation: r : radius ( constant ) Θ : angular position Figure 1.5 The rotation of an object against contant axis that trough O an perpendicular of plane. The relation between xy-coordinate with polar coordinate can be written: x y r = = = r cos Θ r sin Θ x2 + y2 y x tan Θ = KINEMATICS MMA. ARIENTATMI, S.PD - 13 - MODULE PHYSICS XI when a particle moves all along the orbit from x-axis to P point, the angular position is Θ ( rad ) Θ Θ = = s r 2πr rad r = 2 π rad So 1 circle = 360o = 2 π rad 1 rad = 180 o = 57,3 o π 2. Angular Velocity The average angular velocity = ω average = The displacement of angular position time interval ∆θ θ 2 − θ1 = ∆t t 2 − t1 The instantaneous angular velocity = average angular velocity in time interval approaching zero. ω ins = ∆t →0 lim ∆θ dθ = ∆t dt 3. Angular Acceleration ( α ) Average angular acceleration = α average = change of angular velocity time interval ∆ω ω 2 − ω1 = ∆t t 2 − t1 The instantaneous angular acceleration = average angular acceleration in time interval approaching zero. α ins = lim ∆ω dω = ∆t →0 ∆t dt 4. Centripetal Acceleration Tangential acceleration: at = dv dω =r = rα dt dt Centripetal acceleration: as = v2 = ω 2r r KINEMATICS MMA. ARIENTATMI, S.PD - 14 - MODULE PHYSICS XI Total linear acceleration: a= at 2 + as 2 = r 2α 2 + r 2ω 4 = r α 2 + ω 4 KINEMATICS MMA. ARIENTATMI, S.PD - 15 -

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