Tenth Edition
Tenth EditionCHAPTER11VECTOR MECHANICS FOR ENGINEERS:DYNAMICSFerdinand P. BeerE. Russell Johnston, Jr.Phillip J. CornwellLecture Notes:Brian P. SelfKinematics of ParticlesCalifornia Polytechnic State University 2013 The McGraw-Hill Companies, Inc. All rights reserved.
TenthEditionVector Mechanics for Engineers: DynamicsContentsIntroductionRectilinear Motion: Position,Velocity & AccelerationDetermination of the Motion of aParticleSample Problem 11.2Sample Problem 11.3Uniform Rectilinear-MotionUniformly Accelerated RectilinearMotionMotion of Several Particles:Relative MotionSample Problem 11.4Motion of Several Particles:Dependent MotionSample Problem 11.5Graphical Solution of RectilinearMotion ProblemsOther Graphical MethodsCurvilinear Motion: Position, Velocity& AccelerationDerivatives of Vector FunctionsRectangular Components of Velocityand AccelerationMotion Relative to a Frame inTranslationTangential and Normal ComponentsRadial and Transverse ComponentsSample Problem 11.10Sample Problem 11.12 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 2
TenthEditionVector Mechanics for Engineers: DynamicsKinematic relationships are used tohelp us determine the trajectory of agolf ball, the orbital speed of asatellite, and the accelerationsduring acrobatic flying. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 3
TenthEditionVector Mechanics for Engineers: DynamicsIntroduction Dynamics includes:Kinematics: study of the geometry of motion.Relates displacement, velocity, acceleration, and time without referenceto the cause of motion.FthrustFdragFliftKinetics: study of the relations existing between the forces acting ona body, the mass of the body, and the motion of the body. Kinetics isused to predict the motion caused by given forces or to determine theforces required to produce a given motion. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 4
TenthEditionVector Mechanics for Engineers: DynamicsIntroduction Particle kinetics includes: Rectilinear motion: position, velocity, and acceleration of aparticle as it moves along a straight line. Curvilinear motion: position, velocity, and acceleration of aparticle as it moves along a curved line in two or threedimensions. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 5
TenthEditionVector Mechanics for Engineers: DynamicsRectilinear Motion: Position, Velocity & Acceleration Rectilinear motion: particle movingalong a straight line Position coordinate: defined bypositive or negative distance from afixed origin on the line. The motion of a particle is known ifthe position coordinate for particle isknown for every value of time t. May be expressed in the form of afunction, e.g.,23x 6t tor in the form of a graph x vs. t. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 6
TenthEditionVector Mechanics for Engineers: DynamicsRectilinear Motion: Position, Velocity & Acceleration Consider particle which occupies position Pat time t and P’ at t Dt,Dx Average velocityDtDx v limInstantaneous velocityDt 0 Dt Instantaneous velocity may be positive ornegative. Magnitude of velocity is referredto as particle speed. From the definition of a derivative,Dx dxv lim dtDt 0 D te.g., x 6t 2 t 3v 2013 The McGraw-Hill Companies, Inc. All rights reserved.dx 12t 3t 2dt11 - 7
TenthEditionVector Mechanics for Engineers: DynamicsRectilinear Motion: Position, Velocity & Acceleration Consider particle with velocity v at time t andv’ at t Dt,DvInstantaneous acceleration a limDt 0 D t Instantaneous acceleration may be:- positive: increasing positive velocityor decreasing negative velocity- negative: decreasing positive velocityor increasing negative velocity. From the definition of a derivative,Dv dv d 2 xa lim 2dt dtDt 0 Dte.g. v 12t 3t 2dva 12 6tdt 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 8
TenthEditionVector Mechanics for Engineers: DynamicsConcept QuizWhat is true about the kinematics of a particle?a) The velocity of a particle is always positiveb) The velocity of a particle is equal to the slope ofthe position-time graphc) If the position of a particle is zero, then thevelocity must zerod) If the velocity of a particle is zero, then itsacceleration must be zero 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 9
TenthEditionVector Mechanics for Engineers: DynamicsRectilinear Motion: Position, Velocity & Acceleration From our example,x 6t 2 t 3v dx 12t 3t 2dtdv d 2 xa 12 6tdt dt 2 What are x, v, and a at t 2 s ?- at t 2 s, x 16 m, v vmax 12 m/s, a 0 Note that vmax occurs when a 0, and that theslope of the velocity curve is zero at this point. What are x, v, and a at t 4 s ?- at t 4 s, x xmax 32 m, v 0, a -12 m/s2 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 10
TenthEditionVector Mechanics for Engineers: DynamicsDetermination of the Motion of a Particle We often determine accelerations from the forces applied(kinetics will be covered later) Generally have three classes of motion- acceleration given as a function of time, a f(t)- acceleration given as a function of position, a f(x)- acceleration given as a function of velocity, a f(v) Can you think of a physical example of when force is aWhen force is a function of velocity?function of position?a spring 2013 The McGraw-Hill Companies, Inc. All rights reserved.drag11 - 11
TenthEditionVector Mechanics for Engineers: DynamicsAcceleration as a function of time, position, or velocityIf .a a t a a x Kinematic relationshipdt dxdvand a vdtdv a(v)dtdvv a v dx 2013 The McGraw-Hill Companies, Inc. All rights reserved.vtv00 dv a t dtdv a(t )dtv dv a x dxa a v Integratevxv0x0 v dv a x dxvtdv v a v 0 dt0x dx x0vv dv v a v 011 - 12
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.2SOLUTION: Integrate twice to find v(t) and y(t). Solve for t when velocity equals zero(time for maximum elevation) andevaluate corresponding altitude.Ball tossed with 10 m/s vertical velocityfrom window 20 m above ground. Solve for t when altitude equals zero(time for ground impact) and evaluatecorresponding velocity.Determine: velocity and elevation above ground attime t, highest elevation reached by ball andcorresponding time, and time when ball will hit the ground andcorresponding velocity. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 13
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.2SOLUTION: Integrate twice to find v(t) and y(t).dv a 9.81 m s 2dtv t tv t v0 9.81t dv 9.81dtv00v t 10dy v 10 9.81tdty t t dy 10 9.81t dty00m m 9.81 2 ts s y t y0 10t 12 9.81t 2m m y t 20 m 10 t 4.905 2 t 2 s s 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 14
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.2 Solve for t when velocity equals zero and evaluatecorresponding altitude.v t 10m m 9.81 2 t 0s s t 1.019 s Solve for t when altitude equals zero and evaluatecorresponding velocity.m m y t 20 m 10 t 4.905 2 t 2 s s m m y 20 m 10 1.019 s 4.905 2 1.019 s 2 s s y 25.1 m 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 15
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.2 Solve for t when altitude equals zero and evaluatecorresponding velocity.m m y t 20 m 10 t 4.905 2 t 2 0 s s t 1.243 s meaningless t 3.28 sv t 10m m 9.81 2 ts s v 3.28 s 10m m 9.81 2 3.28 s s s v 22.2 2013 The McGraw-Hill Companies, Inc. All rights reserved.ms11 - 16
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.3SOLUTION:a kv Integrate a dv/dt -kv to find v(t). Integrate v(t) dx/dt to find x(t).Brake mechanism used to reduce gunrecoil consists of piston attached to barrelmoving in fixed cylinder filled with oil.As barrel recoils with initial velocity v0,piston moves and oil is forced throughorifices in piston, causing piston andcylinder to decelerate at rate proportionalto their velocity. Integrate a v dv/dx -kv to findv(x).Determine v(t), x(t), and v(x). 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 17
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.3SOLUTION: Integrate a dv/dt -kv to find v(t).vdva kvdttdv v v k 0 dt0lnv t ktv0v t v0 e kt Integrate v(t) dx/dt to find x(t).v t dx v0e ktdtxt00 ktdx ve0 dtt 1 x t v0 e kt k 0x t 2013 The McGraw-Hill Companies, Inc. All rights reserved. v01 e ktk11 - 18
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.3 Integrate a v dv/dx -kv to find v(x).dva v kvdxdv k dxvxv00 dv k dxv v0 kxv v0 kx Alternatively, v01 e ktk withx t andv t v0 e kt or e kt thenv v t x t 0 1 k v0 2013 The McGraw-Hill Companies, Inc. All rights reserved.v t v0v v0 kx11 - 19
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingA bowling ball is dropped from a boat so that itstrikes the surface of a lake with a speed of 5 m/s.Assuming the ball experiences a downwardacceleration of a 10 - 0.01v2 when in the water,determine the velocity of the ball when it strikes thebottom of the lake. yWhich integral should you choose?v(a)(b)t dv a t dtv00xvv dv x dx v a v 00 2013 The McGraw-Hill Companies, Inc. All rights reserved.(c)vxv0x0 v dv a x dxv(d)tdv v a v 0 dt011 - 20
TenthEditionVector Mechanics for Engineers: DynamicsConcept QuestionWhen will the bowling ball start slowing down? yA bowling ball is dropped from a boat so that itstrikes the surface of a lake with a speed of 5 m/s.Assuming the ball experiences a downwardacceleration of a 10 - 0.01v2 when in the water,determine the velocity of the ball when it strikes thebottom of the lake.The velocity would have to be highenough for the 0.01 v2 term to be biggerthan 10 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 21
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingSOLUTION: Determine the proper kinematicrelationship to apply (is accelerationa function of time, velocity, orposition?The car starts from rest and acceleratesaccording to the relationshipa 3 0.001v2 Determine the total distance the cartravels in one-half lap Integrate to determine the velocityafter one-half lapIt travels around a circular track that hasa radius of 200 meters. Calculate thevelocity of the car after it has travelledhalfway around the track. What is thecar’s maximum possible speed? 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 22
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingGiven: a 3 0.001v 2vo 0, r 200 mFind:v after ½ lapMaximum speedChoose the proper kinematic relationshipAcceleration is a function of velocity, andwe also can determine distance. Time is notinvolved in the problem, so we choose:dvv a v dxx dx x0vv dv v a v 0Determine total distance travelledx r 3.14(200) 628.32 m 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 23
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingDetermine the full integral, including limitsxvv dvdx x v a v 00628.32 0vvdv23 0.001v0dx Evaluate the interval and solve for v12 v628.32 ln 3 0.001v 00.002628.32( 0.002) ln 3 0.001v 2 ln 3 0.001(0) ln 3 0.001v 2 1.2566 1.0986 0.15802Take the exponential of each side 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 0.001v 2 e 0.1580211 - 24
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingSolve for v3 0.001v 2 e 0.158023 e 0.15802v 2146.20.0012v 46.3268 m/sHow do you determine the maximum speed the car can reach?a 3 0.001v 2Velocity is a maximum whenacceleration is zero0.001v 2 3This occurs whenvmax 30.001vmax 54.772 m/s 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 25
TenthEditionVector Mechanics for Engineers: DynamicsUniform Rectilinear MotionDuring free-fall, a parachutistreaches terminal velocity whenher weight equals the dragforce. If motion is in a straightline, this is uniform rectilinearmotion.For a particle in uniformrectilinear motion, theacceleration is zero andthe velocity is constant.dx v constantdtxtx00 dx v dtx x0 vtx x0 vtCareful – these only apply touniform rectilinear motion! 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 26
TenthEditionVector Mechanics for Engineers: DynamicsUniformly Accelerated Rectilinear MotionIf forces applied to a bodyare constant (and in aconstant direction), thenyou have uniformlyaccelerated rectilinearmotion.Another example is freefall when drag is negligible 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 27
TenthEditionVector Mechanics for Engineers: DynamicsUniformly Accelerated Rectilinear MotionFor a particle in uniformly accelerated rectilinear motion, theacceleration of the particle is constant. You may recognize theseconstant acceleration equations from your physics courses.dv a constantdtdx v0 atdtxvt dv a dtv00t dx v0 at dtx0dvv a constantdxv v0 atx x0 v0t 12 at 20vxv0x0 v dv a dxv 2 v02 2a x x0 Careful – these only apply to uniformlyaccelerated rectilinear motion! 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 28
TenthEditionVector Mechanics for Engineers: DynamicsMotion of Several ParticlesWe may be interested in the motion of several different particles,whose motion may be independent or linked together. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 29
TenthEditionVector Mechanics for Engineers: DynamicsMotion of Several Particles: Relative Motion For particles moving along the same line, timeshould be recorded from the same startinginstant and displacements should be measuredfrom the same origin in the same direction. x B x A relative position of Bwith respect to AxB x A xB AxBA v B v A relative velocity of Bwith respect to AvB v A vB AvBA a B a A relative acceleration of Bwith respect to AaB a A aB AaBA 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 30
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.4SOLUTION: Substitute initial position and velocityand constant acceleration of ball intogeneral equations for uniformlyaccelerated rectilinear motion. Substitute initial position and constantvelocity of elevator into equation foruniform rectilinear motion.Ball thrown vertically from 12 m levelin elevator shaft with initial velocity of18 m/s. At same instant, open-platformelevator passes 5 m level movingupward at 2 m/s.Determine (a) when and where ball hitselevator and (b) relative velocity of balland elevator at contact. 2013 The McGraw-Hill Companies, Inc. All rights reserved. Write equation for relative position ofball with respect to elevator and solvefor zero relative position, i.e., impact. Substitute impact time into equationfor position of elevator and relativevelocity of ball with respect toelevator.11 - 31
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.4SOLUTION: Substitute initial position and velocity and constantacceleration of ball into general equations foruniformly accelerated rectilinear motion.v B v0 at 18m m 9.81 2 ts s m m y B y0 v0t 12 at 2 12 m 18 t 4.905 2 t 2 s s Substitute initial position and constant velocity ofelevator into equation for uniform rectilinear motion.vE 2ms m y E y0 v E t 5 m 2 t s 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 32
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.4 Write equation for relative position of ball with respect toelevator and solve for zero relative position, i.e., impact.yBE 12 18t 4.905t 2 5 2t 0t 0.39 s meaningless t 3.65 s Substitute impact time into equations for position of elevatorand relative velocity of ball with respect to elevator.y E 5 2 3.65 y E 12.3 mv B E 18 9.81t 2 16 9.81 3.65 vB 2013 The McGraw-Hill Companies, Inc. All rights reserved.E 19.81ms11 - 33
TenthEditionVector Mechanics for Engineers: DynamicsMotion of Several Particles: Dependent Motion Position of a particle may depend on position of oneor more other particles. Position of block B depends on position of block A.Since rope is of constant length, it follows that sum oflengths of segments must be constant.x A 2 x B constant (one degree of freedom) Positions of three blocks are dependent.2 x A 2 x B xC constant (two degrees of freedom) For linearly related positions, similar relations holdbetween velocities and accelerations.dxdx Adx 2 B C 0 or 2v A 2v B vC 0dtdtdtdvdvdv2 A 2 B C 0 or 2a A 2a B aC 0dtdtdt2 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 34
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.5SOLUTION: Define origin at upper horizontal surfacewith positive displacement downward. Collar A has uniformly acceleratedrectilinear motion. Solve for accelerationand time t to reach L. Pulley D has uniform rectilinear motion.Pulley D is attached to a collar whichCalculate change of position at time t.is pulled down at 75 mm/s. At t 0,collar A starts moving down from K Block B motion is dependent on motionsof collar A and pulley D. Write motionwith constant acceleration and zeroinitial velocity. Knowing that velocity relationship and solve for change of blockB position at time t.of collar A is 300 mm/s as it passes L,determine the change in elevation, Differentiate motion relation twice tovelocity, and acceleration of block Bdevelop equations for velocity andwhen block A is at L.acceleration of block B. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 35
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.5SOLUTION: Define origin at upper horizontal surface withpositive displacement downward. Collar A has uniformly accelerated rectilinearmotion. Solve for acceleration and time t to reach L.2v A2 v A 0 2a A x A x A 0 2mm 300 2a A 200 mm s mma A 225 2sv A (v A )0 a At300mmmm 225 2 tss 2013 The McGraw-Hill Companies, Inc. All rights reserved.t 1.333 s11 - 36
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.5 Pulley D has uniform rectilinear motion. Calculatechange of position at time t.x D ( x D ) 0 vD tæ mm öxD - ( xD ) 0 ç 75 (1.333s) 100 mmès ø Block B motion is dependent on motions of collarA and pulley D. Write motion relationship andsolve for change of block B position at time t.Total length of cable remains constant,x A 2 xD xB x A 0 2 xD 0 xB 0 x A x A 0 2 xD xD 0 xB xB 0 0 200 mm 2 100 mm xB xB 0 0xB - (xB )0 - 400mm 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 37
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.5 Differentiate motion relation twice to developequations for velocity and acceleration of block B.x A 2 xD xB constantv A 2v D v B 0mm mm 300 2 75 vB 0s s vB - 450mmmm 450ssa A 2aD aB 0mm 225 2(0) aB 0 2 s aB - 225 2013 The McGraw-Hill Companies, Inc. All rights reserved.mmmm 225s2s211 - 38
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingSlider block A moves to the left with aconstant velocity of 6 m/s. Determine thevelocity of block B.Solution steps Sketch your system and choosecoordinate systemWrite out constraint equationDifferentiate the constraint equation toget velocity 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 39
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingGiven: vA 6 m/s left Find: vBxAThis length is constant nomatter how the blocks moveSketch your system and choose coordinatesyBDefine your constraint equation(s)x A 3 yB constants LDifferentiate the constraint equation toget velocity6 m/s 3vB 0v B 2 m/s Note that as xA gets bigger, yB gets smaller. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 40
Graphical Solution of Rectilinear-Motion ProblemsEngineers often collect position, velocity, and accelerationdata. Graphical solutions are often useful in analyzingthese data.Data Fideltity / Highest Recorded Punch180160Acceleration datafrom a head impactduring a round ofboxing.140Acceleration (g)TenthEditionVector Mechanics for Engineers: 81Time (s) 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 41
TenthEditionVector Mechanics for Engineers: DynamicsGraphical Solution of Rectilinear-Motion Problems Given the x-t curve, the v-t curve isequal to the x-t curve slope. Given the v-t curve, the a-t curve isequal to the v-t curve slope. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 42
TenthEditionVector Mechanics for Engineers: DynamicsGraphical Solution of Rectilinear-Motion Problems Given the a-t curve, the change in velocity between t1 and t2 isequal to the area under the a-t curve between t1 and t2. Given the v-t curve, the change in position between t1 and t2 isequal to the area under the v-t curve between t1 and t2. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 43
TenthEditionVector Mechanics for Engineers: DynamicsOther Graphical Methods Moment-area method to determine particle position attime t directly from the a-t curve:x1 x0 area under v t curve v0t1 v1 t1 t dvv0using dv a dt ,v1x1 x0 v0 t1 t1 t a dtv0v1 t1 t a dt first moment of area under a-t curvev0with respect to t t1 line.x1 x0 v0t1 area under a-t curve t1 t t abscissa of centroid C 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 44
TenthEditionVector Mechanics for Engineers: DynamicsOther Graphical Methods Method to determine particle acceleration from v-x curve:dvdx AB tan BC subnormal to v-x curvea v 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 45
TenthEditionVector Mechanics for Engineers: DynamicsCurvilinear Motion: Position, Velocity & AccelerationThe softball and the car both undergocurvilinear motion. A particle moving along a curve other than astraight line is in curvilinear motion. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 46
TenthEditionVector Mechanics for Engineers: DynamicsCurvilinear Motion: Position, Velocity & Acceleration The position vector of a particle at time t is defined by a vector betweenorigin O of a fixed reference frame and the position occupied by particle. Consider a particle which occupies position P defined by r at time t and P’ defined by r at t Dt, 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 47
TenthEditionVector Mechanics for Engineers: DynamicsCurvilinear Motion: Position, Velocity & AccelerationInstantaneous velocity(vector)Dr drv lim Dt 0 Dtdt 2013 The McGraw-Hill Companies, Inc. All rights reserved.Instantaneous speed(scalar)Ds ds Dt 0 Dtdtv lim11 - 48
TenthEditionVector Mechanics for Engineers: DynamicsCurvilinear Motion: Position, Velocity & Acceleration Consider velocity v of a particle at time t and velocity v at t Dt,Dv dv instantaneous acceleration (vector)Dt 0 Dtdta lim In general, the acceleration vector is not tangentto the particle path and velocity vector. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 49
TenthEditionVector Mechanics for Engineers: DynamicsDerivatives of Vector Functions Let P u be a vector function of scalar variable u, dPDPP u Du P u lim limdu Du 0 Du Du 0Du Derivative of vector sum, d P Q dP dQ dudu duDelete or put in“bonus” slides Derivative of product of scalar and vector functions, d f P dfdP P fdududu Derivative of scalar product and vector product, d P Q dP dQ Q P dududu d P Q dPdQ Q P dududu 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 50
TenthEditionVector Mechanics for Engineers: DynamicsRectangular Components of Velocity & Acceleration When position vector of particle P is given by itsrectangular components, r xi y j zk Velocity vector, dx dy dz v i j k x i y j z kdtdtdt vx i v y j vz k Acceleration vector, d 2 x d 2 y d 2 z a 2 i 2 j 2 k x i y j z kdtdtdt ax i a y j az k 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 51
TenthEditionVector Mechanics for Engineers: DynamicsRectangular Components of Velocity & Acceleration Rectangular components particularly effectivewhen component accelerations can be integratedindependently, e.g., motion of a projectile,a x x 0a y y ga z z 0with initial conditions, vx 0 , v y 0 , vz 0 0x0 y0 z0 0 Integrating twice yieldsv x v x 0x v x 0 t 0v y v y gt0y v y y 12 gt 2vz 0z 0 Motion in horizontal direction is uniform. Motion in vertical direction is uniformly accelerated. Motion of projectile could be replaced by twoindependent rectilinear motions. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 52
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.7SOLUTION: Consider the vertical and horizontal motionseparately (they are independent) Apply equations of motion in y-direction Apply equations of motion in x-directionA projectile is fired from the edgeof a 150-m cliff with an initialvelocity of 180 m/s at an angle of30 with the horizontal. Neglectingair resistance, find (a) the horizontaldistance from the gun to the pointwhere the projectile strikes theground, (b) the greatest elevationabove the ground reached by theprojectile. Determine time t for projectile to hit theground, use this to find the horizontaldistance Maximum elevation occurs when vy 0 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 53
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.7SOLUTION:Given:(v)o 180 m/s(a)y - 9.81 m/s2(y)o 150 m(a)x 0 m/s2Vertical motion – uniformly accelerated:Horizontal motion – uniformly accelerated:Choose positive x to the right as shown 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 54
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.7SOLUTION:Horizontal distanceProjectile strikes the ground at:Substitute into equation (1) aboveSolving for t, we take the positive rootSubstitute t into equation (4)Maximum elevation occurs when vy 0Maximum elevation above the ground 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 55
TenthEditionVector Mechanics for Engineers: DynamicsConcept QuizIf you fire a projectile from 150meters above the ground (seeEx Problem 11.7), what launchangle will give you the greatesthorizontal distance x?a)b)c)d)A launch angle of 45A launch angle less than 45A launch angle greater than 45It depends on the launch velocity 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 56
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingSOLUTION:A baseball pitching machine“ throws ” baseballs with ahorizontal velocity v0. If youwant the height h to be 1 m,determine the value of v0. Consider the vertical and horizontal motionseparately (they are independent) Apply equations of motion in y-direction Apply equations of motion in x-direction Determine time t for projectile to fall to1 m. Calculate v0 0 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 57
TenthEditionVector Mechanics for Engineers: DynamicsGroup Problem SolvingGiven: x 12 m, yo 1.5 m,yf 1 m.Find: voAnalyze the motion inthe y-directiony f y0 (0)t 1 2gt21 23.5 5 gt21 0.5 m (9.81 m/s 2 )t 22Analyze the motion inthe x-directionx 0 (vx )0 t v0t12 m (v0 )(0.3193 s)v0 37.6 m/s 135.3 km/ht 0.3193 s 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 58
TenthEditionVector Mechanics for Engineers: DynamicsMotion Relative to a Frame in TranslationA soccer player must considerthe relative motion of the balland her teammates whenmaking a pass. 2013 The McGraw-Hill Companies, Inc. All rights reserved.It is critical for a pilot toknow the relative motionof his aircraft with respectto the aircraft carrier tomake a safe landing.11 - 59
TenthEditionVector Mechanics for Engineers: DynamicsMotion Relative to a Frame in Translation Designate one frame as the fixed frame of reference.All other frames not rigidly attached to the fixedreference frame are moving frames of reference. Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are rA and rB . r Vector B A joining A and B defines the position ofB with respect to the moving frame Ax’y’z’ and rB rA rB A Differentiating twice, v B v A v B A v B A velocity of B relative to A. a B a A a B A a B A acceleration of B relativeto A. Absolute motion of B can be obtained by combiningmotion of A with relative motion of B with respect tomoving reference frame attached to A. 2013 The McGraw-Hill Companies, Inc. All rights reserved.11 - 60
TenthEditionVector Mechanics for Engineers: DynamicsSample Problem 11.
Motion of Several Particles: Dependent Motion Sample Problem 11.5 Graphical Solution of Rectilinear-Motion Problems Other Graphical Methods Curvilinear Motion: Position, Velocity & Acceleration Derivatives of Vector Functions Rectangular Components of Velocity and Acceleration Motion Relative to a Frame in Translation Tangential and Normal .
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6. Lakhmir Singh’s Science for Class 8 7. Science for Ninth Class (Part 1) PHYSICS 8. Science for Ninth Class (Part 2) CHEMISTRY 9. Science for Tenth Class (Part 1) PHYSICS 10. Science for Tenth Class (Part 2) CHEMISTRY 11. Science for Tenth Class (Part 3) BIOLOGY 12. Rapid Revision in Science (A Question-Answer Book for Class X) 13. Science .
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5. Find the area of the figure. Round to the nearest tenth if necessary. 6. Find the area of the figure. Round to the nearest tenth if necessary. 7. Find the area of a regular octagon with apothem length of 20.5 feet. Round to the nearest tenth if necessary. 8. The coordinates of the vertices of a regular polygon are given.
6. Lakhmir Singh’s Science for Class 8 7. Science for Ninth Class (Part 1) PHYSICS 8. Science for Ninth Class (Part 2) CHEMISTRY 9. Science for Tenth Class (Part 1) PHYSICS 10. Science for Tenth Class (Part 2) CHEMISTRY 11. Science for Tenth Class (Part 3) BIOLOGY 12. Rapid Revision in Science
The Tenth Muse “So long as men can breathe or eyes can see, So long lives this, and this gives life to thee.” ---Shakespeare . Advisor. Mr. Townsend . Student Editors . Sofia Amorim, Romina Generali, and Giuseppe Ceballos . The Tenth Muse. is dedicated to . Romina Generali, Sofia A
