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- 1. VECTOR MECHANICS FOR ENGINEERS: STATICSSTATICS Ninth EditionNinth Edition Ferdinand P. BeerFerdinand P. Beer E. Russell Johnston, Jr.E. Russell Johnston, Jr. Lecture Notes:Lecture Notes: J. Walt OlerJ. Walt Oler Texas Tech UniversityTexas Tech University CHAPTER © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 11 Kinematics of Particles
- 2. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Contents 11 - 2 Introduction Rectilinear Motion: Position, Velocity & Acceleration Determination of the Motion of a Particle Sample Problem 11.2 Sample Problem 11.3 Uniform Rectilinear-Motion Uniformly Accelerated Rectilinear- Motion Motion of Several Particles: Relative Motion Sample Problem 11.4 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 Components Radial and Transverse Components Sample Problem 11.10 Sample Problem 11.12
- 3. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Introduction 11 - 3 • Dynamics includes: - Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion. - Kinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion. • Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line. • Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions.
- 4. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectilinear Motion: Position, Velocity & Acceleration 11 - 4 • Particle moving along a straight line is said to be in rectilinear motion. • Position coordinate of a particle is defined by positive or negative distance of particle from a fixed origin on the line. • The motion of a particle is known if the position coordinate for particle is known for every value of time t. Motion of the particle may be expressed in the form of a function, e.g., 32 6 ttx −= or in the form of a graph x vs. t.
- 5. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectilinear Motion: Position, Velocity & Acceleration 11 - 5 • Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed. • Consider particle which occupies position P at time t and P’ at t+∆t, t x v t x t ∆ ∆ == ∆ ∆ = →∆ 0 lim Average velocity Instantaneous velocity • From the definition of a derivative, dt dx t x v t = ∆ ∆ = →∆ 0 lim e.g., 2 32 312 6 tt dt dx v ttx −== −=
- 6. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectilinear Motion: Position, Velocity & Acceleration 11 - 6 • Consider particle with velocity v at time t and v’ at t+∆t, Instantaneous acceleration t v a t ∆ ∆ == →∆ 0 lim t dt dv a ttv dt xd dt dv t v a t 612 312e.g. lim 2 2 2 0 −== −= == ∆ ∆ = →∆ • From the definition of a derivative, • Instantaneous acceleration may be: - positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity or increasing negative velocity.
- 7. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectilinear Motion: Position, Velocity & Acceleration 11 - 7 • Consider particle with motion given by 32 6 ttx −= 2 312 tt dt dx v −== t dt xd dt dv a 6122 2 −=== • at t = 0, x = 0, v = 0, a = 12 m/s2 • at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0 • at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2 • at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2
- 8. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Determination of the Motion of a Particle 11 - 8 • Recall, motion of a particle is known if position is known for all time t. • Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations. • Three classes of motion may be defined for: - 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)
- 9. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Determination of the Motion of a Particle 11 - 9 • Acceleration given as a function of time, a = f(t): ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )∫∫∫ ∫∫∫ =−=== =−==== tttx x tttv v dttvxtxdttvdxdttvdxtv dt dx dttfvtvdttfdvdttfdvtfa dt dv 0 0 0 0 0 0 0 0 • Acceleration given as a function of position, a = f(x): ( ) ( ) ( ) ( ) ( ) ( )∫∫∫ =−== ===== x x x x xv v dxxfvxvdxxfdvvdxxfdvv xf dx dv va dt dv a v dx dt dt dx v 000 2 02 12 2 1 oror
- 10. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Determination of the Motion of a Particle 11 - 10 • Acceleration given as a function of velocity, a = f(v): ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∫ ∫∫ ∫ ∫∫ =− ==== = ==== tv v tv v tx x tv v ttv v vf dvv xtx vf dvv dx vf dvv dxvfa dx dv v t vf dv dt vf dv dt vf dv vfa dt dv 0 00 0 0 0 0
- 11. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.2 11 - 11 Determine: • velocity and elevation above ground at time t, • highest elevation reached by ball and corresponding time, and • time when ball will hit the ground and corresponding velocity. Ball tossed with 10 m/s vertical velocity from window 20 m above ground. SOLUTION: • Integrate twice to find v(t) and y(t). • Solve for t at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude. • Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity.
- 12. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.2 11 - 12 ( ) ( ) tvtvdtdv a dt dv ttv v 81.981.9 sm81.9 0 0 2 0 −=−−= −== ∫∫ ( ) ttv −= 2 s m 81.9 s m 10 ( ) ( ) ( ) 2 2 1 0 0 81.91081.910 81.910 0 ttytydttdy tv dt dy tty y −=−−= −== ∫∫ ( ) 2 2 s m 905.4 s m 10m20 ttty − += SOLUTION: • Integrate twice to find v(t) and y(t).
- 13. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.2 11 - 13 • Solve for t at which velocity equals zero and evaluate corresponding altitude. ( ) 0 s m 81.9 s m 10 2 = −= ttv s019.1=t • Solve for t at which altitude equals zero and evaluate corresponding velocity. ( ) ( ) ( )2 2 2 2 s019.1 s m 905.4s019.1 s m 10m20 s m 905.4 s m 10m20 − += − += y ttty m1.25=y
- 14. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.2 11 - 14 • Solve for t at which altitude equals zero and evaluate corresponding velocity. ( ) 0 s m 905.4 s m 10m20 2 2 = − += ttty ( ) s28.3 smeaningless243.1 = −= t t ( ) ( ) ( )s28.3 s m 81.9 s m 10s28.3 s m 81.9 s m 10 2 2 −= −= v ttv s m 2.22−=v
- 15. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.3 11 - 15 Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity. Determine v(t), x(t), and v(x). kva −= SOLUTION: • Integrate a = dv/dt = -kv to find v(t). • Integrate v(t) = dx/dt to find x(t). • Integrate a = v dv/dx = -kv to find v(x).
- 16. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.3 11 - 16 SOLUTION: • Integrate a = dv/dt = -kv to find v(t). ( ) ( ) kt v tv dtk v dv kv dt dv a ttv v −=−=−== ∫∫ 00 ln 0 ( ) kt evtv − = 0 • Integrate v(t) = dx/dt to find x(t). ( ) ( ) ( ) t kt t kt tx kt e k vtxdtevdx ev dt dx tv 0 0 0 0 0 0 1 −== == −− − ∫∫ ( ) ( )kt e k v tx − −= 10
- 17. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.3 11 - 17 • Integrate a = v dv/dx = -kv to find v(x). kxvv dxkdvdxkdvkv dx dv va xv v −=− −=−=−== ∫∫ 0 00 kxvv −= 0 • Alternatively, ( ) ( ) −= 0 0 1 v tv k v tx kxvv −= 0 ( ) ( ) 0 0 or v tv eevtv ktkt == −− ( ) ( )kt e k v tx − −= 10 with and then
- 18. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Uniform Rectilinear Motion 11 - 18 For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. vtxx vtxx dtvdx v dt dx tx x += =− = == ∫∫ 0 0 00 constant
- 19. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Uniformly Accelerated Rectilinear Motion 11 - 19 For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. atvv atvvdtadva dt dv tv v += =−=== ∫∫ 0 0 00 constant ( ) 2 2 1 00 2 2 1 00 0 00 0 attvxx attvxxdtatvdxatv dt dx tx x ++= +=−+=+= ∫∫ ( ) ( ) ( )0 2 0 2 0 2 0 2 2 1 2 constant 00 xxavv xxavvdxadvva dx dv v x x v v −+= −=−=== ∫∫
- 20. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Motion of Several Particles: Relative Motion 11 - 20 • For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. =−= ABAB xxx relative position of B with respect to A ABAB xxx += =−= ABAB vvv relative velocity of B with respect to A ABAB vvv += =−= ABAB aaa relative acceleration of B with respect to A ABAB aaa +=
- 21. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.4 11 - 21 Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s. Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact. SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. • Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator.
- 22. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.4 11 - 22 SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. 2 2 2 2 1 00 20 s m 905.4 s m 18m12 s m 81.9 s m 18 ttattvyy tatvv B B − +=++= −=+= • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. ttvyy v EE E +=+= = s m 2m5 s m 2 0
- 23. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.4 11 - 23 • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. ( ) ( ) 025905.41812 2 =+−−+= ttty EB ( ) s65.3 smeaningless39.0 = −= t t • Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. ( )65.325+=Ey m3.12=Ey ( ) ( )65.381.916 281.918 −= −−= tv EB s m 81.19−=EBv
- 24. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Motion of Several Particles: Dependent Motion 11 - 24 • Position of a particle may depend on position of one or more other particles. • Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant. =+ BA xx 2 constant (one degree of freedom) • Positions of three blocks are dependent. =++ CBA xxx 22 constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations. 022or022 022or022 =++=++ =++=++ CBA CBA CBA CBA aaa dt dv dt dv dt dv vvv dt dx dt dx dt dx
- 25. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.5 11 - 25 Pulley D is attached to a collar which is pulled down at 3 in./s. At t = 0, collar A starts moving down from K with constant acceleration and zero initial velocity. Knowing that velocity of collar A is 12 in./s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block A is at L. SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. • Pulley D has uniform rectilinear motion. Calculate change of position at time t. • Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t. • Differentiate motion relation twice to develop equations for velocity and acceleration of block B.
- 26. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.5 11 - 26 SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. ( ) ( )[ ] ( ) 2 2 0 2 0 2 s in. 9in.82 s in. 12 2 == −+= AA AAAAA aa xxavv ( ) s333.1 s in. 9 s in. 12 2 0 == += tt tavv AAA
- 27. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.5 11 - 27 • Pulley D has uniform rectilinear motion. Calculate change of position at time t. ( ) ( ) ( ) in.4s333.1 s in. 30 0 = =− += DD DDD xx tvxx • Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t. Total length of cable remains constant, ( ) ( ) ( ) ( )[ ] ( )[ ] ( )[ ] ( ) ( ) ( )[ ] 0in.42in.8 02 22 0 000 000 =−++ =−+−+− ++=++ BB BBDDAA BDABDA xx xxxxxx xxxxxx ( ) in.160 −=− BB xx
- 28. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.5 11 - 28 • Differentiate motion relation twice to develop equations for velocity and acceleration of block B. 0 s in. 32 s in. 12 02 constant2 =+ + =++ =++ B BDA BDA v vvv xxx s in. 18=Bv 0 s in. 9 02 2 =+ =++ B BDA v aaa 2 s in. 9−=Ba
- 29. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Graphical Solution of Rectilinear-Motion Problems 11 - 29 • Given the x-t curve, the v-t curve is equal to the x-t curve slope. • Given the v-t curve, the a-t curve is equal to the v-t curve slope.
- 30. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Graphical Solution of Rectilinear-Motion Problems 11 - 30 • Given the a-t curve, the change in velocity between t1 and t2 is equal 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 is equal to the area under the v-t curve between t1 and t2.
- 31. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Other Graphical Methods 11 - 31 • Moment-area method to determine particle position at time t directly from the a-t curve: ( )∫ −+= −=− 1 0 110 01 curveunderarea v v dvtttv tvxx using dv = a dt , ( )∫ −+=− 1 0 11001 v v dtatttvxx ( ) =−∫ 1 0 1 v v dtatt first moment of area under a-t curve with respect to t = t1 line. ( )( ) Ct tta-ttvxx centroidofabscissa curveunderarea 11001 = −++=
- 32. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Other Graphical Methods 11 - 32 • Method to determine particle acceleration from v-x curve: == = = BC AB dx dv va θtan subnormal to v-x curve
- 33. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Curvilinear Motion: Position, Velocity & Acceleration 11 - 33 • Particle moving along a curve other than a straight line is in curvilinear motion. • Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle. • Consider particle which occupies position P defined by at time t and P’ defined by at t + ∆t, r r′ = = ∆ ∆ = = = ∆ ∆ = →∆ →∆ dt ds t s v dt rd t r v t t 0 0 lim lim instantaneous velocity (vector) instantaneous speed (scalar)
- 34. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Curvilinear Motion: Position, Velocity & Acceleration 11 - 34 = = ∆ ∆ = →∆ dt vd t v a t 0 lim instantaneous acceleration (vector) • Consider velocity of particle at time t and velocity at t + ∆t, v v ′ • In general, acceleration vector is not tangent to particle path and velocity vector.
- 35. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Derivatives of Vector Functions 11 - 35 ( )uP • Let be a vector function of scalar variable u, ( ) ( ) u uPuuP u P du Pd uu ∆ −∆+ = ∆ ∆ = →∆→∆ 00 limlim • Derivative of vector sum, ( ) du Qd du Pd du QPd += + ( ) du Pd fP du df du Pfd += • Derivative of product of scalar and vector functions, • Derivative of scalar product and vector product, ( ) ( ) du Qd PQ du Pd du QPd du Qd PQ du Pd du QPd ×+×= × •+•= •
- 36. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectangular Components of Velocity & Acceleration 11 - 36 • When position vector of particle P is given by its rectangular components, kzjyixr ++= • Velocity vector, kvjviv kzjyixk dt dz j dt dy i dt dx v zyx ++= ++=++= • Acceleration vector, kajaia kzjyixk dt zd j dt yd i dt xd a zyx ++= ++=++= 2 2 2 2 2 2
- 37. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Rectangular Components of Velocity & Acceleration 11 - 37 • Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile, 00 ==−==== zagyaxa zyx with initial conditions, ( ) ( ) ( ) 0,,0 000000 ==== zyx vvvzyx Integrating twice yields ( ) ( ) ( ) ( ) 0 0 2 2 1 00 00 =−== =−== zgtyvytvx vgtvvvv yx zyyxx • Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions.
- 38. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Motion Relative to a Frame in Translation 11 - 38 • Designate one frame as the fixed frame of reference. All other frames not rigidly attached to the fixed reference frame are moving frames of reference. • Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are .and BA rr • Vector joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and ABr ABAB rrr += • Differentiating twice, =ABv velocity of B relative to A.ABAB vvv += =ABa acceleration of B relative to A. ABAB aaa += • Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A.
- 39. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Tangential and Normal Components 11 - 39 • Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components. • are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and is the angle between them. tt ee ′and ttt eee −′=∆ θ∆ ( ) ( ) θ θ θ θ θ θθ d ed e ee e e t n nn t t = = ∆ ∆ = ∆ ∆ ∆=∆ →∆→∆ 2 2sin limlim 2sin2 00
- 40. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Tangential and Normal Components 11 - 40 tevv =• With the velocity vector expressed as the particle acceleration may be written as dt ds ds d d ed ve dt dv dt ed ve dt dv dt vd a tt θ θ +=+== but v dt ds dsde d ed n t === θρ θ After substituting, ρρ 22 v a dt dv ae v e dt dv a ntnt ==+= • Tangential component of acceleration reflects change of speed and normal component reflects change of direction. • Tangential component may be positive or negative. Normal component always points toward center of path curvature.
- 41. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Tangential and Normal Components 11 - 41 ρρ 22 v a dt dv ae v e dt dv a ntnt ==+= • Relations for tangential and normal acceleration also apply for particle moving along space curve. • Plane containing tangential and normal unit vectors is called the osculating plane. ntb eee ×= • Normal to the osculating plane is found from binormale normalprincipale b n = = • Acceleration has no component along binormal.
- 42. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Radial and Transverse Components 11 - 42 • When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP. r r e d ed e d ed −== θθ θ θ dt d e dt d d ed dt ed rr θθ θ θ == dt d e dt d d ed dt ed r θθ θ θθ −== ( ) θ θ θ θ erer e dt d re dt dr dt ed re dt dr er dt d v r r r rr += +=+== • The particle velocity vector is • Similarly, the particle acceleration vector is ( ) ( ) θ θ θθ θ θθθ θθθ θ errerr dt ed dt d re dt d re dt d dt dr dt ed dt dr e dt rd e dt d re dt dr dt d a r r r r 22 2 2 2 2 ++−= ++++= += rerr =
- 43. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Radial and Transverse Components 11 - 43 • When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors .and,, keeR θ • Position vector, kzeRr R += • Velocity vector, kzeReR dt rd v R ++== θθ • Acceleration vector, ( ) ( ) kzeRReRR dt vd a R +++−== θθθθ 22
- 44. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.10 11 - 44 A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration rate. Knowing that after 8 s the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes are applied. SOLUTION: • Calculate tangential and normal components of acceleration. • Determine acceleration magnitude and direction with respect to tangent to curve.
- 45. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.10 11 - 45 ft/s66mph45 ft/s88mph60 = = SOLUTION: • Calculate tangential and normal components of acceleration. ( ) ( ) 2 22 2 s ft 10.3 ft2500 sft88 s ft 75.2 s8 sft8866 === −= − = ∆ ∆ = ρ v a t v a n t • Determine acceleration magnitude and direction with respect to tangent to curve. ( ) 2222 10.375.2 +−=+= nt aaa 2 s ft 14.4=a 75.2 10.3 tantan 11 −− == t n a a α °= 4.48α
- 46. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.12 11 - 46 Rotation of the arm about O is defined by θ = 0.15t2 where θ is in radians and t in seconds. Collar B slides along the arm such that r = 0.9 - 0.12t2 where r is in meters. After the arm has rotated through 30o , determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm. SOLUTION: • Evaluate time t for θ = 30o . • Evaluate radial and angular positions, and first and second derivatives at time t. • Calculate velocity and acceleration in cylindrical coordinates. • Evaluate acceleration with respect to arm.
- 47. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.12 11 - 47 SOLUTION: • Evaluate time t for θ = 30o . s869.1rad524.030 0.15 2 ==°= = t tθ • Evaluate radial and angular positions, and first and second derivatives at time t. 2 2 sm24.0 sm449.024.0 m481.012.09.0 −= −=−= =−= r tr tr 2 2 srad30.0 srad561.030.0 rad524.015.0 = == == θ θ θ t t
- 48. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.12 11 - 48 • Calculate velocity and acceleration. ( )( ) r r r v v vvv rv srv θ θ θ β θ 122 tan sm270.0srad561.0m481.0 m449.0 − =+= === −== °== 0.31sm524.0 βv ( )( ) ( )( ) ( )( ) r r r a a aaa rra rra θ θ θ γ θθ θ 122 2 2 2 22 2 tan sm359.0 srad561.0sm449.02srad3.0m481.0 2 sm391.0 srad561.0m481.0sm240.0 − =+= −= −+= += −= −−= −= °== 6.42sm531.0 γa
- 49. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth Edition Sample Problem 11.12 11 - 49 • Evaluate acceleration with respect to arm. Motion of collar with respect to arm is rectilinear and defined by coordinate r. 2 sm240.0−== ra OAB

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