Seventh Edition CHAPTER 11 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Kinematics of Particles Lecture Notes: J. Walt Oler Texas Tech University © 2004 The McGraw-Hill Companies, Inc. All rights reserved. Seventh Edition Vector Mechanics for Engineers: Dynamics Introduction • Rigid Body Dynamics : distance between points is fixed, no size or shape changes, idealization of flexible body - particle: mass concentrated at one point, center of mass, rotation of points w/r to c.m. is neglected - rigid body: rotation of points w/r to c.m. is taken into account • Example - Ski jumper : – model as a particle: if interested in finding how far she jumps – model as a rigid body: if interested in finding the position of the head w/r to torso • Example - Football player kicking a ball : – model as particles: if interested in finding how high/far the ball goes – model as rigid bodies: if interested in finding the forces on the knee © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 49 1 Seventh Edition Vector Mechanics for Engineers: Dynamics Introduction • 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, i.e. forces are not considered. - 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. • Newton’s laws: 1. F=0 2. F = ma 3. Action-reaction © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Derivatives of Functions • Let s(u) be a scalar function of scalar variable u, ∆s ds s (u + ∆u ) − s (u ) = lim = lim du ∆u → 0 ∆u ∆u → 0 ∆u r • Let P(u ) be a vector function of scalar variable u, r r r r dP P(u + ∆u ) − P(u ) ∆P = lim = lim du ∆u →0 ∆u ∆u →0 ∆u © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 49 2 Seventh Edition Vector Mechanics for Engineers: Dynamics Position, Velocity & Acceleration • Consider a particle moving along a certain path • 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. r r′ • Consider particle which occupies position P defined r r by r at time t and P’ defined by r ′ at t + ∆t, r r ∆r dr r r r r v = lim = ∆r = r ′ − r dt ∆t →0 ∆t = instantaneous velocity (vector) v = lim ∆s ∆t →0 ∆t = ds dt = instantaneous speed (scalar) •Velocity vector is always tangent to particle path. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Position, Velocity & Acceleration r • Consider velocity v of particle at time t and velocity r v′ at t + ∆t, r r ∆v dv r a = lim = dt ∆t →0 ∆t = instantaneous acceleration (vector) • In general, acceleration vector is not tangent to particle path. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 6 - 49 3 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion of a particle along a straight line • The motion of a particle is known if position is known for all time t. • If path is a straight line we have rectilinear motion, we can describe the motion in terms of a, v and x. • 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) © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 7 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion of a particle along a straight line • Acceleration given as a function of time, a = f(t): v (t ) t dv = a = f (t ) dv = f (t ) dt dv = ∫ f (t ) dt ∫ dt 0 v dx = v(t ) dt x (t ) dx = v(t ) dt t v(t ) − v0 = ∫ f (t ) dt 0 0 t ∫ dx = ∫ v(t ) dt x0 t x(t ) − x0 = ∫ v(t ) dt 0 0 • Acceleration given as a function of position, a = f(x): v= dx dx or dt = dt v v dv = f ( x )dx a= v( x ) dv dv or a = v = f ( x ) dt dx x ∫ v dv = ∫ f ( x )dx v0 x0 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 1 v ( x )2 2 − 12 v02 = x ∫ f ( x )dx x0 8 - 49 4 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion of a particle along a straight line • Acceleration given as a function of velocity, a = f(v): dv = a = f (v ) dt v (t ) dv = dt f (v ) dv t = ∫ dt ∫ v0 f (v ) 0 dv =t ∫ v0 f (v ) v v (t ) dv = a = f (v ) dx dx = v (t ) v dv x(t ) − x0 = ∫ v f (v ) v dv f (v ) x (t ) v (t ) v dv ∫ dx = ∫ f (v ) x0 v0 0 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Uniform Rectilinear Motion For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. dx = v = constant dt x t x0 0 ∫ dx = v ∫ dt x − x0 = vt x = x0 + vt © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 10 - 49 5 Seventh Edition Vector Mechanics for Engineers: Dynamics Uniformly Accelerated Rectilinear Motion For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. dv = a = constant dt v t v0 0 ∫ dv = a ∫ dt v − v0 = at v = v0 + at dx = v0 + at dt x= x t x0 0 ∫ dx = ∫ (v0 + at )dt x − x0 = v0 t + 12 at 2 x0 + v0 t + 12 at 2 v dv = a = constant dx v 2 = v02 v x v0 x0 ∫ v dv = a ∫ dx 1 2 (v 2 − v02 ) = a(x − x0 ) + 2 a ( x − x0 ) © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 11 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Graphical Solution of Rectilinear-Motion Problems • 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 49 6 Seventh Edition Vector Mechanics for Engineers: Dynamics Graphical Solution of Rectilinear-Motion Problems • 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 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. Ball tossed with 10 m/s vertical velocity from window 20 m above ground. • Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity. 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 14 - 49 7 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 SOLUTION: • Integrate twice to find v(t) and y(t). dv = a = −9.81 m s 2 dt v (t ) t v(t ) − v0 = −9.81t ∫ dv = − ∫ 9.81 dt v0 0 v(t ) = 10 dy = v = 10 − 9.81t dt y (t ) t ∫ dy = ∫ (10 − 9.81t )dt y0 0 m ⎛ m⎞ − ⎜ 9.81 2 ⎟ t s ⎝ s ⎠ y (t ) − y0 = 10t − 12 9.81t 2 m⎞ ⎛ m⎞ ⎛ y (t ) = 20 m + ⎜10 ⎟t − ⎜ 4.905 2 ⎟t 2 ⎝ s⎠ ⎝ s ⎠ © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 • Solve for t at which velocity equals zero and evaluate corresponding altitude. v(t ) = 10 m⎞ m ⎛ − ⎜ 9.81 2 ⎟ t = 0 s ⎝ s ⎠ t = 1.019 s • Solve for t at which altitude equals zero and evaluate corresponding 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 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 16 - 49 8 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 • Solve for t at which altitude equals zero and evaluate corresponding 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 s v(t ) = 10 m ⎛ m⎞ − ⎜ 9.81 2 ⎟ t s ⎝ s ⎠ v(3.28 s ) = 10 m ⎛ m⎞ − ⎜ 9.81 2 ⎟ (3.28 s ) s ⎝ s ⎠ v = −22.2 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. m s 17 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 SOLUTION: 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 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. • Integrate a = v dv/dx = -kv to find v(x). Determine v(t), x(t), and v(x). © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 18 - 49 9 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 SOLUTION: • Integrate a = dv/dt = -kv to find v(t). v (t ) t v(t ) dv dv = −kv = −k ∫ dt ln = − kt a= ∫ v0 v dt v 0 0 v(t ) = v0 e − kt • Integrate v(t) = dx/dt to find x(t). dx v(t ) = = v0 e − kt dt t x (t ) t ⎡ 1 − kt ⎤ − kt ( ) dx v e dt x t v e = = − ∫ 0∫ 0⎢ ⎥⎦ ⎣ k 0 0 0 ( v x(t ) = 0 1 − e − kt k © 2004 The McGraw-Hill Companies, Inc. All rights reserved. ) 19 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 • Integrate a = v dv/dx = -kv to find v(x). a=v dv = − kv dx dv = − k dx v x v0 0 ∫ dv = −k ∫ dx v − v0 = − kx v = v0 − kx • Alternatively, with and then ( v x(t ) = 0 1 − e − kt k ) v(t ) v(t ) = v0 e − kt or e − kt = v0 v0 ⎛ v(t ) ⎞ ⎟ x(t ) = ⎜⎜1 − k ⎝ v0 ⎟⎠ © 2004 The McGraw-Hill Companies, Inc. All rights reserved. v = v0 − kx 20 - 49 10 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion of Several Particles: Dependent Motion • 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. 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 hold between velocities and accelerations. dx dx A dx + 2 B + C = 0 or 2v A + 2v B + vC = 0 dt dt dt dv A dv B dvC 2 +2 + = 0 or 2a A + 2a B + aC = 0 dt dt dt 2 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 21 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.5 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. Pulley D is attached to a collar which Calculate change of position at time t. is pulled down at 3 in./s. At t = 0, collar A starts moving down from K • Block B motion is dependent on motions of collar A and pulley D. Write motion with constant acceleration and zero relationship and solve for change of block initial velocity. Knowing that B position at time t. velocity of collar A is 12 in./s as it passes L, determine the change in • Differentiate motion relation twice to elevation, velocity, and acceleration develop equations for velocity and of block B when block A is at L. acceleration of block B. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 22 - 49 11 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.5 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. v 2A = (v A )02 + 2a A [x A − ( x A )0 ] 2 ⎛ in. ⎞ ⎜12 ⎟ = 2a A (8 in.) ⎝ s ⎠ aA = 9 in. s2 v A = (v A )0 + a At 12 in. in. =9 2t s s t = 1.333 s © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 23 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.5 • Pulley D has uniform rectilinear motion. Calculate change of position at time t. x D = ( x D )0 + v D t ⎛ in. ⎞ x D − ( x D )0 = ⎜ 3 ⎟(1.333 s ) = 4 in. ⎝ s ⎠ • 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, x A + 2 x D + x B = ( x A )0 + 2( x D )0 + ( x B )0 [x A − ( x A )0 ]+ 2[x D − ( x D )0 ]+ [x B − (x B )0 ] = 0 (8 in.) + 2(4 in.) + [x B − ( x B )0 ] = 0 x B − ( x B )0 = −16 in. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 24 - 49 12 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.5 • Differentiate motion relation twice to develop equations for velocity and acceleration of block B. x A + 2 x D + x B = constant v A + 2v D + v B = 0 ⎛ in. ⎞ ⎛ in. ⎞ ⎜12 ⎟ + 2⎜ 3 ⎟ + v B = 0 ⎝ s ⎠ ⎝ s ⎠ v B = 18 in. s a A + 2a D + a B = 0 ⎛ in. ⎞ ⎜ 9 2 ⎟ + vB = 0 ⎝ s ⎠ in. a B = −9 2 s © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 25 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Rectangular Components of Velocity & Acceleration • When position vector of particle P is given by its rectangular components, r r r r r = xi + y j + zk • Velocity vector, r r r r dx r dy r dz r v = i + j + k = x&i + y& j + z&k dt dt dt r r r = vx i + v y j + vz k • Acceleration vector, r r r r d 2 xr d 2 y r d 2 z r a = 2 i + 2 j + 2 k = &x&i + &y& j + &z&k dt dt dt r r r = ax i + a y j + az k © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 26 - 49 13 Seventh Edition Vector Mechanics for Engineers: Dynamics Rectangular Components of Velocity & Acceleration • Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile, a x = &x& = 0 a y = &y& = − g a z = &z& = 0 with initial conditions, (v x )0 , (v y ) , (v z )0 = 0 x0 = y 0 = z 0 = 0 0 Integrating twice yields v x = (v x )0 x = (v x )0 t ( ) ( )0 v y = v y − gt 0 y = v y y − 12 gt 2 vz = 0 z=0 • Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 27 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.7 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 28 - 49 14 Seventh Edition Vector Mechanics for Engineers: Dynamics Tangential and Normal Components • 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. r r • et and et′ are tangential unit vectors for the particle path at P and P’. When drawn with r r r respect to the same origin, ∆et = et′ − et and ∆θ is the angle between them. ∆et = 2 sin (∆θ 2 ) r ∆et sin (∆θ 2 ) r r lim en = en = lim ∆θ →0 ∆θ ∆θ →0 ∆θ 2 r de r en = t dθ © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 29 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Tangential and Normal Components r r • With the velocity vector expressed as v = vet the particle acceleration may be written as r r r de dv r de dθ ds r dv dv r a= = et + v = et + v dt dt dt dt dθ ds dt but r det r ds = en ρ dθ = ds =v dt dθ After substituting, dv v2 r dv r v 2 r a = et + en at = an = ρ ρ dt dt • 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 30 - 49 15 Seventh Edition Vector Mechanics for Engineers: Dynamics Tangential and Normal Components • Relations for tangential and normal acceleration also apply for particle moving along space curve. r dv r v 2 r a = et + en ρ dt at = dv dt an = v2 ρ • Plane containing tangential and normal unit vectors is called the osculating plane. • Normal to the osculating plane is found from r r r eb = et × en r en = principal normal r eb = binormal • Acceleration has no component along binormal. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 31 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.10 SOLUTION: • Calculate tangential and normal components of acceleration. • Determine acceleration magnitude and direction with respect to tangent to curve. 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 32 - 49 16 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.10 SOLUTION: • Calculate tangential and normal components of acceleration. ∆v (66 − 88) ft s ft = −2.75 2 at = = ∆t 8s s an = 60 mph = 88 ft/s 45 mph = 66 ft/s v2 ρ = (88 ft s )2 2500 ft = 3.10 ft s2 • Determine acceleration magnitude and direction with respect to tangent to curve. ft a = at2 + an2 = (− 2.75)2 + 3.10 2 a = 4.14 2 s α = tan −1 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. an 3.10 = tan −1 at 2.75 α = 48.4° 33 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.11 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 34 - 49 17 Seventh Edition Vector Mechanics for Engineers: Dynamics Radial and Transverse Components • 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 r = re r r de r r = eθ dθ • The particle velocity vector is r de dr r dr r dθ r r d r v = ( re r ) = e r + r r = e r + r eθ dt dt dt dt dt r r = r& er + rθ& eθ r deθ r = − er dθ r r der der dθ r dθ = = eθ dt dt dθ dt r r deθ deθ dθ r dθ = = − er dt dt dθ dt • Similarly, the particle acceleration vector is dθ r ⎞ r d ⎛ dr r a = ⎜ er + r eθ ⎟ dt ⎝ dt dt ⎠ r r d 2 r r dr der dr dθ r d 2θ r dθ deθ = 2 er + + eθ + r 2 eθ + r dt dt dt dt dt dt dt dt r r = &r& − rθ& 2 er + (rθ&& + 2r&θ& )eθ ( ) © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 35 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Radial and Transverse Components • When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit r r r vectors eR , eθ , and k . • Position vector, r r r r = R e R +z k • Velocity vector, r r r r dr & r v= = R eR + Rθ& eθ + z& k dt • Acceleration vector, r r r dv && − Rθ& 2 erR + (Rθ&& + 2 R& θ& )erθ + &z& k = R a= dt ( © 2004 The McGraw-Hill Companies, Inc. All rights reserved. ) 36 - 49 18 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 SOLUTION: • Evaluate time t for θ = 30o. • Evaluate radial and angular positions, and first and second derivatives at time t. 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. • Calculate velocity and acceleration in cylindrical coordinates. • Evaluate acceleration with respect to arm. 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 37 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 SOLUTION: • Evaluate time t for θ = 30o. θ = 0.15 t 2 = 30° = 0.524 rad t = 1.869 s • Evaluate radial and angular positions, and first and second derivatives at time t. r = 0.9 − 0.12 t 2 = 0.481 m r& = −0.24 t = −0.449 m s &r& = −0.24 m s 2 θ = 0.15 t 2 = 0.524 rad θ& = 0.30 t = 0.561 rad s θ&& = 0.30 rad s 2 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 38 - 49 19 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 • Calculate velocity and acceleration. vr = r& = −0.449 m s vθ = rθ& = (0.481m )(0.561rad s ) = 0.270 m s v β = tan −1 θ v = vr2 + vθ2 vr v = 0.524 m s β = 31.0° ar = &r& − rθ &2 = −0.240 m s 2 − (0.481m )(0.561rad s )2 = −0.391m s 2 aθ = rθ&& + 2r&θ& ( ) = (0.481m ) 0.3 rad s 2 + 2(− 0.449 m s )(0.561rad s ) = −0.359 m s a = ar2 + aθ2 2 a γ = tan −1 θ ar a = 0.531 m s γ = 42.6° © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 39 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 • Evaluate acceleration with respect to arm. Motion of collar with respect to arm is rectilinear and defined by coordinate r. a B OA = &r& = −0.240 m s 2 © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 40 - 49 20 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion Relative to a Frame in Translation • 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 r r the fixed frame of reference Oxyz are rA and rB . r • Vector rB A joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and r r r rB = rA + rB A • Differentiating twice, r r r r vB = v A + vB A vB r r r a B = a A + aB A A = velocity of B relative to A. r aB A = acceleration of B relative to A. • 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 41 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Motion of Several Particles: Relative Motion • 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. = x B − x A = relative position of B with respect to A xB = x A + xB A xB A = v B − v A = relative velocity of B with respect to A vB = v A + vB A vB A = a B − a A = relative acceleration of B with respect to A aB = a A + aB A aB A © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 42 - 49 21 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.4 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. 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. • 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. © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 43 - 49 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.4 SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. v B = v0 + at = 18 m ⎛ m⎞ − ⎜ 9.81 2 ⎟t s ⎝ 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 of elevator into equation for uniform rectilinear motion. vE = 2 m s ⎛ m⎞ y E = y0 + v E t = 5 m + ⎜ 2 ⎟t ⎝ s⎠ © 2004 The McGraw-Hill Companies, Inc. All rights reserved. 44 - 49 22 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 11.4 • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. yB E ( ) = 12 + 18t − 4.905t 2 − (5 + 2t ) = 0 t = −0.39 s (meaningless ) t = 3.65 s • Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. y E = 5 + 2(3.65) vB E y E = 12.3 m = (18 − 9.81t ) − 2 = 16 − 9.81(3.65) vB © 2004 The McGraw-Hill Companies, Inc. All rights reserved. E = −19.81 m s 45 - 49 23