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Ch. 9: Plane of Rigid Bodies 473 9.0 Outline 473

 Introduction 474  Rotation 478  Absolute 486  Relative 508  Instantaneous Center of Zero Velocity 531  Relative 544  Motion Relative to Rotating Axes 566

9.0 Outline Ch. 9: Plane Kinematics of Rigid Bodies 474 9.1 Introduction Kinematics of rigid bodies involves both linear and angular quantities.

Usage 1. In designing the machines to perform the desired motion. 2. To determine the motion resulting from the applied .

A A system of particles for which the distance between the particles remain unchanged. Thus there will be no change in the position vector of any particle measured from the body-fixed coordinate system.

9.1 Introduction Ch. 9: Plane Kinematics of Rigid Bodies 475

9.1 Introduction Ch. 9: Plane Kinematics of Rigid Bodies 476 Plane motion of a rigid body all parts of the body move in parallel planes. The body then can be treated as a thin slab with motion confined to the plane of motion; plane that contains the center.

Translation motion in which every line in the body remains parallel to its original position at all . That is there is no rotation of any line in the body. The motion of the body is completely specified by the motion of any point in the body, since all points have same motion. Rectilinear translation all points in the body move in parallel straight lines of the same distance Curvilinear translation all points move on parallel curves of the same distance

9.1 Introduction Ch. 9: Plane Kinematics of Rigid Bodies 477 Rotation motion in which all particles move in circular paths about the axis of rotation. All lines in the body which are perpendicular to the axis of rotation rotate through the same angle in the same time. of a point helps describe the rotating motion.

General plane motion combination of translation and rotation. Principle of relative motion helps describe the general motion.

Approaches 1. Direct calculation of the absolute , velocity, and acceleration from the geometry 2. Use the principle of relative motion

9.1 Introduction Ch. 9: Plane Kinematics of Rigid Bodies 478 9.2 Rotation Rotation of a rigid body is described by its angular motion, which is dictated by the change in the angular position (specified by angle θ measured from any fixed line) of any line attached to the body.

θθβ21= +

∆=∆θθ21  θθ21=   θθ21= All lines on a rigid body in its plane of motion have the same , the same , and the same

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 479 Angular motion relations

Pick up any line in the plane of motion and associated it with the angular position coordinate θ. The angular velocity ω and accelerationα of a rigid body in plane rotation are defined. ωθ=  αωθ= =  ωωd= αθ d or θθ d = θθ  d Analogies between the linear and angular motion

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 480 Rotation about a fixed axis

All points other than those on the rotation axis move in concentric circles about the fixed axis. Curvilinear motion of a point A is related to the angular motion of the rigid body by the familiar n-t coordinate kinematic relationship. v= rω , r = constant

at = vr = α v2 ar=ω 2 = n r

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 481 Rotation about a fixed axis

v = r =ω ×= r by vector differentiation, r constant ω = angular velocity of the vector r = angular velocity of the rigid body a=v= ωω××( r) + α × r

arn =××ωω( )

art =α ×

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 482 P. 9/1 The two V-belt pulleys form an integral unit and rotate about the fixed axis at O. At a certain instant, point A on the belt

of the smaller pulley has a velocity vA = 1.5 m/s, and point B 2 on the belt of the larger pulley has an acceleration aB = 45 m/s as shown. For this instant determine the magnitude of

the acceleration aC of point C and sketch the vector in your solution.

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 483 P. 9/1

find ωpulley from vA [v==×=ωωω r] 1.5 0.075, 20 rad/s CW

find αpulley from a B 2 [at = rα] 45 =×= 0.4 αα , 112.5 rad/s CCW acceleration at C a= rω 2 a = 0.36 ×= 2022 144 m/s nCn a= rα a =×= 0.36 112.5 40.5 m/s2 [ tC] t a =a22 += a 149.6 m/s2 C CCnt

aCn aCt

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 484 P. 9/2 A V-belt -reduction drive is shown where pulley A drives the two integral pulleys B which in turn drive pulley C. If A starts from rest at time t = 0 and is given a constant angular

acceleration α1, derive expressions for the angular velocity of C and the magnitude of the acceleration of a point P on the belt, both at time t.

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 485 P. 9/2

2 dω rr  α= ω = αω = 11 αω= α  A1B t,  1C t,  1t dt r22 r 2 rr11  αA1B= αα, =  αα 1C, =  α 1 rr22  4 r 2 ar=ωα2 a = r1 ( t) n P2n  1 r2 2 r a= rαα a = r 1 [ t] P2t  1 r2

4 rr2  a= aa22 +=11αα 1 + 24t P PPnt 1 1 rr22

9.2 Rotation Ch. 9: Plane Kinematics of Rigid Bodies 486 9.3 Absolute Motion It is an approach to the kinematics analysis. It starts with the geometric relations that define the configuration involved. Then, the time derivatives of the relations are done to obtain and . The +/- sense must be kept consistent throughout the analysis.

If the geometric configuration is too complex, resort to the principle of relative motion is recommended.

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 487 P. 9/3 A wheel of radius r rolls on a flat surface w/o slipping. Determine the angular motion of the wheel in terms of the of its center O. Also determine the acceleration of a point on the rim of the wheel as the point comes into contact with the surface on which the wheel rolls.

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 488 P. 9/3 For ROLLING w/o SLIPPING only: displacement of the center O= arc length along the rim of the wheel distance s= arc C'A →= s rθ

time derivatives: vOO= rωα and a= r

vO = velocity of the center O of the wheel rolling w/o slipping ω = angular velocity of the wheel rolling w/o slipping DO NOT mix up the rolling motion with the rotation about a fixed axis

Set up a fixed coordinate system originated at the point of contact between the rim and the ground. Trajectory of a point C on the rim is a cycloidal path. x-y coordinates of point C at arbitrary position C' is x=−=− s rsinθθ r( sin θ) y =−=− r rcos θ r( 1 cos θ) time derivatives:  x =−=− rθ( 1 cos θ) vOO( 1 cos θ) y = r θθ sin = v sin θ  2 x=−+ v  O( 1cosθ) v OO θθ sin =−+ a( 1cos θ) r ω sin θ  2 y=+=+ vsin  OOθ v θ cos θ asin O θω r cos θ when point C comes to contact, θ = 0 x= 0, y  = 0, x = 0, and y = rω 2 9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 489 P. 9/4 The load L is being hoisted by the pulley and cable arrangement shown. Each cable is wrapped securely around its respective pulley so it does not slip. The two pulleys to which L is attached are fastened together to form a single rigid body. Calculate the velocity and acceleration of the load L and the corresponding angular velocity ω and angular acceleration α of the double pulley under the following conditions:

Case (a) Pulley 1: ω1 = 0, α1 = 0 (pulley at rest)

Pulley 2: ω2 = 2 rad/s 2 α2 = -3 rad/s

Case (b) Pulley 1: ω1 = 1 rad/s 2 α1 = 4 rad/s Pulley 2: ω2 = 2 rad/s 2 α2 = -2 rad/s

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 490 P. 9/4

Tangential displacement, velocity, and acceleration of a point on the rim of the wheel equal the corresponding motion of the points on the wrapped no-slip, inextensible cable.

(a) During time dt, line AB (on the integral pulley) moves to AB' through angle dθ . New ( tangential) position of point A and B are determined from the motion induced by the cable. = =ωα = = = = − 2 vB v D22 r 0.2 m/s and ( aB)t a D22 r 0.3 m/s dsA = 0 left cable does not move =θω = = α dsBBBAB d v AB ( a )t AB AB' maintains straight and point O has absolute motion in vertical direction: dsO= AO dθωα v OO= AO a = AO ∴=ωα0.2 / 0.3 = 2 / 3 rad/s CCW, =− 0.3/ 0.3 =− 1 rad/s2 CW 2 vOO= 0.1 × 2 / 3 = 0.0667 m/s and a= 0.1 ×− 1 =− 0.1 m/s

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 491 P. 9/4 (b) During time dt, line AB moves to A'B' through angle dθ . New ( tangential) position of point A and B are determined from the motion induced by the cable. = =ωα = = = = 2 vA v C11 r 0.1 m/s and ( aA)t a C11 r 0.4 m/s = =ωα = = = = − 2 vvBDr22 0.2 m/s and ( aB)t a D r 22 0.2 m/s − =θω −= − = α dsBA dsAB d vBA vAB ( aBA)tt( a ) AB − =θω −= − = α dsOA dsAO d vOA vAO aOA( a )t AO ∴=ωα(0.2 − 0.1) / 0.3 = 1/ 3 rad/s CCW, =−−( 0.2 0.4) / 0.3 =− 2 rad/s2 CW 2 vO =+× 0.1 0.1 1/ 3 = 0.133 m/s and aO = 0.4 + 0.1 ×− 2 = 0.2 m/s

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 492 P. 9/5 The telephone-cable reel rolls w/o slipping on the horizontal surface. If point A on the cable has a velocity vA = 0.8 m/s to the right, compute the velocity of the center O and the angular velocity ω of the reel. (Be careful not to make the mistake of assuming that the reel rolls to the left.)

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 493 P. 9/5

rolling w/o slipping: velocity at the contact point= 0 no slippage at the inner hub: velocity of the rim= velocity of the wrapped cable OC remains straight and determines the angular motion v 0.8 O = , v= 1.2 m/s → 0.9 0.6 O 1.2 [v=ωω r] = = 1.333 rad/s CW O 0.9

Vo Va = 0.8 m/s

C 9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 494 P. 9/6 The cable from drum A turns the double wheel B, which rolls on its hubs w/o slipping. Determine the angular velocity ω and angular acceleration α of drum C for the instant when the angular velocity and angular acceleration of A are 4 rad/s and 3 rad/s2, respectively, both in the CCW direction.

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 495 P. 9/6

[v= rω] vA = 0.2 ×= 4 0.8 m/s → =α = ×=2 → [atA r] ( a)t 0.2 3 0.6 m/s 0.8 0.6 ωα= = 4 / 3 rad/s CW = = 1 rad/s2 CW BB0.6 0.6

vC = 4 / 3 ×= 0.2 0.267 m/s ← =×=2 ← (aC )t 1 0.2 0.2 m/s 0.267 0.2 ωα= = 4 / 3 rad/s CCW = = 1 rad/s2 CCW CC0.2 0.2

Va

Vc 9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 496 P. 9/7 The rod OB slides through the collar pivoted to the rotating link at A. If OA has an angular velocity ω = 3 rad/s for an interval of motion, calculate the angular velocity of OB when θ= 45°.

β y

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 497 P. 9/7

given: θθ=°=− 45 ,  3 rad/s, CA = 0.2 m/s geometric relation: ysinβ−CA sin θ = 0 and CA cos θβ += ycos 0.4 solve for y= 0.2947 m and β = 28.675 ° diff w.r.t. time: ysin βββ+− ycosCA θ cos θ = 0 −CAθsin θ +− ycos β y ββ sin = 0 ∴y =−=0.576 m/s and β − 0.572 rad/s

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 498

P. 9/8 Show that the expressions v = rω and at = rα hold for the motion of the center O of the wheel which rolls on the concave or convex circular arc, where ω and α are the absolute angular velocity and acceleration, respectively, of the wheel. (Hint: Follow the sample problem and allow the wheel to roll a small distance. Be very careful to identify the correct absolute angle through which the wheel turns in each case in determining its angular velocity and angular acceleration.)

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 499 P. 9/8 concave arc β geometry: θ = angular motion of the wheel rolling distance= Rβ = r( θβ + ) ( 1) motion of the center O= s =( R − r) β θ β rearrange ( 1) to give ( R−= r) βθ r vs==−== ( Rr) βθω r r  at ==−== v ( Rr) βωα r r

convex arc geometry: θ = angular motion of the wheel θ rolling distance= Rβ = r( θβ − ) ( 2) β motion of the center O= s =( R + r) β rearrange ( 2) to give ( R+= r) βθ r  vs==+== ( Rr) βθω r r β  at ==+== v ( Rr) βωα r r independent of the curvature, R, of the terrain! 9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 500 P. 9/9 The Geneva wheel is a mechanism for producing intermittent rotation. Pin P in the integral unit of wheel A and locking plate B engages the radial slots in wheel C thus turning wheel C one-fourth of a revolution for each revolution of the pin. At the engagement position shown, θ= 45°. For a constant CW angular velocity ω1 = 2 rad/s of wheel A, determine the corresponding CCW angular velocity ω2 of the wheel C for θ= 20°. (Note that the motion during engagement is governed by the geometry of triangle O1O2P with changing θ.)

β

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 501 P. 9/9

triangle O12 O P: 1 sinθ OPsinθ tanβ = 1 = 2 − θ 1 OO12 OP 1 cos 1− cosθ 2 1  1 11 1− cosθθθθθθ  × cos −× sin sin  2 2  2 22 ββsec = 2 1 1− cosθ 2 given: θθ=°=− 20 ,  2 rad/s, θ = 0 ββ=35.783 °=− ,  1.923 rad/s

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 502 P. 9/10 The rod AB slides through the pivoted collar as end A moves along the slot. If A starts from rest at x = 0 and moves to the right with a constant acceleration of 0.1 m/s2, calculate the angular acceleration α of AB at the instant when x = 150 mm. 0.15 m

θ

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 503 P. 9/10 x= 0.1 m/s2 constant x = 0.1t x = 0.05t2

geometry: x= 0.2 tanθ x = 0.2θθ sec2 x= 0.2θ sec22 θ +× 0.2 θ ( 2 θ  sec θθ tan ) at x= 0.15 m, t = 3 sec →= x 0.1 3 and x = 0.1 at that posture, tanθθ= 3/ 4 sec = 5 / 4 ∴=θ 0.554 rad/s and θ = − 0.1408 rad/s2

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 504 P. 9/11 The punch is operated by a simple harmonic oscillation of the pivoted sector given by

θθ= O sin 2 π t where the amplitude is

θπO = /12 rad ( 15°) and the time for one complete oscillation is 1 second. Determine the acceleration of the punch when (a) θ= 0 and (b) θ= π/12.

y

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 505 P. 9/11   2 θθ= OOsin 2 πθ t, = 2 πθ cos 2 πθ t, = −( 2 πθ) O sin 2 π t 0.122=+− y 0.14 2 0.28ycosθ 0=−+ 2yy 0.28ycosθ 0.28y θθ sin 0=+− 2y2 2yy  0.28ycos θ + 0.28y  θθ sin +++ 0.28yθθθθθθ sin 0.28y sin 0.28y 2 cos

  when θ= 0, t = 0, θ = 2 πθO , θ = 0 y= 0.24, 0.04 y= 0, y = − 0.909 m/s2

  2 when θπ= /12, t = 1/ 4, θ = 0, θ = −( 2 πθ) O y= 0.2284, 0.042 y= 0, y = 0.918 m/s2

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 506 P. 9/12 One of the most common mechanisms is the

slider-crank. Express the angular velocity ωAB and angular acceleration αAB of the connecting rod AB in terms of the crank angle θ for a given

constant crank speed ωO. Take ωAB and αAB to be positive counterclockwise.

β

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 507 P. 9/12   given: θω=O , θ = 0 geometry: lr sinβθ= sin diff w.r.t. time: lrβ cos βθ= cos θ rrωcos θω cosθ β =OO = llcos β r 2 1− sin2 θ l 2 lβcos ββ−=− l 22 sin β rr θ  cos θθ  sin θ r 2 22 2 −1 lrβsin βθ− sin θrω 2 βθ = = O sin l llcos β 2 3/2 r 2 1− 2 sin θ l

9.3 Absolute Motion Ch. 9: Plane Kinematics of Rigid Bodies 508 9.4 Principle of relative motion is another way to solve the kinematics problems. This method is usually suitable to the complex motion as it is more scalable.

Velocity propagation in the rigid body Refer to Chapter 2, the relative velocity equation using the non- is

vA= vv B + A/B Let the two points, A and B, are on the same rigid body. The consequence of this choice is that the motion of one point as seen by an observer translating with the other point must be circular since the radial distance to the observed point from the reference point does not Changed. 9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 509

Movement of the rigid body is partitioned into two parts: translation and rotation. In the figure, after the translation of the rigid body, expressed by the motion of B, the body appears to undergo fixed-axis rotation about B with A executing circular motion as shown in (b). Hence the relationship for circular motion describes the relative portion of A’s motion. 9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 510 With B as the reference point, total displacement of A is

∆rA =∆ rr B +∆ A/B

∆rrA/B = −∆ B/A has the magnitude r∆θθ as ∆ → 0

** Relative linear motion ∆rA/B is accompanied by the absolute angular motion as seen from the translating axes x'-y'. vA =v B+ v A/B

If distance r between A and B is constant, vA/B is the velocity of the circular motion. That is vrA/B =×=⊥ω A/B v A/B r A/Bω v A/B AB ω = absolute angular velocity of rigid body

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 511

general translational rotational motion motion motion

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 512 Velocity propagation among rigid bodies Refer to Chapter 7, the relative velocity equation using the non-rotating reference frame is

vA= vv B + A/B This time point A and B are coincident points on different rigid bodies for the instant. In this case, the distance between two points are not constrained to be fixed, even it is zero at the instant.

Point A is on the red linkage. Point B is on the blue linkage. They are, at the instant, coincided and are at the position of the pin in the slot.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 513 Methods in solving the relative velocity equation

vA= vv B + A/B 1. Vector algebra approach 2. Graphical analysis approach 3. Vector/Graphic approach

** Sketch of the vector polygon representing the vector equation is helpful!

Vector algebra approach Write each term in terms of i- and j-components  two scalar equations  at most two unknowns.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 514 Graphical analysis approach Known vectors are constructed using a convenient scale. Unknown vectors which complete the polygon are then measured directly from the drawing. This is suitable when the vector terms result in an awkward math expression.

Vector/Graphic approach Scalar component equations may be written by projecting the vectors along convenient directions. Simultaneous equations may be avoided by a careful choice of the projections.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 515 P. 9/13 The wheel of radius r = 300 mm rolls to the

right w/o slipping and has a velocity vo = 3 m/s of its center O. Calculate the velocity of point A on the wheel for the instant represented.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 516 P. 9/13

[vA =v O+ v A/O ]

[vrA/O =ω × A/O ]

[ωω=vO / r] = 3/ 0.3 = 10 rad/s CW

[vA =v C+ v A/C ]

v0C = for rolling w/o slipping wheel

[vrA/C =ω × A/C ]

vA = 4.36 m/s  23.4°

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 517 P. 9/14 The screw turns at a speed that gives the threaded collar C a velocity of 0.25 m/s vertically down. Determine the angular velocity of the slotted arm when θ= 30°.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 518 P. 9/14 point A on the slotted arm, point B on the collar A and B coincide at θ = 30 °

Because of the contact constraint from the slot,

vA/B has the direction along the slot (away from O for CCW ω ).

[vA =v BA+ v /B] v A = 0.25cos30 = 0.217

ω =vA /OA = 0.417 rad/s CCW

Imagine B immobile, A will be seen moving radially outward from O along the slot.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 519 P. 9/15 The rotation of the gear is controlled by the horizontal motion of end A of the rack AB. If the piston rod has a constant velocity 300 mm/s during a short interval of motion, determine the angular velocity of the gear and the angular velocity of AB at the instant when x = 800 mm.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 520 P. 9/15 C is the point on the rack AB at where the line AB made a contact with the pitch of the gear ∴point C is constrained to have the same velocity as the coincident point on the gear ( but not acceleration)

∴⊥vC OC −1 [vvvC=+==° A C/A ] θ sin( 0.2 / 0.8) 14.48

vC==×= 0.3cosθω OO 0.2 ω 1.45 rad/s CW

22 vC/A = 0.3sinθω =×−AB 0.8 0.2 ωAB = 0.0968 rad/s CCW

C vA θ vC/A θ vC A O

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 521 P. 9/16 The flywheel turns CW with a constant speed of 600 rev/min, and the connecting rod AB slides through the pivoted collar at C. For the position of θ= 45°, determine the angular velocity of AB by using the relative velocity relations. (Choose a point D on AB coincident with C as a reference point whose direction of velocity is known.)

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 522 P. 9/16

D= point on linkage AB that coincides with pivot point C on collar

[vvvD=+= C D/C] vv D D/C along the slot and pointing out of point A

[vvvA=+ D A/D] v A =×× 0.2( 600 2π / 60) = 12.566 m/s

vA/D= v A sin 59.63= 10.84 =ωAB × 0.56

ωAB =19.36 rad/s CW vA

vA/D A 0.56 m 0.2 m v 14.63° D O 0.4 m D

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 523 P. 9/17 The Geneva mechanism is shown again here. By relative motion principles, determine the angular velocity of wheel C for θ= 20°. Wheel A has a constant CW angular velocity ω1 = 2 rad/s.

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 524 P. 9/17 P is the point on wheel A at the knob Q is the point on the wheel C, coincident with P when θ = 20 ° = + vvvP Q P/Q v P/Q along the slot pointing to O2

vP =×= 2( 0.2 / 2) 0.283 m/s

vQP= v sin 34.355 =×=ωω2 0.083 2 1.924 rad/s CCW

P, Q 0.083 m

20°vQ 35.6° O1 O2

34.35° v vP/Q P

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 525 P. 9/18 At the instant represented, a = 150 mm and b = 125 mm, and the distance a+b between A and C is decreasing at the rate of 0.2 m/s. Determine the common velocity v of points B and D for this instant.

53.13° 67.38°

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 526 P. 9/18

[vvvB= A + B/A ]

[vD= vv C + D/C ]

constraint: vvBD= a +=−→ b 0.2 block A and C become closer and from the mechanism, A moves to the right and B to the left

∴−=vvAC0.2 i m/s from the velocity diagram vv BB+=0.2, vj= 0.0536 m/s tan 22.62 tan 36.87 B

vD/C vB/A

vB, vD

vC vA 9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 527 P. 9/19 The wheel rolls w/o slipping. For the instant portrayed, when O is directly under point C, link OA has a velocity v = 1.5 m/s to the right and θ= 30°. Determine the angular velocity ω of the slotted link.

vQ 23.78° 124 mm

vP/Q

vP 15°

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 528 P. 9/19

[vO==×=ω r] 1.5 ωω OO 0.1, 15 rad/s CW

vPO=ω ××( 2 0.1cos15) = 2.9 m/s P is the point at the pin on the disk Q is the coincident point on the slotted arm = + vvvP Q P/Q v P/Q directs along the slot outward point C

vQP= v cos( 15 += 23.78) 2.26 m/s

ωC = 2.26 / 0.124= 18.23 rad/s CCW

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 529 P. 9/20 Ends A and C of the connected links are controlled by the vertical motion of the piston rods of the hydraulic cylinders. For a short interval of motion, A has an upward velocity of 3 m/s, and C has a downward velocity of 2 m/s. Determine the velocity of B for the instant when y = 150 mm.

60° 36.87°

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 530 P. 9/20 =+=+ vvvB C B/C vv A B/A vB/A 96.87° from the velocity diagram v vA vB/C B 5 vv =B/C = B/A 3 m/s sin60 sin 96.87 sin 23.13

vB/C = 5.732 m/s and vB/A = 2.268 m/s 222 2 m/s (vB/C sin 23.13) ++( 3 vB/A cos83.13) = vB vC vB = 3.97 m/s directed along the dotted arrow 23.13°

9.4 Relative Velocity Ch. 9: Plane Kinematics of Rigid Bodies 531 9.5 Instantaneous Center of Zero Velocity (ICZV) Principle of relative motion find the velocity of a point on a rigid body by adding the relative velocity, due to rotation about a reference point, to the velocity of the reference point. If the reference point has zero velocity momentarily, the body may be considered to be in pure rotation about an axis, normal to the plane of motion, passing through this point. This point is called ICZV, which aids in visualizing and analyzing velocity in plane motion.

ICZV has zero velocity but not acceleration  not ICZA

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 532

Point A and B, instantaneously, have absolute circular motion about point C C = ICZV = instantaneous center of rotation C may not be on the body physically, visualized as lying on the extended body ICZV is not a fixed point in the body nor a fixed point in the plane

Ω= vA/rA = vB/rB, vB = (rB/rA)vA If the body translates only, ICZV is at infinity along the perpendicular line to the velocity 9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 533 P. 9/21 Vertical oscillation of the spring-loaded plunger F is controlled by a periodic change in pressure in the vertical hydraulic cylinder E. For the position θ= 60°, determine the angular velocity of AD and the velocity of the roller A in its horizontal guide if the plunger F has a downward velocity of 2 m/s.

vA

ICZV

vB

vD vF 9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 534 P. 9/21

vB is forced to move downward by the hydraulic

vA is forced to move horizontally by the guide

∴ICZV is at point C →⊥vD CD plunger moves down 2 m/s

∴vertical component of vD is 2 m/s downward

vD = 2 / cos30 =ωAD ×( 2×=0.1cos30) , ωAD 13.33 rad/s CW

vA=×=ω AD ( 0.2sin 60) 2.309 m/s, to the right

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 535 P. 9/22 Determine the angular velocity ω of the ram head AE of the rock crusher in the position for which θ= 60°. The crank OB has an angular speed of 60 rev/min. When B is at the bottom of its circle, D and E are on a horizontal line through F, and lines BD and AE are vertical. The dimension are OB = 100 mm, BD = 750 mm, and AE = ED = DF = 375 mm. Carefully construct the configuration graphically, and use the method of Art. 5/5.

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 536 P. 9/22

l1 vB

α ICZVBD ICZVDE l3 l2 θ

l4 ξ β

vE vD 9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 537 P. 9/22

100

475

850 750

375

375 375

Home position of the machine

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 538 P. 9/22 Graphical method is the best way for this problem! Here we show the vector/graphic approach

from the initial and current posture, between point O and F: 100sin 60−+= 750cosαβ 375cos 375 −100cos60 ++ 750sinα 375sinβ = 850 by the help of complex exponent, αβ= 85.85 °= , 23.9 °

between point O and A: 375cosθξα++ 375cos 750cos − 100sin 60 = 375 −+375sinθξα 375sin + 750sin − 100cos60 = 475 by the help of complex exponent, θξ= 81.04 °= , 21.49°

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 539 P. 9/22 find distance from ICZV to point of interest

ll12sin 30=−+− 100cos60 750sinαβ sin

ll12cos30=+− cosβα 750cos 100sin 60

ll12= 773.63 mm, = 768.2 mm

ll34cosβθ= cos + 375cos ξ

ll44sinβθ= sin − 375sin ξ

ll34= 435.8 mm, = 317.8 mm

vB =×=+× 60 100(l1 100) ωωBD , BD = 6.868 rev/min CW

vD=ω BD ×=ll 2 5276 mm rev/min=×3ωω DE , DE = 12.1 rev/min CW

vE=ωω DE ×=l 4 AE ×375, ωAE = 10.26 rev/min = 1.07 rad/s CW

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 540 P. 9/23 The shaft at O drives the arm OA at a clockwise speed of 90 rev/min about the fixed bearing at O. Use the method of ICZV to determine the of gear B (gear teeth not shown) if (a) ring gear D is fixed and (b) ring gear D rotates CCW about O with a speed of 80 rev/min.

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 541 P. 9/23 Motion of gear A is constrained by 1. motion of link OA at point A: v= v AAgear link 2. motion of the leftmost point on the rim: v= v CCgearA gearB

3. motion of the rightmost point on the rim: vDD= v gearA gearD 90a

(a) ring gear D is fixed

vA = 90a ↓

vC=×=× 2 90aωω BB a/2, = 360 rev/min CW 120a (b) ring gear D rotates 80 rev/min CCW 3a vAD=↓= 90a , v 80 = 120a ↑ 2 10 / 7 90a v=( 90a) =×=ωω a/2, 600 rev/min CW C3/7 BB

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 542 P. 9/24 The large roller bearing rolls to the left on its outer race with a velocity of its center O of 0.9 m/s. At the same time the central shaft and inner race rotate CCW with an angular speed of 240 rev/min. Determine the angular velocity of each of the rollers.

0.9 m/s

0.3566 m/s

0.18 m/s x

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 543 P. 9/24

vO = 0.9 m/s →

ωπi = 240 rev/min= 8 rad/s CCW 0.9 [v= ω r] ICZV of the inner race is m lower to point O 8π velocity of the point on the roller contacting w/ inner race 0.9 =×−=8π  0.05 0.3566 m/s → 8π

0.9 ω = = 7.2 rad/s CCW O 0.125 velocity fo the point on the roller contacting w/ the outer race =×=7.2 0.025 0.18 m/s ← 0.18 x by similar triangle, = , x= 16.77 mm 0.3566 0.05− x

ωroller =0.18/ x = 10.732 rad/s CW

9.5 ICZV Ch. 9: Plane Kinematics of Rigid Bodies 544 9.6 Relative Acceleration Relative acceleration relationship with nonrotating reference axes can be obtained from differentiating the relative velocity relation w.r.t. time:

aA= aa B + A/B

aA/B is the acceleration that A appears to have to a nonrotating observer moving with B.

If A and B are two points on the same rigid body in plane motion, the distance between them remains constant so that the observer moving with B perceives A to have circular motion about B.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 545 Therefore, relative acceleration term can be partitioned into the normal component, directed from A toward B due to the change of direction of v A/B , and the tangential component, perpendicular to AB due to change in magnitude of v A/B . =++ aaA B( a A/B)nt( a A/B ) =××ωω =ω 22 = (arA/B )nn( ), ( aA/B ) r vA/B /r =×==α α (arA/B )tt, ( aA/B ) r v A/B

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 546

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 547 Methods in solving the relative acceleration equation

aA= aa B + A/B 1. Vector algebra approach 2. Graphical analysis approach 3. Vector/Graphic approach

** Sketch of the vector polygon representing the vector equation is helpful!

Vector algebra approach Write each term in terms of i- and j-components  two scalar equations  at most two unknowns.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 548 Graphical analysis approach Known vectors are constructed using a convenient scale. Unknown vectors which complete the polygon are then measured directly from the drawing. This is suitable when the vector terms result in an awkward math expression.

Vector/Graphic approach Scalar component equations may be written by projecting the vectors along convenient directions. Simultaneous equations may be avoided by a careful choice of the projections.

Because an depend on velocities, normally it is required to solve for the velocities before the acceleration calculation can be made. 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 549 P. 9/25 If the wheel in each case rolls on the circular surface w/o slipping, determine the acceleration of the point C on the wheel momentarily in contact with the circular surface. The wheel has an angular velocity ω and an angular acceleration α.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 550 P. 9/25

vC = 0 because no slipping =+==ωα [aaaC O C/O ] and vO r ia , ( O )t r i =++ + (a) aaC( O)tn( a O) ( a C/O) t( a C/O ) n 22 rRω 22 = rαi+ j− rr αω i+ j= r ω j Rr−−Rr

22R aC = rrωω> Rr−

=++ + (b) aaC( O)tn( a O) ( a C/O) t( a C/O ) n 22 r ω 22R = rαij − − rrαωi+ j= r ω j Rr+ Rr+

22R aC = rrωω< Rr+

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 551 P. 9/26 The simplified clam-shell bucket is shown. With the block at O considered fixed and with the constant velocity of the control cable at C equal to 0.5 m/s, determine the angular acceleration α of the right-hand bucket jaw when θ= 45° as the bucket jaws are closing.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 552 P. 9/26 50.345° 0.0782

22.5°

(a ) B t 0.2604

(aB/C)t ° 50.345 sketch diagram 574.24 mm 62.155° 67.5°

ICZV

correct diagram

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 553 P. 9/26 velocity analysis use ICZV locate ICZV of the right jaw: CD= CO tan 50.345 = 692.78 mm

ωBC =0.5/CD = 0.7212 rad/s CW closing

vB=×ω BC (CO / cos50.345 −=BO) 0.2165 m/s

ωBO=v B /BO = 0.361 rad/s CCW acceleration analysis =+ += + [aB aaC B/C] ( a B)nt( a B) ( a B/C) n( a B/C ) t =ωω2222 = = = (a B/C )nnBC BC 0.2604 m/s ( a B ) BO BO 0.0782 m/s horizontal direction: +− − = 0.2604cos22.5( aB/C )tt sin 22.5( aB ) cos50.345 0.0782sin 50.345 0 vertical direction: − −+ = 0.2604sin22.5( aB/C )t cos 22.5( aB )t sin 50.345 0.0782cos50.345 0 =−=22 (aB/C )tt 0.049 m/s ( wrong direction) ( aB ) 0.2532 m/s 2 αBC =0.049 / 0.5 = 0.098 rad/s CW 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 554 P. 9/27 The mechanism where the flexible band F attached to the sector at E is given a constant velocity of 4 m/s as shown. For the instant when BD is perpendicular to OA, determine the angular acceleration of BD.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 555 P. 9/27 point E moves in circular path about O with velocity 4 m/s 4 ωOA = = 20 rad/s CCW 0.2 vA

vA=×=ω OA 0.125 2.5 m/s ↑ (4) velocity analysis vD/A

[vvvD=+== A D/A] v D / v A 3/ 4, vD 1.875 m/s (5)

vD/A / v A = 5/ 4, vD/A = 3.125 m/s

ωBD=v D / 0.25 = 7.5 rad/s CCW, ωAD= v D/A / 0.25 = 12.5 rad/s CCW vD (3) acceleration analysis

22 [aaaD=+ A D/A] a A = 0.125 ×= 20 50 m/s ← =×=22 (aD/A )n 0.25 12.5 39.0625 m/s =×=22 ↑ (aD )n 0.25 7.5 14.0625 m/s ⊥⊥ (aaD/A )ttAD and ( D ) BD from the polygon, 34 39.0625×−( a) ×− 14.0625 = 0, ( a) = 11.72 m/s2 55D/A ttD/A 43 39.0625×+( a) ×+( a) − 50 = 0, ( a) = 11.72 m/s2 55D/A ttD D t (aa) ( ) αα=D/A tt = 46.875 rad/s22 CCW, =D = 46.875 rad/s CW AD 0.25 BD 0.25 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 556 P. 9/28 Elements of the switching device are shown. If the velocity v of the control rod is 0.9 m/s and is slowing down at the rate of 6 m/s2 when θ= 60°, determine the magnitude of the acceleration of C.

ICZV

vA

vB

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 557 P. 9/28

60°

acceleration diagram

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 558 P. 9/28 velocity analysis by ICZV of linkage ABC 0.9 ω = =13.856 rad/s CCW AB 0.075cos30 acceleration analysis

[aA= aa B + A/B ] see the acceleration diagram vertical direction: +− = 6 14.4sin 30( aA/B )t sin 60 0 horizontal direction: = + aA 14.4cos30( aA/B )t cos60 = 22= (aA/B )t 15.242 m/s aA 20.09 m/s (a ) α =A/B t = 203.227 rad/s2 CW AB AB =+==22 [aaaC B C/B] ( a C/B )tn 15.242 m/s ( aC/B ) 14.4 m/s

aC =(14.4cos30 + 15.242cos60)i+( 6 −+ 14.4sin 30 15.242sin 60) j 2 =20.09 i+ 12.0j, aC = 23.4 m/s 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 559 P. 9/29 An oil pumping rig is shown in the figure. The flexible pump rod D is fastened to the sector at E and is always vertical as it enters the fitting below D. The link AB causes the beam BCE to oscillate as the weighted crank OA revolves. If OA has a constant CW speed of 1 rev every 3 s, determine the acceleration of the pump rod D when the beam and the crank OA are both in the horizontal position shown.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 560 P. 9/29

vB

vA

ICZV

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 561 P. 9/29

78.1° 16.7° = 11.9° = 16.7°

acceleration diagram

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 562 P. 9/29

ωπOA =1/ 3 rev/s = 2 / 3 rad/s CW const velocity analysis

vA=ωπ OA ×=OA 0.4 m/s ↑ find the ICZV of linkage AB ( see the figure)

ωπAB=v A /10.1 = 0.04 rad/s CW

vB ωAB ×9.92 ωCE = = = 0.398 rad/s CW 0.922++ 3 0.9 22 3 acceleration analysis

[aaaB= A + B/A ] see the acceleration polygon horizontal direction: −− −− = 2.632 0.046cos78.1( aB/A )tt cos11.9( aB ) sin16.7 0.496cos16.7 0 vertical direction: −+ − + = 0.046sin 78.1( aB/A )tt sin11.9( aB ) cos16.7 0.496sin16.7 0 = 22= (a B/A )tt2.036 m/s ( aB ) 0.54 m/s

(a B )t 2 αCE = = 0.1724 rad/s CW 0.922+ 3 = =×=α 2 ↓ aDE( a)t 3.3CE 0.569 m/s 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 563 P. 9/30 An intermitten-drive mechanism for perforated tape F consists of the link DAB driven by the crank OB. The trace of the motion of the finger at D is shown by the dotted line. Determine the acceleration of D at the instant shown when both OB and CA are horizontal if OB has a constant CW rotational velocity of 120 rev/min.

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 564 P. 9/30

acceleration diagram

9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 565 P. 9/30 2π ωπ=×=120 4 rad/s CW const OB 60 from the velocity direction of point A and B, ICZV is at ∞

∴linkage AB translates downward with velocity =ωOB ×=OB 0.628 m/s v ω =A = 5.0265 rad/s CW AC AC v2 [a=+== aa] a7B .896 m/s2 ← A B A/B B OB v2 (a ) =A =3.158 m/s2 ← A n AC = (aA/B )n 0  no rotation at this =−=2 (aA/B )t ( 7.896 3.158) / cos14.4775 4.893 m/s (a ) α =A/B t = 24.467 rad/s2 CW AB AB =+= [aD aa B D/B] ( a D/B )n 0  no rotation at this moment =α × = 2 (a D/B)t AB BD 7.34 m/s

aD =−+( 7.896 7.34cos14.4775)i+( 7.34sin14.4775) j 22 =ij−+0.789 1.835 m/s aD = 1.997 m/s 9.6 Relative Acceleration Ch. 9: Plane Kinematics of Rigid Bodies 566 9.7 Motion Relative to Rotating Axes

So far, v A/B and a A/B are measured from nonrotating reference axes. However, there are many situations where motion is generated within or observed from a system that itself is rotating. In these cases, the solution is greatly facilitated by the use of rotating reference axes. An example is the motion of the fluid particle along the curved vane of a rotating pump.

Consider two particles A and B moving independently in plane motion. Motion of A is observed from a moving reference frame x-y that goes with B, and that rotates with an angular velocity ωθ= .

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 567

rA=++ r B r A/B =r B (xy i+ j) ij and are not constant since their directions change ddij =×ωωi=ωω j and =×−j= i dt dt

∴rA =r B +×ω r A/B +(xy i+ j)

or vvA= B +×ω r A/B + v rel

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 568 Relative Velocity

vvA= B +×ω r A/B + v rel

vA = velocity of the particle A

vB = velocity of the particle B

vA/B = velocity of A relative to B

=×+≠ω rvA/B rel velocity of A as seen from the observer

vrel = xy i+ j = velocity of A as seen from an observer at anywhere fixed to the rotating x-y axes

ω ×=rvA/B( rel observed from nonrotating x-y)

− (vrel observed from rotating x-y)

If we use nonrotating axes, there will be no ω ×rA/B term.

This makes vvrel= A/B , which means the velocity seen by the observer is the velocity of A relative to B.

If B coincides with A, rA/B = 0 . This makes vvrel= A/B , which means the velocity seen by the observer is the velocity of A relative to B, even the observer is rotating. 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 569 Visualization of Relative Velocity Equation Physically, there are two particles A and B, which may be on different rigid bodies. Imagine there is a rotating plate on which particle B is located. On this plate, imagine the virtual point P currently coincident with point A. Relationship between

vP and vB is indicated by vvrvrP= B +×ωω P/B = B +× A/B

Since point A is on the different body, vA ≠ vP. The velocity of A as seen from P (actually from any point

fixed to the rotating plate) is vrel and is tangent to the path (slot) fixed in the rotating plate. 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 570 Visualization of Relative Velocity Equation In other words, the slot is the trajectory seen by the observer fixed to the rotating plate. It is not the absolute path of A (which must be measured by the fixed observer).

The magnitude of vrel, or the relative speed, is s .

From the velocity of vA and vP, it is concluded that vvrel= A/P that is, the velocity of A seen by the rotating B or P is the same, and is equal to the velocity of A relative to P (not to B) 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 571 Summary of Relative Velocity

v= v +×ω r+v AB    rel

vvA= B + v A/B v= v + v +v A  B   P/B A/P

vvvA = P + A/P

Observers moving with different velocities (different vB) on the same rotating x-y see the same vrel. Observers moving with same velocity, but are on different rotating x-y frames (different ω), see different vrel. Nonrotating observer sees the resultant of circular motion plus relative velocity. 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 572 Vector differentiation Change of a vector w.r.t. time as seen from a general reference frame depends on the intrinsic change of the vector itself and the change of the vector induced by the motion of the reference frame, whether it be the translation or rotation. Here the interested frame is constrained not to translate, but rotate around its origin.

An arbitrary vector V= VV xy i+ j has the time derivative dV   =(VVxyi+ j) + ( VV xy i+ j) dt XY ddVV  =  +×ω V dtXY  dt xy

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 573 Vector differentiation This expression means the time derivative of V when measured in the fixed frame (total time derivative) is equal to the time derivative of V as measured in the rotating frame plus the compensation due to rotation of the reference frame.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 574 Vector differentiation More insights can be seen from the vector diagram. Vector V changes in both direction and magnitude to V’. x-y changes by rotating with angular velocity ω while X-Y is fixed. During time dt, the observer in rotating x-y see the change in magnitude of V, dV , plus the change in direction, Vd β , due to relative rotation of V to x-y.

The change the observer recognized is (dV)xy. What he did not notice is the rotation of V induced by the rotation of x-y, Vdθ . Imagine that V is fixed to x-y. Hence its direction changes by the rotation of x-y, which is not known to the observer rotating together.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 575 Relative Acceleration Differentiating the relative velocity equation results in the relative acceleration equation:

aaA= B +×ωω r A/B +× r A/B + v rel V Recall the vector differentiation relation, ω

rvA/B= rel +×ω r A/B

va rel= rel +×ω vrel dω  dd ωω  ω = +× ωω, ∴  =  dt xy dtXY  dt xy →=angular acceleration observed in fixed frame that observed in the rotating frame

Note: rA/B = xy i+ j vrel = xy i+  j arel = xy i+  j

(varel and rel = velocity and acceleration seen by the rotating observer)

∴aaA = B +×ω r A/B +× ωω( × rA/B ) +2 ω × varel + rel

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 576 Visualization of Relative Acceleration Equation

Imagine a plate rotating with the same rate as x-y frame. P is the point on the plate coincident with A. Therefore P is seen to have circular motion about nonrotating observer at B. 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 577 Visualization of Relative Acceleration Equation aaP= B +×ω r A/B +× ωω( × rA/B ) aA=+ aa P A/P, ∴ a A/P =×+ 2ω vrel a rel ≠ a rel and ≠ v rel arel = acceleration of A seen by the rotating observer as moving along the relative path (slot)

= rate of change of vrel as observed in the rotating frame

Treat the relative path as absolute, the direction of arel will always point to the side where the center of curvature of the relative path is located. The n-t coordinate system seems most appropriate. = (arel )t  s tangent to the relative path = 2 ρ (arel)n v rel / toward the center of curvature of the relative path

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 578 Visualization of Relative Acceleration Equation

2ω ×=vrel Coriolis acceleration = difference between the acceleration of A relative to P as measured from nonrotating and from rotating axes =×+ωω =×+ (aA/P )XY 22 vrel( a A/P )xy varel rel = = = + = +×ω = + Note: (vA/P )XY ( vA/P)xy v rel ( v A v P v A/P v P 0+vrel v P( v A/P )xy ) ≠== (aA/P )XY ( aA/P)xy av rel(  rel )xy ≠ =×+ωω= =×+ (vA/P )XY ( aA/P )XY ( aA/P )XY 2 varel rel but ( vA/P )XY vrel varel rel

∴avrel,  rel , and aA/P are all three different quantities avrel , the simplest to visualize, is the change of rel observed in rotating frame v rel , one step more difficult, is the change of vvrel observed in fixed frame -- differ by ω × rel aA/P , the most difficult to imagine, is the acceleration of A relative to P -- differ by 2ω × vrel

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 579 Visualization of Relative Acceleration Equation

One ω × vvrel comes from change of rel due to rotation of x-y. ωω× ×= The other vrel is from change of rA/B due to ( rvA/B)xy rel

∴If A is seen moving (vrel ) in the rotating coordinate frame (ω ) , there will be Coriolis acceleration that the observer will be unaware of.

** Any rotating observer will see A moving with acceleration arel . ** Nonrotating observer will see the resultant of normal, tangential (circular motion), Coriolis, and relative acceleration.

Summary of Acceleration Equations aa= +×ω r +× ωω( × r) +2 ω × va + A B A/B A/B    rel   rel aa=++ a a ABP/B A/P

aA = aaP + A/P

aA= aa B + A/B

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 580 P. 9/31 The crank OA revolves CW with a constant angular velocity of 10 rad/s within a limited arc of its motion. For the position θ= 30°, determine the angular velocity of the slotted link CB and the acceleration of A as measured relative to the slot in CB.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 581 P. 9/31

[vA= vv P + A/P] v A = 0.2 ×= 10 2 m/s

vP ==×× 2cos30 2 0.2cos30ωωCB , CB = 5 rad/s CW

vrel = 2sin 30 = 1 m/s

[aA=+×× a Cωω CB CB r+ A/C ω CB × r+ A/C2 ω CB × v+a rel rel ] v2 ωω××rv =8.66, 2ω ×= 10, a ==A 20 CB CB A/C CB rel A OA 2 from diagram, arel = 20cos30−= 8.66 8.66 m/s along the slot towards C

ωCB×r A/C =20cos60 −= 10 0, ωCB = 0

vP vA/P

vA = 2 30°

velocity diagram

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 582 P. 9/31

|ωxωxr| = 8.66 30°

|2xωxv| = 10

aA

arel

acceleration diagram

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 583

P. 9/32 Determine the angular acceleration α2 of wheel C for the instant when θ= 20°. Wheel A has a constant CW angular velocity of 2 rad/s.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 584 P. 9/32

141.4 mm 82.7 mm 20° α geometry analysis: 200 sin 20 tanαα= 2 , = 35.78 and PO = 82.7 mm 200 2 200− cos 20 2 velocity analysis:

[vA=+=×= vv P rel] v A 2 0.2 / 2 0.283 m/s

vPA= v cos55.78 ==×= 0.159ωω2 0.0827, 2 1.923 rad/s CCW

vA/P= v A sin 55.78= 0.234 m/s acceleration analysis: a= a +××ωω r+ ω × r+2 ω × v+a A O2 2 2 A/O 2 A/O 2 rel rel

consider in ω2×r A/O direction 2 ω2×=+r A/O 0.9 0.5656cos34.22, ω2 = 16.54 rad/s CCW 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 585

P. 9/32 35.78° 20° vP velocity diagram vA

vA/P arel

acceleration diagram

|2ωxv| = 0.9

|αxr|

20° 35.78° 34.22°

aA = 0.5656 |ωxωxr| = 0.306 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 586 P. 9/33 The shuttle A is in an equatorial circular orbit of 240-km altitude and is moving from west to east. Determine the velocity and acceleration which it appears to have to an observer B fixed to and rotating with the earth at the equator as the shuttle passes overhead. Use R = 6378 km for the radius of the earth. Also use Fig. 1/1 for the appropriate value of g and carry out your calculations to 4-figure accuracy.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 587 P. 9/33

[vA= v B +×ω e r A/B +v rel ]

from appendix, ωωe= 0.7292E-4 rad/s ∴=× vBe 6378E3 = 465.084 m/s ←

vvA =465.084 + 0.7292E-4 ×+ 240E3 rel ←← normal acceleration of the shuttle is g towards the center of the earth

2 2 R vA 2 (aAA) = g =, g = 9.814 m/s , ∴= v 7766.79 m/s ← n Rh++ Rh

∴=vrel 7284.205 m/s = 26223 km/h ←

[aA=+×× a Bωω e e r+ A/B ω e × r+ A/B2 ω e × v+a rel rel ] →= assume the shuttle orbits with constant velocity in circular path aaAA( )n the observer's acceleration is governed by the rotation of the earth = ×=ωω2 ↓ = = (aB)nt 6378E3 e 33.9E-3 ( aBe) 0   0 2 9.115= 33.9E-3 +( 0.7292E-4) × 6378E3 +× 2 0.7292E-4 × 7284.205 +arel ↓↓ ↓↓ 2 arel = 8.018 m/s ↓ 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 588 P. 9/34 The figure shows the vanes of a centrifugal-pump impeller which turns with a constant CW speed of 200 rev/min. The fluid particles are observed to have an absolute velocity whose component in the r-direction is 3 m/s at discharge from the vane. Furthermore, the magnitude of the velocity of the particles measured relative to the vane is increasing at the rate of 24 m/s2 just before they leave the vane. Determine the magnitude of the total acceleration of a fluid particle an instant before it leaves the impeller. The radius of curvature ρof the vane at its end is 8 inch.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 589 P. 9/34

(vA)r [vA= vv P + rel ] (vA)θ 2π v= 200 ×× 0.15 =π rad/s ↓ , ( v) = 3 m/s vrel PA60 r 45° v= 3 2 m/s and ( v) =(π − 3) m/s rel A θ vP

[aAP=+× a2ω vrel +a rel ] 2 2 velocity diagram 20ππ 20 (32) aA =0.15 × +× 2 × 3 2 + 24 + 3 3 0.2 ←   = 13.187ij− 45.04 m/s2 2 aA = 46.93 m/s

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 590 P. 9/35 The mechanism shown is a device to produce high in the shaft at O. The gear unit, pivoted at C, turns the right- handed screw at a constant speed N = 100 rev/min in the direction shown which advances the threaded collar at A along the screw toward C.

Determine the time rate of change ω  AO of the angular velocity of AO as it passes the vertical position shown. The screw has 3 single threads per centimeter of length.

403 mm

29.745°

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 591 P. 9/35 ° Rotation of the screw causes relative translation of collar A. 29.745 vA 100 3 rev= 1 cm →= vA/P cm/min = 1/180 m/s v 3 P vA/P [vA= vv P + rel ] 1/180 v = =×=ωω0.2, 0.032 rad/s CCW velocity diagram A cos 29.745 OA OA

vvPP = tan 29.745, vP = 0.0032 m/s, ωPC = = 0.00788 rad/s CCW vrel CP

[aA=+× a P2ω PC v rel +a rel ]

screw turns at constant rate →=a0rel =×=ω 22 (aA )n 0.2OA 204.8E-6 m/s =×=ω 22 (a A )n AC PC 25.41E-6 m/s 2 2ωPC×=v rel 87.556E-6 m/s in vertical direction: = × +×+ 204.8E-6 25.41E-6 sin29.745 87.556E-6 cos29.745( aP )t cos 29.745 = 2 (aP )t 133.8E-6 m/s in horizontal direction: =ω ×=− × + + (aA)t  OA 0.2 25.41E-6 cos29.745( 87.556E-6 133.8E-6) sin 29.745

ωOA = 438.8E-6 rad/s CW 9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 592 P. 9/35

(aP)n = 25.41E-6

29.745° |2ωxv| = 87.556E-6

(aA)n = 204.8E-6 (aP)t

(aA)t

acceleration diagram

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 593 P. 9/36 Determine the angular acceleration of link EC in the position shown, where ωβ =  = 2 rad/s and β  = 6 rad/s 2 when θβ = = 60 ° . Pin A is fixed to link EC. The circular slot in link DO has a radius of curvature of 150 mm. In the position shown, the tangent to the slot at the point of contact is parallel to AO.

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 594 P. 9/36

[vA=+ vv P rel] v P =×= 2 0.15 0.3 m/s

vrel vP =tan 60, vrel = 0.5196 m/s and vA = = 0.6 m/s vP cos60

ωAC =0.6 / 0.15 = 4 rad/s CW

[aA=+× a P2ω OA v rel +a rel ] refer to the diagram

unknown: (aA )tt and (a rel ) vP consider the direction normal to ( a rel )t + += + (aA )t cos60 2.4cos30 0.9 2.0784 1.8 = = αα= 2 60° (aA )t 1.8 0.15EC , EC 12 rad/s CCW

30° vrel

vA

velocity diagram

9.7 Motion Relative to Rotating Axes Ch. 9: Plane Kinematics of Rigid Bodies 595 P. 9/36

(aP)t = 0.9

30°30° 30°

(aP)n = 0.6

|2ωxvrel| = 2.0784

(aA)t

30° (arel)t

60°

30° (arel)n = 1.8 acceleration diagram

9.7 Motion Relative to Rotating Axes