Ch. 9: Plane Kinematics of Rigid Bodies
473
9.0 Outline
473
 Introduction
 Rotation
 Absolute Motion
 Relative Velocity
 Instantaneous Center of Zero Velocity
 Relative Acceleration
 Motion Relative to Rotating Axes
474
478
486
508
531
544
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 force.
A rigid body 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 mass center.
Translation motion in which every line in the body
remains parallel to its original position at all time. 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. Circular
motion 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 displacement,
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.
θ 2= θ1 + β
∆θ 2 =
∆θ1
θ2 = θ1
θ2 = θ1
All lines on a rigid body in its plane of motion have
the same angular displacement, the same angular velocity,
and the same angular acceleration
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.
ω = θ
α= ω= θ
dθ or θdθ θdθ
=
ω dω α=
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.
=
ω , r constant
v r=
a t= v= rα
2
v
ω2
=
a n r=
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 = ω × (ω × r ) + α × r
a n =ω × (ω × r )
a t= α × r
9.2 Rotation
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/1
482
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
on the belt of the larger pulley has an acceleration aB = 45 m/s2
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 v A
ω r ] 1.5 =
ω × 0.075,
[v =
ω=
20 rad/s CW
find α pulley from a B
rα ]
[a t =
45 =
0.4 × α , α =
112.5 rad/s 2 CCW
acceleration at C
a n = rω 2  a Cn = 0.36 × 202 = 144 m/s 2
[a t =rα ] a Ct =0.36 ×112.5 =40.5 m/s 2
aC =
a C2 n + a C2 t = 149.6 m/s 2
aCn
aCt
9.2 Rotation
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/2
484
A V-belt speed-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
 r1 
 r1 
dω 

=
ωA α=
  α=
  α1 t
1 t, ωB
1 t, ωC
α dt =
 r2 
 r2 
2
 r1 
 r1 
=
α A α=
  α=
  α1
1, αB
1, αC
r
 2
 r2 
4
r 
2
a n r=
=
ω 2  a Pn r2  1  (α1t )
 r2 
2
 r1 
=
α ] a Pt r2   α1
[a t r=
 r2 
4
aP =
 r1  2 4
r12
2
2
α 1 1 +   α1 t
a Pn + a Pt =
r2
 r2 
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 velocities and accelerations. 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
P. 9/3
487
A wheel of radius r rolls on a flat surface w/o
slipping. Determine the angular motion of the
wheel in terms of the linear motion 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:
=
v O r=
ω and a O rα
v O = 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 − rsinθ =−
r (θ sin θ )
y=
r − rcosθ =−
r (1 cosθ )
time derivatives:
x =rθ (1 − cos θ ) =v O (1 − cos θ ) y =rθsinθ =v O sin θ

x =v O (1 − cos θ ) + v Oθ sin θ =a O (1 − cos θ ) + rω 2 sin θ

y =v sin θ + v θ cos θ =a sin θ + rω 2 cos θ
O
O
O
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
P. 9/4
489
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
α2 = -3 rad/s2
Case (b) Pulley 1: ω1 = 1 rad/s
α1 = 4 rad/s2
Pulley 2: ω2 = 2 rad/s
α2 = -2 rad/s2
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.
v B = v D = r2ω2 = 0.2 m/s and
( a B )t =
a D = r2α 2 = −0.3 m/s 2
ds A = 0  left cable does not move
ds B
=
ω ( a B ) t ABα
dθ v B AB=
AB
AB' maintains straight and point O has absolute motion in vertical direction:
ds O
=
=
ω a O AOα
AO
dθ v O AO
∴ω =
−0.3 / 0.3 =
−1 rad/s 2 CW
0.2 / 0.3 =
2 / 3 rad/s CCW, α =
v=
0.1× 2 / =
3 0.0667 m/s and a =
0.1× −=
1 −0.1 m/s 2
O
O
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.
r1α=
0.4 m/s 2
(aA =
)t a=
C
1
v B = v D = r2ω2 = 0.2 m/s and ( a B ) t = a D = r2α 2 = −0.2 m/s 2
ABω ( a B ) t −=
ds B=
− ds A ABdθ v=
( a A )t ABα
B − vA
ds O =
− ds A AOdθ v O=
− v A AOω a O − =
( a A )t AOα
∴ω =
( 0.2 − 0.1) / 0.3 =1/ 3 rad/s CCW, α =−
( 0.2 − 0.4 ) / 0.3 =−2 rad/s 2 CW
v=
v=
r1ω=
0.1 m/s and
A
C
1
v O = 0.1 + 0.1× 1/ 3 = 0.133 m/s and a=
0.4 + 0.1× −=
2 0.2 m/s 2
O
9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/5
492
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 O 0.8
= =
, v O 1.2 m/s →
0.9 0.6
1.2
ω = 1.333 rad/s CW
vO ω r ] =
[=
0.9
Vo
Va = 0.8 m/s
C
9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/6
494
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=
[a t =
rω ]
rα ]
v A = 0.2 × 4 = 0.8 m/s →
( a A )t =
0.2 × 3= 0.6 m/s 2 →
0.8
0.6
= 4 / 3 rad/s CW α
=
= 1 rad/s 2 CW
B
0.6
0.6
vC = 4 / 3 × 0.2 = 0.267 m/s ←
ω
=
B
1× 0.2 =
0.2 m/s 2
( a C )t =
ω
=
C
←
0.267
= 4 / 3 rad/s CCW
0.2
α=
C
0.2
= 1 rad/s 2 CCW
0.2
Va
Vc
9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/7
496
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: θ =°
45 , θ =
−3 rad/s, CA =
0.2 m/s
geometric relation:
ysinβ − CA
sin θ 0 and CA cos θ +=
ycosβ 0.4
=
solve=
for y 0.2947 m and
=
β 28.675°
diff w.r.t. time:
 β + yβ cosβ − CAθ cos θ =
ysin
0
 β − yβ sinβ =
0
−CAθ sin θ + ycos
−0.576 m/s and β =
−0.572 rad/s
∴ y =
9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/8
498
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
geometry: θ = angular motion of the wheel
= R=
β r (θ + β )
rolling distance
β
(1)
(R − r) β
rearrange (1) to give ( R − r ) β =
rθ
v=
s =
rθ =
rω
( R − r ) β =
a t ==
v ( R − r ) β =
rω =
rα
motion of the center O= s=
θ β
convex arc
geometry: θ = angular motion of the wheel
= R=
β r (θ − β )
rolling distance
θ
( 2)
(R + r) β
rearrange ( 2 ) to give ( R + r ) β =
rθ
v=
s =
rθ =
rω
( R + r ) β =
a t ==
v ( R + r ) β =
rω =
rα
β
motion of the center O= s=
β
independent of the curvature, R, of the terrain!
9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/9
500
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 O1O 2 P:
1
sin θ
O1 P sin θ
2
=
tanβ =
O1O2 − O1 P cos θ 1 − 1 cos θ
2
1
1

 1 
 1 

−
×
−
×
θ
θ
θ
θ
θ sin θ 
1
cos
cos
sin



2
2
2

 2

β sec 2 β = 
2
1


−
1
cos θ 

2


−2 rad/s, θ =
given: θ =°
20 , θ =
0
−1.923 rad/s
β=
35.783°, β =
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/s 2 constant
=
x 0.1t
=
x 0.05t 2
geometry: x = 0.2 tan θ
x = 0.2θ sec 2 θ
(

=
x 0.2θsec 2 θ + 0.2θ × 2θ sec 2 θ tan θ
)

at
=
x 0.15 m,
=
t
3 sec →
=
x 0.1 3 and=
x 0.1
at that posture,
=
tanθ 3=
/ 4 secθ 5 / 4
∴θ =
0.554 rad/s and θ = −0.1408 rad/s 2
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
θ = θ O sin 2π t where the amplitude is
=
θ O π /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
θ = θ O sin 2π t, θ = 2πθ O cos 2π t, θ = − ( 2π ) θ O sin 2π t
0.12 =
y 2 + 0.142 − 0.28ycosθ
 θ + 0.28yθsinθ
0=
2yy − 0.28ycos
 − 0.28ycos
 θ + 0.28y θsinθ
0 = 2y 2 + 2yy
+ 0.28y θsinθ + 0.28yθsinθ + 0.28yθ 2 cos θ
θ 0,=
θ 2πθ O , θ = 0
when=
t 0,=
y = 0.24, 0.04
y = 0, 
y = −0.909 m/s 2
2
when θ = π /12, t = 1/ 4, θ = 0, θ = − ( 2π ) θ O
y = 0.2284, 0.042

=
y 0,=
y 0.918 m/s 2
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
 0
=
θ ω=
given:
O, θ
geometry: l sin β = r sin θ
diff w.r.t. time: l β cos β = rθ cos θ
rωO cos θ rωO
β =
=
l cos β
l
cos θ
r2
1 − 2 sin 2 θ
l
l β cos β − l β 2 sin β =rθ cos θ − rθ 2 sin θ
β
r2
−1
2
l
l β 2 sin β − rθ 2 sin θ rωO2
sin θ
=
3/ 2
l cos β
l
 r2

2
1 − 2 sin θ 
 l

9.3 Absolute Motion
Ch. 9: Plane Kinematics of Rigid Bodies
508
9.4 Relative Velocity
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-rotating reference frame is
v=
v B + v A/B
A
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 = ∆rB + ∆rA/B
∆rA/B = −∆rB/A has the magnitude r∆θ as ∆θ → 0
** Relative linear motion ∆rA/B is accompanied by the
absolute angular motion as seen from the translating axes x'-y'.
v A = v B + v A/B
If distance r between A and B is constant, v A/B is the velocity
of the circular motion. That is
v A/B =
rA/Bω
v A/B =
v A/B ⊥ AB
ω × rA/B
ω = absolute angular velocity of rigid body
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
general
motion
translational
motion
511
rotational
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
v=
v B + v A/B
A
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
v=
v B + v A/B
A
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
[ v A = v O + v A/O ]
[ v A/O= ω × rA/O ]
=
ω
[ω vO / r ]=
3=
/ 0.3 10 rad/s CW
[ v A = v C + v A/C ]
v C = 0 for rolling w/o slipping wheel
[ v A/C=
vA
ω × rA/C ]
4.36 m/s  23.4°
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/14
517
The power 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 sliding contact constraint from the slot,
v A/B has the direction along the slot (away from O for CCW ω ).
vA
[ v A = v B + v=
A/B ]
=
0.25cos
30 0.217
=
0.417 rad/s CCW
ω v=
A / OA
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
P. 9/15
519
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 )
∴ v C ⊥ OC
v A + v C/A ]
[ vC =
θ=
sin −1 ( 0.2 / 0.8 ) =
14.48°
vC =
0.3cos θ =
ωO × 0.2
ωO =
1.45 rad/s CW
v C/A =
0.3sin θ =
ωAB × 0.82 − 0.22
ωAB =
0.0968 rad/s CCW
C
θ
θ
O
vA
vC/A
vC
A
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/16
521
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
v C + v D/C ]
[ vD =
[ v A =v D + v A/D ]
vD =
v D/C along the slot and pointing out of point A
v A =0.2 × ( 600 × 2π / 60 ) =12.566 m/s
v A/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
vD
14.63°
O
0.4 m
D
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/17
523
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°
 v=
v Q + v P/Q 
P
(
v P/Q along the slot pointing to O 2
)
vP =
2 × 0.2 / 2 =
0.283 m/s
ω2 × 0.083 ω2 =
vQ =
v P sin 34.355 =
1.924 rad/s CCW
P, Q
0.083 m
20° vQ
35.6°
O1
O2
34.35°
vP/Q
vP
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
v A + v B/A ]
[ v=
B
v C + v D/C ]
[ v=
D
constraint: v B = v D
a + b =−0.2 → block A and C become closer
and from the mechanism, A moves to the right and B to the left
∴ vA − vC =
0.2i m/s
from the velocity diagram
vB
vB
+
=
0.2, v B = 0.0536 j m/s
tan 22.62 tan 36.87
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.
23.78°
124 mm
vQ
vP/Q
15°
vP
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/19
15 rad/s
ω r ] 1.5 =
ωO × 0.1, ωO =
[ vO =
v P = ωO × ( 2 × 0.1cos15 ) = 2.9 m/s
528
CW
P is the point at the pin on the disk
Q is the coincident point on the slotted arm
 v=
v Q + v P/Q  v P/Q directs along the slot outward point C
P
v Q= v P cos (15 + 23.78=
) 2.26 m/s
=
=
/ 0.124 18.23 rad/s CCW
ωC 2.26
9.4 Relative Velocity
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/20
529
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
v B =v C + v B/C =v A + v B/A
vB/A
from the velocity diagram
v B/C
v B/A
5
= =
sin60 sin 96.87 sin 23.13
v B/C 5.732
m/s and v B/A 2.268 m/s
=
( vB/C sin 23.13) + ( 3 + vB/A cos83.13)
2
2
v B2
=
v B = 3.97 m/s directed along the dotted arrow
vB/C
vB
96.87°
vA
3 m/s
2 m/s
vC
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
P. 9/21
533
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
v B is forced to move downward by the hydraulic
v A is forced to move horizontally by the guide
∴ ICZV is at point C → v D ⊥ CD
plunger moves down 2 m/s
∴ vertical component of v D is 2 m/s downward
=
v D 2 / cos=
30 ωAD × ( 2 × 0.1cos 30 ) , ωAD =
13.33 rad/s CW
vA =
ωAD × ( 0.2sin 60 ) =
2.309 m/s, to the right
9.5 ICZV
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/22
535
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
ICZVBD
l2
ICZVDE
α
l3
θ
l4
vE
ξ
β
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 − 750 cos α + 375cos β =
375
−100 cos 60 + 750sin α + 375sin β = 850
by the help of complex exponent, α = 85.85°, β = 23.9°
between point O and A:
375cos θ + 375cos ξ + 750 cos α − 100sin 60 =
375
−375sin θ + 375sin ξ + 750sin α − 100 cos 60 =
475
by the help of complex exponent, θ= 81.04°, ξ= 21.49°
9.5 ICZV
Ch. 9: Plane Kinematics of Rigid Bodies
539
P. 9/22
find distance from ICZV to point of interest
l1 sin 30 =
−100 cos 60 + 750sin α − l2 sin β
l1 cos 30 =
l2 cos β + 750 cos α − 100sin 60
l1 773.63
=
=
mm, l2 768.2 mm
l3=
cos β l4 cos θ + 375cos ξ
l4=
sin β l4 sin θ − 375sin ξ
l3 435.8
=
=
mm, l4 317.8 mm
v B =60 × 100 =( l1 + 100 ) × ωBD , ωBD =6.868 rev/min CW
v D = ωBD × l2 = 5276 mmrev/min = l3 × ωDE , ωDE = 12.1 rev/min CW
v E = ωDE × l4 = ωAE × 375, ωAE = 10.26 rev/min= 1.07 rad/s CW
9.5 ICZV
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/23
540
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 rotational speed 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 Agear = v Alink
2. motion of the leftmost point on the rim: v CgearA = v CgearB
3. motion of the rightmost point on the rim: v DgearA = v DgearD
(a )
90a
ring gear D is fixed
=
v A 90a ↓
v C =×
2 90a =
360 rev/min CW
ωB × a/2, ωB =
120a
( b ) ring gear D rotates 80 rev/min CCW
 3a 
v A =90a ↓, v D =80   =120a ↑
 2
10 / 7
v C = ( 90a ) =
600 rev/min CW
ωB × a/2, ωB =
3/ 7
90a
9.5 ICZV
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/24
542
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
=
v O 0.9 m/s →
=
=
rev/min 8π rad/s CCW
ωi 240
0.9
m lower to point O
8π
velocity of the point on the roller contacting w/ inner race
[ v = ωr]
ICZV of the inner race is
0.9 

8π ×  0.05 −
0.3566 m/s →
=
=
8
π


0.9
= 7.2 rad/s CCW
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
=
=
/ x 10.732 rad/s CW
ωroller 0.18
ωO =
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:
a=
a B + a A/B
A
a A/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 .
aA =
a B + ( a A/B )n + ( a A/B ) t
( a A/B )n =ω × (ω × r ) , ( a A/B )n =rω 2 =v 2A/B / r
rα =
v A/B
α × r, ( a A/B ) t =
( a A/B )t =
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
a=
a B + a A/B
A
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
P. 9/25
549
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
v C = 0 because no slipping
a O + a C/O ] and v O =
rω i, ( a O ) t =
rα i
[a C =
( a ) aC = ( aO )t + ( aO )n + ( aC/O )t + ( aC/O )n
r 2ω 2
 R 
j − rα i + rω 2 j = rω 2 
= rα i +
j
R −r
R
r
−


 R 
2
a C rω 2 
r
ω
=
>

R −r
(b)
a C = ( a O ) t + ( a O )n + ( a C/O ) t + ( a C/O )n
r 2ω 2
 R 
j − rα i + rω 2 j = rω 2 
= rα i −
j
R+r
R+r
 R 
2
=
a C rω 2 
 < rω
R+r
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/26
551
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°
(aB)t
(aB/C)t
sketch diagram
50.345°
574.24 mm
0.2604
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
=
=
/ CD 0.7212 rad/s CW closing
=
=
ωBC 0.5
(
)
v B =ωBC × CO / cos 50.345 − BO =0.2165 m/s
0.361 rad/s CCW
=
ωBO v=
B / BO
acceleration analysis
a C + a B/C ] ( a B )n + ( a B ) t =
( aB/C )n + ( aB/C )t
[a B =
2
2
a B/C )n BC
=
0.2604 m/s 2 =
=
ωBC
ωBO
( a B )n BO
(=
0.0782 m/s 2
horizontal direction:
0.2604cos22.5 + ( a B/C ) t sin 22.5 − ( a B ) t cos 50.345 − 0.0782sin 50.345 =
0
vertical direction:
0
0.2604sin22.5 − ( a B/C ) t cos 22.5 − ( a B ) t sin 50.345 + 0.0782 cos 50.345 =
−0.049 m/s 2 ( wrong direction ) ( a B ) t =
0.2532 m/s 2
( a B/C )t =
/ 0.5 0.098 rad/s 2 CW
=
α BC 0.049
=
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/27
554
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
= 20 rad/s CCW
0.2
v A =ωOA × 0.125 =2.5 m/s ↑
ω=
OA
vA
(4)
velocity analysis
v A + v D/A ]
[ vD =
vD / vA =
3 / 4, v D =
1.875 m/s
vD/A
(5)
v D/A / v A 5=
/ 4, v D/A 3.125 m/s
=
7.5 rad/s CCW,
=
=
ωAD v=
12.5 rad/s CCW
ωBD v=
D/A / 0.25
D / 0.25
acceleration analysis
vD
(3)
[aD =a A + a D/A ] a A =0.125 × 202 =50 m/s 2 ←
( a D/A )n =0.25 ×12.52 =39.0625 m/s 2
( a D )n = 0.25 × 7.52 = 14.0625 m/s 2 ↑
( a D/A )t ⊥ AD and ( a D )t ⊥ BD
from the polygon,
3
4
39.0625 × − ( a D/A ) t × − 14.0625
= 0, =
( a D/A )t 11.72 m/s 2
5
5
4
3
39.0625 × + ( a D/A ) t × + ( a D )=
− 50 0, =
( a D )t 11.72 m/s 2
t
5
5
( a D/A )t
( a D )t
α AD
α BD =
46.875 rad/s 2 CW
=
= 46.875 rad/s 2 CCW, =
0.25
0.25
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/28
556
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
= 13.856 rad/s CCW
0.075cos 30
acceleration analysis
=
ωAB
[a=
A
a B + a A/B ]
see the acceleration diagram
vertical direction:
6 + 14.4sin 30 − ( a A/B ) t sin 60 =
0
horizontal direction:
=
a A 14.4 cos 30 + ( a A/B ) t cos 60
=
=
m/s 2 a A 20.09 m/s 2
( a A/B )t 15.242
α AB
=
( a A/B )t
= 203.227 rad/s 2 CW
AB
15.242 m/s 2
a B + a C/B ] ( a C/B ) t =
[a C =
aC =
14.4 m/s 2
( a C/B )n =
(14.4 cos 30 + 15.242 cos 60 ) i + ( 6 − 14.4sin 30 + 15.242sin 60 ) j
= 20.09i + 12.0 j,
a C = 23.4 m/s 2
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/29
559
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°
= 16.7°
= 11.9°
acceleration diagram
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
562
P. 9/29
=
ωOA 1/=
3 rev/s 2π / 3 rad/s CW const
velocity analysis
v A = ωOA × OA = 0.4π m/s ↑
find the ICZV of linkage AB ( see the figure )
0.04π rad/s CW
=
ωAB v=
A /10.1
vB
ωAB × 9.92
=
= 0.398 rad/s CW
2
2
0.9 + 3
0.92 + 32
acceleration analysis
=
ωCE
[a=
B
a A + a B/A ]
see the acceleration polygon
horizontal direction:
2.632 − 0.046cos78.1 − ( a B/A ) t cos11.9 − ( a B ) t sin16.7 − 0.496 cos16.7 =
0
vertical direction:
−0.046sin 78.1 + ( a B/A ) t sin11.9 − ( a B ) t cos16.7 + 0.496sin16.7 =
0
=
=
m/s 2 ( a B ) t 0.54 m/s 2
( a B/A )t 2.036
( a B )t
0.1724 rad/s 2 CW
=
2
2
0.9 + 3
a D =( a E ) t =3.3 × α CE =0.569 m/s 2 ↓
α CE
=
9.6 Relative Acceleration
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/30
563
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
= 4π rad/s CW const
60
from the velocity direction of point A and B, ICZV is at ∞
ωOB = 120 ×
∴ linkage AB translates downward with velocity = ωOB × OB = 0.628 m/s
vA
= 5.0265 rad/s CW
AC
v 2B
7.896 m/s 2 ←
a B + a A/B ] a B =
=
[a A =
OB
v 2A
( a A=
)n = 3.158 m/s 2 ←
AC
( a A/B )n = 0  no rotation at this moment
ω=
AC
4.893 m/s 2
( a A/B )t =
( 7.896 − 3.158) / cos14.4775 =
( a A/B )t
2
α AB
=
= 24.467 rad/s CW
AB
0  no rotation at this moment
a B + a D/B ] ( a D/B )n =
[a D =
( a D/B=
)t α AB × BD = 7.34 m/s 2
aD =
( −7.896 + 7.34 cos14.4775) i + ( 7.34sin14.4775) j
= −0.789i + 1.835 j m/s 2
aD =
1.997 m/s 2
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 ω = θ.
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
567
rA =
rB + rA/B = rB + ( xi + yj)
i and j are not constant since their directions change
di
dj
=ω × i = ω j and
=ω × j = −ω i
dt
dt
∴ rA = rB + ω × rA/B + ( x i + y j)
or v A = v B + ω × rA/B + v rel
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
568
Relative Velocity
v A = v B + ω × rA/B + v rel
v A = velocity of the particle A
v B = velocity of the particle B
v A/B = velocity of A relative to B
=ω × rA/B + v rel ≠ velocity of A as seen from the observer
v rel = x i + y j
= velocity of A as seen from an observer
at anywhere fixed to the rotating x-y axes
ω × rA/B =
( v rel observed from nonrotating x-y )
− ( v rel observed from rotating x-y )
If we use nonrotating axes, there will be no ω × rA/B term.
This makes v rel = v 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 v rel v 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
v P = v B + ω × rP/B = v B + ω × rA/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
v rel = v 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 A = v B + ω × r + v rel

v=
vB +
A
v A/B
v=
v B + v P/B + v A/P
A



=
vA
vP
+ v 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 = Vx i + Vy j has the time derivative
 dV 
 i+V
 j) + V i + V j
=
V
(
x
y
x
y


 dt XY
 dV 
 dV 
=



 +ω ×V
 dt XY  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, Vdβ , 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:
a A = a B + ω × rA/B + ω × rA/B + v rel
Recall the vector differentiation relation,
rA/B = v rel + ω × rA/B
v rel = a rel + ω × v rel
V
ω
 dω 
 dω 
 dω 
ω 
ω
ω
=
+
×
∴
=
,





 dt  xy
 dt XY  dt  xy
→ angular acceleration observed in fixed frame =
that observed in the rotating frame
=
Note: rA/B x=
i + yj v rel x=
i + y j a rel 
xi + 
yj
( v rel
and a rel = velocity and acceleration seen by the rotating observer )
∴ a A = a B + ω × rA/B + ω × (ω × rA/B ) + 2ω × v rel + a 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
a P = a B + ω × rA/B + ω × (ω × rA/B )
a A =a P + a A/P , ∴ a A/P =2ω × v rel + a rel ≠ a rel and ≠ v rel
a rel = acceleration of A seen by the rotating observer
as moving along the relative path (slot)
= rate of change of v rel as observed in the rotating frame
Treat the relative path as absolute, the direction of a rel 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.
( a rel )t = s tangent to the relative path
( a rel )n = v 2rel / ρ 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 × v rel =
Coriolis acceleration
= difference between the acceleration of A relative to P
as measured from nonrotating and from rotating axes
( a A/P )XY = 2ω × v rel + ( a A/P )xy = 2ω × v rel + a rel
( v A/P )XY = ( v A/P )xy = v rel ( v A = v P + v A/P = v P + ω × 0 + v rel = v P + ( v A/P )xy )
a rel ( v rel ) xy
( aA/P )XY ≠ ( a A/P )xy ==
( v A/P )XY ≠ ( a A/P )XY  ( a A/P )XY =2ω × v rel + a rel but ( v A/P )XY =v rel =ω × v rel + a rel
Note:
∴ a rel , v rel , and a A/P are all three different quantities
a rel , the simplest to visualize, is the change of v rel observed in rotating frame
v rel , one step more difficult, is the change of v rel observed in fixed frame -- differ by ω × v rel
a A/P , the most difficult to imagine, is the acceleration of A relative to P -- differ by 2ω × v rel
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
579
Visualization of Relative Acceleration Equation
One ω × v rel comes from change of v rel due to rotation of x-y.
The other ω × v rel is from change of ω × rA/B due to ( rA/B ) xy =
v rel
∴ If A is seen moving ( v rel ) 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 a rel .
** Nonrotating observer will see the resultant of normal, tangential (circular motion),
Coriolis, and relative acceleration.
Summary of Acceleration Equations
a A = a B + ω × rA/B + ω × (ω × rA/B ) + 2ω × v rel + a rel




 
aA =
aB +
a P/B



+
a A/P
aA =
+
a A/P
aP
a=
a B + a A/B
A
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/31
580
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 =
v P + v A/P ]
v A = 0.2 ×10 = 2 m/s
vP =
2 cos 30 =
2 × 0.2 cos 30 × ωCB , ωCB =
5 rad/s CW
v rel 2sin
30 1 m/s
=
=
[aA =aC + ωCB × ωCB × rA/C + ω CB × rA/C + 2ωCB × v rel + arel ]
v 2A
ω CB × ω CB × rA/C = 8.66, 2ω CB × v rel = 10, a A =
= 20
OA
2
8.66 m/s along the slot towards C
from diagram, a rel = 20 cos 30 − 8.66 =
 CB 0
20 cos 60 −=
10 0, ω=
ω CB × r=
A/C
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
P. 9/32
583
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
20°
geometry analysis:
200
sin 20
2
=
=
=
tanα
, α 35.78
and PO2 82.7 mm
200
200 −
cos 20
2
velocity analysis:
v P + v rel ]
[ vA =
82.7 mm
α
vA =
2 × 0.2 / 2 =
0.283 m/s
ω2 × 0.0827, ω2 =
vP =
v A cos 55.78 =
0.159 =
1.923 rad/s CCW
=
v A/P v=
0.234 m/s
A sin 55.78
acceleration analysis:
a A = a O2 + ω 2 × ω 2 × rA/O + ω 2 × rA/O + 2ω 2 × v rel + a rel 
consider in ω 2 × rA/O direction
ω 2 × rA/O =0.9 + 0.5656 cos 34.22, ω 2 =16.54 rad/s 2 CCW
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/32
585
35.78°
20°
vP
velocity diagram
vA
vA/P
arel
acceleration diagram
|2ωxv| = 0.9
|αxr|
20°
34.22°
aA = 0.5656
35.78°
|ωxωxr| = 0.306
9.7 Motion Relative to Rotating Axes
Ch. 9: Plane Kinematics of Rigid Bodies
P. 9/33
586
The space 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 × rA/B + v rel ]
from appendix, ωe = 0.7292E-4 rad/s ∴ v B = ωe × 6378E3 = 465.084 m/s ←
v A = 465.084 + 0.7292E-4 × 240E3 + v rel
←
←
normal acceleration of the shuttle is g towards the center of the earth
2
v 2A
 R 
=
=
=
g
, g 9.814 m/s 2 , ∴
v A 7766.79 m/s ←
( a=
A )n

R+h
R+h
m/s 26223 km/h ←
=
∴ v rel 7284.205
=
[aA = aB + ωe × ωe × rA/B + ω e × rA/B + 2ωe × v rel + arel ]
assume the shuttle orbits with constant velocity in circular path → a A =
( a A )n
the observer's acceleration is governed by the rotation of the earth
( a B=
)n
2
6378E3 × ω
=
33.9E-3 ↓
e
( a B=
)t
e 0
0 ω
=
9.115
= 33.9E-3 + ( 0.7292E-4 ) × 6378E3 + 2 × 0.7292E-4 × 7284.205 + a rel
2
↓
↓
↓
↓
=
a rel 8.018 m/s 2 ↓
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
[ v=
A
v P + v rel ]
(vA)r
2π
× 0.15 = π rad/s ↓, ( v A )r = 3 m/s
60
v rel
= 3 2 m/s and ( v A )=
(π − 3) m/s
θ
v P = 200 ×
[aA =aP + 2ω × v rel + arel ]
20π
 20π 
2
a A= 0.15 × 
+
×
×3

3
 3 
←

2
3 2)
(
2 + 24 +
2
(vA)θ
vrel
45°
vP
velocity diagram
0.2


= 13.187i − 45.04 j m/s 2
a A = 46.93 m/s 2
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 torque in the shaft at O. The gear unit, pivoted at C, turns the righthanded 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
vA
Rotation of the screw causes relative translation of collar A.
100
3 rev = 1 cm → v A/P =
cm/min = 1/180 m/s
3
v P + v rel ]
[ v=
A
1/180
vA =
0.032 rad/s CCW
=
ωOA × 0.2, ωOA =
cos 29.745
vP
vP
= tan 29.745, =
= 0.00788 rad/s CCW
v P 0.0032 m/s, ω=
PC
v rel
CP
vP
29.745°
vA/P
velocity diagram
a P + 2ω PC × v rel + a rel ]
[a A =
0
screw turns at constant rate → a rel =
2
=204.8E-6
( a A )n =0.2 × ωOA
( a A )n = AC × ωPC2 = 25.41E-6
m/s 2
m/s 2
2ω PC × v rel =
87.556E-6 m/s 2
in vertical direction:
= 25.41E-6 × sin29.745 + 87.556E-6 × cos29.745 + ( a P ) t cos 29.745
204.8E-6
( a P )t = 133.8E-6
m/s 2
in horizontal direction:
−25.41E-6 × cos29.745 + ( 87.556E-6 + 133.8E-6 ) sin 29.745
ω OA × 0.2 =
( a A )t =
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 ω= β= 2 rad/s and β = 6 rad/s 2 when θ= β= 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
v P + v rel ]
[ vA =
vP =
2 × 0.15 =
0.3 m/s
v rel
vP
=
tan 60,=
v rel 0.5196 m/s and=
vA =
0.6 m/s
vP
cos 60
/ 0.15 4 rad/s CW
ωAC 0.6
=
=
a P + 2ω OA × v rel + a rel ]
[a A =
unknown: ( a A ) t nd
a ( a rel ) t
refer to the diagram
consider the direction normal to ( a rel ) t
= 2.0784 + 1.8
( a A )t cos 60 + 2.4 cos 30 + 0.9
= 0.15α EC , α EC
= 12 rad/s 2 CCW
( a A=
)t 1.8
vP
60°
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
(arel)t 30°
60°
(arel)n = 1.8
30°
acceleration diagram
9.7 Motion Relative to Rotating Axes