1
KINEMATICS OF RIGID BODIES IN PLANE AND 3-D MOTION
15.90 (Beer & Johnston)
W DISK  500 r / min k
W DISK 
500(2)
k rad / s
60
(cons tan t )
The disk shown has a constant angular velocity of 500 r/min
counterclockwise. Keeping that rod BD is 250 mm long,
determine the acceleration of collar D when (a)  = 90, (b)  =
180.
(a)  = 90
r D / B  (0.15  0.05)i  (0.25) 2  (0.15  0.03) 2 j  (0.10i  0.229 j)m
Velocity Analysis
 B rotates about a fixed axis through A 
500(2)
V B  DISK  r B / A 
k  (0.05i)  2.62 j m / s
60
 D translates  BD is in general plane motion (B is chosen as reference point)
V D  V B   BD  r D / B
 V D j  2.62 j   BD kx (0.10i  0.229 j)
 V D j  2.62 j  0.1 BD j  0.229 BD i
i – components: 0.229 BD = 0  BD = 0
Acceleration Analysis
 500(2) 
 0.05i   137.08i m / s 2
 

60


2
a B   Disk  r B / A  
2
Disk B / A
r
a D  a B   BD  r D / B  2BD r D / B
a D j  137.08i   BD k  (0.1i  0.229 j)
 237.08i  0.1 BD j  0.229 BD i
2
i-components: 137.08 + 0.229BD = 0  BD =
 137.08
= -598.6 rad/s2
0.229
2
j-components: aD = 0.1BD = -0.1 (-598.6) = 59.9 m/s2  a D  59.9 j m / s
Vector Polygon
aD/Bt
aD
0.1BDj
Not to scale!
aB
0.2299BDi
b)  = 180
r D / B  0.15i  (0.25) 2  (0.15) 2 j  (0.15i  0.2 j)m
Velocity Analysis
500(2)
k  (0.05 j)  2.62i m / s 2
60

V B   Disk  r B / A 

V D  V B  BD  r D / B
  VD j  2.62i   BD k  (0.15i  0.2 j)
 VD j  2.62i  0.15 BD j  0.2 BD i
i-components: 2.62 + 0.2 BD = 0  BD =
 2.62
= -13.1 rad/s
0 .2
Acceleration Analysis
 500(2) 
a B   Disk  r B / A   2Disk r B / A   
(0.05 j)  137.08 j m / s 2

 60 
2
a D  a B   BD  r D / B  BD r D / B
2
a D j  137.08 j   BD k  (0.15i  0.2 j)  (13.1) 2 (0.15i  0.2 j)
 137.08 j  0.15 BD j  0.2 BD i  25.74i  34.32 j
3
 25.74
= -128.7 rad/s2
0 .2
2
j-components: aD = 137.08 – 0.15BD + 34.32 = 190.7 m/s2  a D  190.7 j m / s
i-components: 0 = 0.2BD + 25.74  BD =
Vector Polygon
aD/Bn
35.32i
25.74 i
0.15BDj
0.2BDi
aD/Bt
Not to scale!
aB
aD
Example: In the four-bar linkage shown, control link OA
has a counterclockwise angular velocity 0 = 10
rad/s during a short interval of motion. When link
CB passes the vertical position shown, point A
has coordinates x = -60 mm and y = 80 mm.
Determine, by means of vector algebra, the
angular velocity of AB and BC.
 Link AO is in rotation about a fixed axis through 0
 V A  o  r A / O  10k  (60i  80 j)   600 j  800i  100(8i  6 j) mm / s
 Link CB is in rotation about a fixed axis through C
 V B   BC  r B / C   BC k  180 j  180 BC i mm / s
 Link AB is in general plane motion  V B  V A  AB  r B / A
 180  BC i  100(8i  6 j)   AB k  (240i  100 j)
 180  BC i  800i  600 j  240 AB j  100 AB i
j-components: 0 = -600 + 240AB
4
AB  2.5k rad / s  2.5 rad / s
AB = 600/240 = 2.5 rad/s
i-components: -180WBC = -800 - 100AB
180BC = 800 + 100(2.5)
BC = 1050/180 = 5.83 rod/s
BC  5.83k rad / s  5.83 rad / s
15.93
A
aC/An
0.15m
aB/An
B
aC/At
0.075m
j
60(2)
(k ) rad / s
60
 2k rad / s
(cons tan t )
 AB 
VB/At
i
aB/At
AB rotates with a constant angular velocity of 60 r/min clockwise. Knowing that gear A does
not rotate, determine the acceleration of the tooth of gear B which is in contact with gear A.
Velocity Analysis
 B rotates about a fixed axis through A
 V B   AB  r B / A  2k  0.225 j  1.414i m / s
 Gear A does not rotate  V C  0
 C is the instantaneous center of rotation of gear B
V
1.414
B  B 
 18.85 rad / s (clockwise )
rB / C 0.075
Acceleration Analysis
a B   AB  r B / A  2AB r B / A  (2) 2 (0.225 j)  8.88 j
a C  a B  a B  r C / B  2B r C / B  8.88 j  (18.85) 2 (0.075 j)  8.88 j  26.65 j
a C  17.77 j
5
Note: Gear B is in general plane motion; B is chosen as reference point.
Vector Polygon
aC
aC/Bn
Not to scale!
aB/An
RATE OF CHANGE OF A VECTOR WITH RESPECT TO A ROTATINT FRAME OF
REFERENCE
Y
A
Y1
IA/B

IA
X1
B
IB
X
 XY frame is fixed
 xy frame rotates with angular velocity  about he z-axis (i.e. perpendicular to plane of
screen)
r A  r B  r A / B  rB  i x   j y 
i, j not fixed since xy rotating.
r A  r B  i x   j y   x 
Evaluation of
dj
di
and
dt
dt
dj
di
 y
dt
dt
6
di/dt = jd/dt
dj/dt = -id/dt
j
i
d
d
di = j d
di = (1) d
Introduce cross-product
j
dk
 k  k  0
dt
i
k
d di
   i
dt dt
d d j
k  j  i   i

  j
dt dt
k  i  j   j
r A  r B  rx   jy     ix     jy
 r B  r A / B    r A / B
d
r A XY  d r B XY  d r A xy    r A xy
dt
dt
dt
Generalization
For any vector A
 dA 
 dA 

 
   A
 dt  XY  dt  xy
Background
|dA|XY
X11
Y
|dA|x1y1
Y1
Ad
dA=d(|A|)
Y11

dA
d



X1
Ad
X
7
Vector A swings to A1 in time dt observer attached to frame xy (i.e. rotating frame) sees that
 dA 

 consists of two components.
 dt  xy
- A dB/dt due to rotation of A through d/B in xy.
- dA/dt due to change in magnitude of A.
Part of absolute rate of change is A not seen by rotating observer is A
A is magnitude of vector A.
Plan motion in a rotating frame
VA  VB   r A / B  VA / B
Acceleration

  r A / B    r A / B  V
aA  aB  
A/B
V A / B  ix   jy 
dj
di

 
V
x   y 
A / B  ix  jy 
dt
dt
 a A / B    ix     jy 
 a A/ B   VA/ B
  rA/B   rA/B



 dA 
 dA 
 recall 


    A 
 dt  fixed  dt  rot


d
 A .
dt
8
 rA/B  
d
ix   jy)
dt
dj 

di
   ix   jy   x   y
dt
dt 

   VA / B    r A / B 
a A  a B    r A / B  x(  r A / B )  2  V A / B  a A / B
Y
y1
A
tangential
acceleration
due to
angular
acceleration
of rotating
frame
x1
B
normal or
centripetal
acceleration
due to
rotation of
rotating
frame
X
2VAB – CORIOLIS ACCELERATION
a P  a A   Y  r P / A  y  (y  r P / A )  2y  V P / A  a P / A
a A   Y  r A / O  y  (y  r A / O )
a A  15 j  (15 j  3i)
y  (y  r P / A )  15 j  (15 j  1j)
V P / A  A  r P / A  45i  1j
2y  V P / A  2(15 j)  (45i  1 j)


a P / A   A  r P / A   A   A  r P / A   45i  45i  1j
A  A  r A / IC
9
 Consider a rotating disk with a radial slot
 A small particle A is confined to slide in the slot
 Let  = constant and Vrel = constant
 The velocity of A has two components:
x (due to rotation of the disk)
vrel (due to motion of A in the slot)
 Consider the rate of change of the velocity of A:
- no change in magnitude of Vrel since Vrel = constant.
- change in direction of Vrel is x d
- change in magnitude of x is dx
- change in direction of x is xd
Rates of change are: x
x
d
dx
, 
dt
dt
and x
d
dt
d
dx
and 
are in the (+) y-direction
dt
dt
x
d
is in the (-) x-direction
dt
 
 Total rate of change of VA: a A  x   x j  x 2  i   x2 i  2x  j
(normal) (Coriolis)
a rel  0 since Vrel = constant and slot has no curvature
  r A / O  0 since  is constant
w
10
XY : Fixed Frame
xy : Rotating Frame
 Recall for a fixed frame: V A  V B  V A / B
 Now for a rotating frame: V A / B  V rel    r A / B  V A  V B    r A / B  V rel
V A  V B V P / B V A/ P
VA 
VP
V A/ P
V A  V B V A/ B
XY : Fixed Frame
Xy : Rotating Frame
  (  r A / B ) : normal acceleration of a point (P) fixed in the rotating frame
  r A / B : tangential acceleration of a point (P) fixed in the rotating frame
a rel : acceleration of point A in the rotating frame
2  V rel : Coriolis acceleration brought about by the rotating () of the rotating frame and
relative motion (Vrel) in the rotating frame
11
15.119
The motion of pin P is guided by slots cut in rods
AE and BD. Knowing that the rods rotate with the
constant angular velocity A = 4 rad/s ↓ and B =
5 rad/s ↓, determine the velocity of pin P for the
position shown.
 Pin P moves in BD and AE both of which rotate 
relative motion in a rotating frame
 V P  V A  A  r P / A  V P / AE
(for AE)
 0.25 
 i  VP / AE (i)
V P  4(k )  
 cos30 
V P  (1.1547 j  VP / AE i) m / s

V P  V B  B  r P / B  V P / BD (for BD)
 5(k )  (0.25 tan 30)(  j*)  VP / BD ( j*)

 (0.722i * VP / BD j*) m / s
Equateand    1.1547 j  VP / AE i  0.722i * VP / BD j *
Coordinate transformation:
j*
i*  i cos 30  j sin 30
j
30o
30o
j*  i sin 30  j cos 30
i*
i


  1.1547 j  VP / AE i  0.722 i cos 30  j sin 30  VP / BD (i sin 30  j cos 30)
j(1.1547  0.722 sin 30  VP / BD cos 30)  i(VP / AE  0.722 cos 30  VP / BD sin 30)
j-component: -1.1547 + 0.722 sin 30 + VP/BD cos 30 = 0
1.1547  0.722 sin 30
 0.916 m / s
cos 30
i-components: -VP/AE + 0.722 cos 30 - VP/BD sin 30 = 0
VP / BD 
VP / AE  0.722 cos 30  0.916 sin 30  0.167 m / s
V P  (1.155 j  0.167i) m / s
Vp
-0.167i
-1.155j
12
or
V P  0.722i * 0.916 j * m / s
Vp
-0.916j*
-0.722i*
15.123 At the instant shown the length of the boom is
being decreased at the constant rate of 150 mm/s and the
boom is being lowered at the constant rate of 0.075
rad/s. Knowing that  = 30, determine (a) the velocity,
(b) the acceleration of point B.
 There is relative motion of B in the rotating x-y
frame
(a) V B  V A    r B / A  V Brel
A is fixed
V B  (0.075)(k)  6i  0.15i
V B  (0.45 j  0.15i) m / s
(b) a B  a A  a  r B / A  2 r B / A  2  V B rel  a B rel
-0.45j
VB
-0.15i
( is cons tan t ) (VBrel is cons tan t )
a B  (0.075) (6i)  2(0.075)(k)  0.15(i)
2
a B  (0.034i  0.023 j m / s 2
-0.034i
0.023j
aB
The vertical shaft and attached clevis rotate about the zaxis at the constant rate  = 4 rad/s. Simultaneously,
the shaft B revolves about its axis OA at the constant
rate 0 = 3 rad/s, and the angle  is decreasing at the
constant rate of /4 rad/s. Determine the angular
velocity  and the magnitude of the angular
acceleration  of shaft B when  = 30. The x-y-z axes
are attached to the clevis and rotate with it.
  k  o cos  j  o sin  k   i

 4k  3(0.866) j  3(0.5)k  i
4
 (2.5k  2.898 j  0.78i) rad / s
13

d
 k  k  
 o cos  j  o ( sin  )  j  o cos  j  
 o sin  k o (cos  )  k  o sin k   i   i
()  
dt
  k  o sin  j  o cos  j  o cos  k  o sin  k   i
i    i , j    j , k    k
   i  (o sin  )i  (o cos ) j  o cos  k
    (  i)  o  sin  j  o cos  (  j)  o  cos  k
   ( j)  o  sin  j  o  cos i)  o  cos  k




(4) j  3  sin 30 j  (3)( 4) cos 30i  3  cos 30k
4
4
4
  11.44 m / s 2
1. The circular plate and rod are rigidly connected and
rotate about the ball-and-socket joint ( ) with an angular
velocity  =  i +  j +  k. Knowing that VA = -(540
mm/s)i + 350 mm/s)j + (r4)2k and ij = 4 rad/s. Determine
(a) the angular velocity of the assembly, (b) the velocity of
point B.
2. A disk of radius r rotates at a constant rate 2 with
respect to the are ( ), which itself rotates at a constant
rate 1 about the Y axis. Determine (a) the angular
velocity and angular acceleration of the disk, (b) the
velocity and acceleration of point A on the rim of the
disk.
14
3. The bent rod ABC rotates at a constant rate 1. Knowing
that the collar D moves downward along the rod at a constant
relative speed u, determine for the position shown (a) the
velocity of D, (b) the acceleration of D.
4. A disk of radius r spins at the constant rate 2 about
an axle held by a fork-ended horizontal rod which
rotates at the constant rate 1. Determine the
acceleration of point I for an arbitrary value of the
angle .