Chapter 15
Kinematics of
Rigid Bodies
1
Introduction
• Kinematics of rigid bodies: relations between
time and the positions, velocities, and
accelerations of the particles forming a rigid
body.
• Classification of rigid body motions:
- translation:
- rectilinear translation
- curvilinear translation
- rotation about a fixed axis
- general plane motion
- motion about a fixed point
- general motion
2
Translation
• Consider rigid body in translation:
- direction of any straight line inside the
body is constant,
- all particles forming the body move in
parallel lines.
• For any two particles in the body,

 
rB  rA  rB A
• Differentiating with respect to time,

 

rB  rA  rB A  rA


vB  v A
All particles have the same velocity.
• Differentiating with respect to time again,
rB  rA  rB A  rA


aB  a A
All particles have the same acceleration.
3
Rotation About a Fixed Axis. Velocity
• Consider rotation of rigid body about a
fixed axis AA’


• Velocity vector v  dr dt of the particle P is
tangent to the path with magnitude v  ds dt
s  BP   r sin  
v
ds

 lim r sin  
 r sin 
dt t 0
t
• The same result is obtained from

 dr  
v
 r
dt




   k   k  angular velocity
4
Rotation About a Fixed Axis. Acceleration
• Differentiating to determine the acceleration,

 dv d  
a
   r 
dt dt


d   dr

r  
dt
dt

d   

r  v
 dt
d 
   angular acceleration
•
dt





  k   k   k
• Acceleration of P is combination of two
vectors,
     
a    r     r
 
  r  tangential acceleration component
  
    r  radial acceleration component
5
Rotation About a Fixed Axis. Representative Slab
• Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
• Velocity of any point P of the slab,
 
  
v    r  k  r
v  r
• Acceleration of any point P of the slab,
     
a    r     r
 

  k  r   2r
• Resolving the acceleration into tangential and
normal components,
 

at  k  r
a t  r


an   2 r
an  r 2
6
Equations Defining the Rotation of a Rigid Body About a Fixed Axis
• Motion of a rigid body rotating around a fixed axis is
often specified by the type of angular acceleration.
• Recall  
d
dt
or
dt 
d

d d 2
d

 2 
dt
d
dt
• Uniform Rotation,  = 0:
   0  t
• Uniformly Accelerated Rotation,  = constant:
  0  t
   0   0t  12  t 2
 2   02  2    0 
7
General Plane Motion
• General plane motion is neither a translation nor
a rotation.
• General plane motion can be considered as the
sum of a translation and rotation.
• Displacement of particles A and B to A2 and B2
can be divided into two parts:
- translation to A2 and B1
- rotation of B1 about A2 to B2
8
Absolute and Relative Velocity in Plane Motion
• Any plane motion can be replaced by a translation of an
arbitrary reference point A and a simultaneous rotation
about A.



vB  v A  vB A
 

vB A   k  rB A
vB A  r
 


vB  v A   k  rB A
9
Absolute and Relative Velocity in Plane Motion
• Assuming that the velocity vA of end A is known, wish to determine the
velocity vB of end B and the angular velocity  in terms of vA, l, and .
• The direction of vB and vB/A are known. Complete the velocity diagram.
vB
 tan 
vA
v B  v A tan 
vA = vA =
cos
v B A l
=
vA
l cos
10
Absolute and Relative Velocity in Plane Motion
• Selecting point B as the reference point and solving for the velocity vA of end A
and the angular velocity  leads to an equivalent velocity triangle.
• vA/B has the same magnitude but opposite sense of vB/A. The sense of the
relative velocity is dependent on the choice of reference point.
• Angular velocity  of the rod in its rotation about B is the same as its rotation
about A. Angular velocity is not dependent on the choice of reference point.
11
Sample Problem 15.3
SOLUTION:
• Will determine the absolute velocity of
point D with



vD  vB  vD B

• The velocity v B is obtained from the
given crank rotation data.

•
The
directions
of
the
absolute
velocity
v
The crank AB has a constant clockwise
D

and the relative velocity v D B are deterangular velocity of 2000 rpm.
mined from the problem geometry.
For the crank position indicated,
• The unknowns in the vector expression
determine (a) the angular velocity of
are the velocity magnitudes vD and vD B
the connecting rod BD, and (b) the
which may be determined from the
velocity of the piston P.
corresponding vector triangle.
• The angular velocity of the connecting
rod is calculated from v D B .
12
Sample Problem 15.3
7.6 cm
SOLUTION: (Graphical Method)
• Will determine the absolute velocity of point D with



vD  vB  vD B

• The velocity vB is obtained from the crank rotation data.
rev  min  2π rad 


ωAB   2000

  209.4 rad s
min
60
s
rev




v B  AB ωAB  7.6 cm.209.4 rad s   1591.44cm/s
The velocity direction is as shown.

• The direction of the absolute velocity vD is horizontal.

The direction of the relative velocity vD B is
perpendicular to BD. Compute the angle between the
horizontal and the connecting rod from the law of sines.
sin40
sinβ

20.3cm. 7.6 cm.
β  13.95
13
Sample Problem 15.3
• Determine the velocity magnitudes vD and vD B
from the vector triangle.
v D B 1591.44cm. s
vD


sin53.95 sin50
sin76.05
vB =1591.44 cm/s

v D/B  1256.16cm. s v D/B 1256.16cm./s  76.05o

v D  1325.78cm. s v D 1325.78cm./s 



vD  vB  vD B
v P  v D  1325.78cm./s 
v D B  lω BD
vD B
1256.16cm. s
l
20.3 cm.
 61.88 rad s
ω BD 



ωBD  61.88 rad s k
14
Sample Problem 15.3
Solution (Vector Method)
sin40
sinβ

20.3cm. 7.6 cm.
β  13.95
rev  min  2π rad 


ω AB   2000

  209.4 rad s
min  60 s  rev 





v B  v A  v B/A ; v A  0




v B  v B/A  ω AB  rB/A




o
o
v B  209.44k  0.076cos40 i  0.076sin40 j



v B  10.23 i  12.19 j


15
Sample Problem 15.3






v D  v B  v D/B  v B  ω BD  rD/B






v D i  10.23 i  12.19 j   ω BD k  0.203cos13.92 i  0.203sin13.92 j 





v D i  10.23 i  12.19 j   0.197ω BD j  0.0488ω BD i
Consider in each components

i ; v D  10.23  0.0488ω BD

j ; 0  12.19  0.197ω BD
ω BD  61.88 rad/s
v D  10.23  0.0488  61.88  13.25 m/s
16
Instantaneous Center of Rotation in Plane Motion
• Plane motion of all particles in a slab can always be
replaced by the translation of an arbitrary point A and
a rotation about A with an angular velocity that is
independent of the choice of A.
• The same translational and rotational velocities at A are
obtained by allowing the slab to rotate with the same
angular velocity about the point C on a perpendicular to
the velocity at A.
• The velocity of all other particles in the slab are the
same as originally defined since the angular velocity
and translational velocity at A are equivalent.
• As far as the velocities are concerned, the slab seems to
rotate about the instantaneous center of rotation C.
17
Instantaneous Center of Rotation in Plane Motion
• If the velocity at two points A and B are known, the
instantaneous center of rotation lies at the intersection
of the perpendiculars to the velocity vectors through A
and B .
• If the velocity vectors are parallel, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
• If the velocity vectors at A and B are perpendicular to
the line AB, the instantaneous center of rotation lies at
the intersection of the line AB with the line joining the
extremities of the velocity vectors at A and B.
• If the velocity magnitudes are equal, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
18
Instantaneous Center of Rotation in Plane Motion
• The instantaneous center of rotation lies at the intersection of
the perpendiculars to the velocity vectors through A and B .
v
v
v
v B   BC   l sin   A
 A  A
l cos
AC l cos
 v A tan 
• The velocities of all particles on the rod are as if they were
rotated about C.
• The particle at the center of rotation has zero velocity.
• The particle coinciding with the center of rotation changes
with time and the acceleration of the particle at the
instantaneous center of rotation is not zero.
• The acceleration of the particles in the slab cannot be
determined as if the slab were simply rotating about C.
• The trace of the locus of the center of rotation on the body
is the body centrode and in space is the space centrode.
19
Sample Problem 15.5
SOLUTION:
• Determine the velocity at B from the
given crank rotation data.
The crank AB has a constant clockwise
angular velocity of 2000 rpm.
For the crank position indicated,
determine (a) the angular velocity of
the connecting rod BD, and (b) the
velocity of the piston P.
• The direction of the velocity vectors at B
and D are known. The instantaneous
center of rotation is at the intersection of
the perpendiculars to the velocities
through B and D.
• Determine the angular velocity about the
center of rotation based on the velocity
at B.
• Calculate the velocity at D based on its
rotation about the instantaneous center
of rotation.
20
Sample Problem 15.5
SOLUTION:
• From Sample Problem 15.3,





v B  1023 i  1219j cm. s 

v B  1591.44cm. s
50o
β  13.95
• The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities through B and D.
 B  40    53.95
 D  90    76.05
BC
CD
20.3 cm.


sin76.05 sin53.95 sin50 
BC  25.7 cm. CD  21.4 cm.
• Determine the angular velocity about the center of
rotation based on the velocity at B.
v B  BC ω BD
ω BD 
v B 1591.44cm. s

BC
25.7 cm.
ωBD  61.88 rad s
• Calculate the velocity at D based on its rotation about
the instantaneous center of rotation.
v D  CD ωBD  21.4 cm.61.88 rad s 
v P  v D  1325cm. s
21
Absolute and Relative Acceleration in Plane Motion
• Absolute acceleration of a particle of the slab,



aB  a A  aB A

• Relative acceleration a B A associated with rotation about A includes
tangential and normal components,
 

a B A    k  rB A a B A   r
t
aB A n   2 rB A
t
a B A n  r 2
22
Absolute and Relative Acceleration in
Plane Motion


• Given a A and v A ,


determine a B and  .



aB  a A  aB A



 a A  a B A   a B
n

• Vector result depends on sense of a A and the
relative magnitudes of a A and a B A 
n
• Must also know angular velocity .
23

A t
Absolute and Relative Acceleration in
Plane Motion



• Write a B  a A  a B

 x components:

A
in terms of the two component equations,
0  a A  l 2 sin   l cos
y components:  a B  l 2 cos  l sin 
• Solve for aB and .
24
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the
connecting rod BD and the acceleration
of point D will be determined from






aD  aB  aD B  aB  aD B  aD B

t 
n
• The acceleration of B is determined from
the given rotation speed of AB.
Crank AG of the engine system has a
constant clockwise angular velocity of
2000 rpm.
• The directions of the accelerations



are
a D , a D B , and a D B
t
n
determined from the geometry.
For the crank position shown,
determine the angular acceleration of
the connecting rod BD and the
acceleration of point D.
• Component equations for acceleration
of point D are solved simultaneously for
acceleration of D and angular
acceleration of the connecting rod.




25
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the connecting rod BD and
the acceleration of point D will be determined from






aD  aB  aD B  aB  aD B  aD B

t 
n
• The acceleration of B is determined from the given rotation
speed of AB.
ω AB  2000 rpm  209.4 rad s  constant
α AB  0
7.6 cm

aB

aB

aB

aB

aB


 a A  a B/A





 a A  a B/A t  a B/A n ; a A  a B/A t  0



 ω 2AB rB/A  209.442 0.076cos40o i  0.076sin40o j


 2553.8 i  2142.89j

 3333.74 m/s 2

40o
26
Sample Problem 15.7




Assign a D  a D i and α BD  α BD k




a D  a B  a D/B t  a D/B n





a D  a B  α BD  rD/B  ω 2BD rD/B



a D  2553.8 i  2142.89j



o
o
 α BD k  0.203cos13.92 i  0.203sin13.92 j


2
o
o
 61.88 0.203cos13.92 i  0.203sin13.92 j







a D i  2553.8 i  2142.89j  0.197α BD j  0.0488α BD i  754.34 i  186.86j




27
Sample Problem 15.7







a D i  2553.8 i  2142.89j  0.197α BD j  0.0488α BD i  754.34 i  186.86j
Consider in each components

i ; a D  2553.8  0.0488α BD  754.34
a D  0.0488α BD  3308.14

j ; 0  2142.89  0.197α BD  186.86
α BD  9929.09 rad/s 2
a D  2823.6 m/s 2 , 2823.6 m/s 2
28
Sample Problem 15.8
SOLUTION:
• The angular velocities are determined by
simultaneously solving the component
equations for



vD  vB  vD B
In the position shown, crank AB has a
constant angular velocity 1 = 20 rad/s
counterclockwise.
• The angular accelerations are determined
by simultaneously solving the component
equations for



aD  aB  aD B
Determine the angular velocities and
angular accelerations of the connecting
rod BD and crank DE.
29
Sample Problem 15.8
SOLUTION:
• The angular velocities are determined by simultaneously
solving the component equations for



vD  vB  vD B








v D  v E  ω DE  rD/E  ω DE k   17 i  17 j


 17ω DE i  17ω DE j







v B  v A  ωAB  rB/A  20k  8 i  14 j


 280 i  160j

 



v D B  ωBD  rD B  ωBD k  12 i  3 j


 3ωBD i  12ωBD j





x components:
 17ωDE  280  3ωBD
y components:
 17ωDE  160  12ωBD


ωBD  29.33rad sk


ωDE  11.29rad sk
30
Sample Problem 15.8
• The angular accelerations are determined by
simultaneously solving the component equations for



aD  aB  aD B




a D  α DE  rD/E  ω 2DE rD/E





2
 α DE k   17 i  17 j  11.29   17 i  17 j




 17α DE i  17α DE j  2170 i  2170 j










2
2 
a B  α AB  rB/A  ω AB rB/A  0  20 8 i  14 j


 3200 i  5600j





a D B  α BD  rB D  ω 2BD rB D

 
 
2
 α BD k  12 i  3 j  29.33 12 i  3 j




 3α BD i  12α BD j  10,320 i  2580j

x components:
y components:



 17α DE  3α BD  15,690
 17α DE  12α BD  6010




2
2
α BD   645 rad s k
α DE  809 rad s k




31
Problem 1
ก้าน AB หมุนรอบจุด A ด้วยความเร็ วรอบคงที่ 900 rpm ในทิศทาง
ตามเข็มนาฬิกา จงหาความเร่ งของจุด P เมื่อ = 60o
32
Problem 1

ขั้นตอนการแก้ปัญหา

rB / A
 1. หาเวคเตอร์ ระบุตาแหน่ ง


v B และความเร็ วเชิงมุม AB
 2. หาความเร็ ว

 3. หาความเร่ ง
aB

 4. หาเวคเตอร์ ระบุตาแหน่ ง rD / B


 5. หาความเร็ ว v และความเร็ วเชิ งมุม BD
D

 6. หาความเร่ ง a
D
33
Problem 1



rB / A
ขั้นตอนที่ 1 : หาเวคเตอร์ระบุตาแหน่ง

o
o
rB / A  0.051sin60 i  0.051cos60 j

vB
ขั้นตอนที่ 2 : หาความเร็ ว

900  2 π 

ω AB  
k  94.25k  cons tan t, α AB  0
60
 

v B  v A  v B/A




v B  v A  ω AB  rB/A




o
o
v B  0   94.25k  0.051sin 60 i  0.051 cos 60 j 



v B  4.16 j  2.4 i m/s  4.8 m/s
60 o


34
Problem 1


ขั้นตอนที่ 3 : หาความเร่ ง a B

aB

aB

aB

aB

aB


 a A  a B/A





 a A  a B/A t  a B/A n ; a A  a B/A t  0


2
2 
o
o
 ω AB rB/A  94.25 0.051sin 60 i  0.051cos 60 j


 392.36 i  226.52 j

 453.06 m/s 2

30o
35
Problem 1


ขั้นตอนที่ 4 : หาเวคเตอร์ระบุตาแหน่ง rD / B
15.2
5 .1

sin 60 o sin 
D

  16.89 o
   180 o  60 o  16.89 o
15.2 cm
  103.11o
  103.11  30
o
  73.11o

o
60o

B
5.1 cm
A



o
o
rD / B  0.152 cos 73.11 i  0.152 sin 73.11 j
36
Problem 1


v
ขั้นตอนที่ 5 : หาความเร็ ว และความเร็
วเชิงมุม
D

BD
 


Assign BD  BD k, v D  v D j


 


v D  v B  v D/B  v B  ω BD  rD/B






o
o
v D j  2.4 i  4.16 j  ω BD k  -0.152 cos 73.11 i  0.152 sin 73.11 j





v D j  2.4 i  4.16 j  0.044ω BD j  0.145ω BD i



Consider in each components

i ; 0  2.4  0.145ω BD
ω BD  16.506
rad/s

j ; v D  4.16  0.044ω BD
 4.16  0.04416.506
 4.886
m/s
 4.886
m/s
37
Problem 1

ขั้นตอนที่ 6 : หาความเร่ ง

aD




Assign a D  a D j and α BD  α BD k




a D  a B  a D/B t  a D/B n





a D  a B  α BD  rD/B  ω 2BD rD/B



a D  392.36 i  226.52 j



o
o
 α BD k  -0.152 cos 73.11 i  0.1523 sin 73.11 j


2
o
o
 16.506  -0.152 cos 73.11 i  0.1523 sin 73.11 j







a D j  392.36 i  226.52 j -0.044 α BD j -0.1454 α BD i  11.9877 i -39.6139 j




38
Problem 1







a D j  392.36 i  226.52 j -0.044 α BD j -0.1454 α BD i  11.9877 i -39.6139 j
Consider in each components

i ; 0  392.3649  0.1454α BD  11.9877
α BD  2616.07 rad/s 2
 2616.07
rad/s 2

j ; a D  226.5181  0.044α BD  39.6139
 226.5181  0.044 2616.07 -39.6139
 151.0249 m/s 2
 151.0249 m/s 2
39
Problem 2

กลไกดังรู ปถูกใช้ในเครื่ องยนต์ของเรื อดาน้ า ซึ่งประกอบด้วยก้านเดี่ยว AB
ก้านต่อ BC และ BD จงหาความเร็ วของจุด C ที่ตาแหน่งดังรู ป โดยก้าน
AB หมุนด้วยความเร็ วเชิงมุมคงที่ 5 rad/s ในทิศทางตามเข็มนาฬิกา
40
Problem 3

จากชุดกลไกที่แสดงดังรู ป กาหนดให้กา้ น AB เคลื่อนที่ในทิศทางหมุนตาม
เข็มนาฬิกา ด้วยความเร็ วเชิงมุมคงที่ 6 rad/s จงหาความเร่ งของจุด D
41
Problem 4

ก้าน BD ยาว 300 mm และก้าน AB หมุนด้วยความเร็ วรอบคงที่ 300 rpm ใน
ทิศทางทวนเข็มนาฬิกา จงหาความเร่ งของจุด D เมื่อ  = 90o
42