Dynamics

Spring 2023

Ch 15 Kinematics of Rigid Bodies

Dr. Jyun Rong Zhuang

2023/ 03/ 27

E-mail: jrzhuang@nchu.edu.tw

Office: P426

1
 Objectives
• Classify the various types of rigid-body planar motion

• Investigate rigid-body translation and annular motion about

a fixed axis

• Study planar motion using an absolute motion analysis

• Provide a relative motion analysis of velocity and

acceleration using a translating frame of reference

• Find the instantaneous center of zero velocity and

determine the velocity of a point on a body using this
method

• Provide a relative-motion analysis of velocity and

acceleration using a rotating frame of reference

2
 Contents
• Introduction
• Translation
• Rotation About a Fixed Axis: Velocity
• Rotation About a Fixed Axis: Acceleration
• Rotation About a Fixed Axis: Representative Slab
• Equations Defining the Rotation of a Rigid Body About a Fixed Axis
• Analyzing General Plane Motion
• Absolute and Relative Velocity in Plane Motion
• Instantaneous Center of Rotation
• Absolute and Relative Acceleration in Plane Motion
• Analysis of Plane Motion in Terms of a Parameter
• Rate of Change With Respect to a Rotating Frame
• Plane Motion Relative to a Rotating Frame 3
 Introduction
• Kinematics of rigid bodies:
-relations between time t and the positions s, velocities v, and accelerations a
of the particles forming a rigid body.
Ø Classification of rigid body motions:
1. Translation: 2. rotation about a fixed axis

rectilinear curvilinear
translation. translation.
3. general plane motion. 4. motion about a fixed point.

5. general motion. 4
 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.
  
rB = rA + rB A à1st Differentiating with respect to time,

   
rB = rA + rB A = rA
 
vB = v A ∵ rigid body ∴ %⃗!/#
̇ =0

à2nd Differentiating with respect to time,

  
 

rB = rA + rB A = rA
 
aB = a A ∵ rigid body ∴ %⃗!/#
̈ =0

When a rigid body is in translation, all the points of the body

have the same velocity and the same acceleration at any given instant. 5
 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
àis tangent to the path with magnitude
% sin 0
Δ" = $% Δ& = ' sin + Δ&

." Δ&
-= = lim ' sin + = '&̇ sin +
./ )*→, Δ/

• The same result is obtained from

.'⃗
-⃗ = = 0×'⃗
./
̇ = 456784' -98:;</=
0 = 02 = &2
6
 Rotation About a Fixed Axis. Acceleration
.'⃗ àDifferentiating to determine the acceleration,
-⃗ = = 0×'⃗ 
./  dv d  
a= = (w ´ r )
dt dt
 
d w   dr
= ´r +w´
dt dt

dw   
= ´r +w´v
dt
.0
= @⃗ = angular acceleration,
./
= @2 = 02̇ = &2 ̈
• Acceleration of P is combination of two vectors,

"⃗ = $×
⃗ &⃗ + (×((×&)
⃗
tangential radial
acceleration acceleration
component component 7
 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 = w ´ r = wk ´ r
v = rw
• Acceleration of any point P of the slab,
6
3⃗
"⃗ = $×
⃗ &⃗ + (×((×&)⃗
= $+×&⃗ − (. &⃗ 12 −%⃗
7 5
• Resolving the acceleration into
tangential and normal components,
Vector Magnitude

"⃗/ = $+×&⃗ "/ = &$

"⃗0 = −(. &⃗ "0 = &(. 8

 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.
.& ."
0= -=
./ ./
.0 .-
@= 4=
./ ./
.8& .0 .-
@= 8 =0 4=-
./ .& ."
• Uniform Rotation, a = 0:
q = q0 + w t
• Uniformly Accelerated Rotation,
àa = constant:
w = w0 + a t
q = q 0 + w0t + 12 a t 2
w 2 = w02 + 2a (q - q 0 ) 9
 Sample Problem 15.3 1

Cable C has a constant acceleration of 225 mm/s2 and an initial

velocity of 300 mm/s, both directed to the right.
Determine (a) the number of revolutions of the pulley in 2 s,
(b) the velocity and change in position of the load B after 2 s,
and (c) the acceleration of the point D on the rim of the inner
pulley at t = 0.
• The tangential velocity and acceleration of D are equal to the velocity and acceleration of C.

• Apply the relations for uniformly accelerated rotation to determine velocity and angular
position of pulley after 2 s.

( )
q = w 0t + 12 at 2 = (4 rad s )(2 s ) + 12 3 rad s 2 (2 s )2
= 14 rad
æ 1 rev ö
N = (14 rad )ç ÷ = number of revs N = 2.23rev
è 2p rad ø
10
Sample Problem 15.3 3

• Evaluate the initial tangential and normal acceleration

components of D.

Magnitude and direction of the total acceleration,

aD = (aD )t2 + (aD )2n

2 2
= (225) + (1200) = 1221 mm s 2
aD = 1.221 m s2

f = 79.4°
11
 Example – General Plane Motion
• Planetary gear systems are used in applications requiring a large
reduction ratio and a high torque-to-weight ratio.
• The small gears undergo general plane motion: the center of the
gear is translating while the gear rotates about its center.

12
• ©Plus Pix / Agefotostock
 Analyzing General Plane Motion
• General plane motion is neither a translation nor a rotation.

Plane motion = Translation with A + Rotation about A

• General plane motion can be considered as
à the sum of a translation and rotation.

Displacement of particles A and B to A2

and B2 can be divided into two parts:

1. translation to A2 and $1 9

2. rotation of B1¢ about A2 to B2.

13
 Absolute and Relative Velocity in Plane
Motion 1

Plane motion = Translation with A + Rotation about A

• 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 -⃗:/< = 02×'⃗:/<
-:/< = '0
   
vB = v A + w k ´ rB A
14
 Absolute and Relative Velocity in Plane
Motion 2

B B

θ θ
l l

A A

B B

θ θ
l l

A
A

• vA/B has the same magnitude but opposite sense of vB/A.

àThe sense of the relative velocity is dependent on the choice of reference point.
• Angular velocity w of the rod in its rotation about B is the same as its rotation
about A. Angular velocity w is not dependent on the choice of reference point.
15
 Sample Problem 15.6 1

• The double gear rolls on the stationary lower rack: the

velocity of its center is 1.2 m/s.
• Determine (a) the angular velocity of the gear, and (b) the
velocities of the upper rack R and point D of the gear.

• The displacement of the gear center in one revolution = the outer circumference.

• For xA > 0 (moves to right), w < 0 (rotates clockwise).

xA q
=- x A = - r1q
2p r 2p

• Differentiate to relate the translational and angular velocities.

v A = -r1w
vA 1.2 m s
w=- =-   
r1 0.150 m w = w k = - ( 8 rad s ) k

16
 Sample Problem 15.6 3

     
• For any point P on the gear, vP = v A + vP A = v A + w k ´ rP A

• Velocity of the upper rack is equal to velocity of point B:

3⃗$ = 3⃗! = 3⃗# + 12×%⃗!/#
m rad
= 1.2 C⃗ − 8 2× 0.10 m H⃗
s s
m m 
= 1.2 C⃗ + 0.8 C⃗ 
s s vR = ( 2 m s ) i

• Velocity of the point D:

3⃗% = 3⃗# + 12×%⃗%/#

m rad   
= 1.2 C⃗ − 8 2× −0.150 m C⃗ vD = (1.2 m s ) i + (1.2 m s ) j
s s
vD = 1.697 m s 17
 Sample Problem 15.7 1

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.
• Will determine the absolute velocity of point D with,
  
vD = vB + vD B #⃗! is obtained from the
crank rotation data.
æ rev ö æ min ö æ 2p rad ö
w AB = ç 2000 ÷ø ç ÷ çè ÷ø = 209.4 rad s
è min è 60s ø rev
vB = ( AB ) w AB = ( 75 mm )( 209.4 rad s ) = 15, 705 mm/s

• The direction of the absolute velocity #⃗" is horizontal.

• The direction of the relative velocity #⃗"/! is

perpendicular to BD

sin 40° sin b

= b = 13.95°
200 mm 75 mm
18
Sample Problem 15.7 3

B B

D
D

• Determine the velocity magnitudes vD and vD B

from the vector triangle.
#" #$/! 15,705 00/2
= =
sin 5 3.95° sin 5 0° sin76.05°

vD = 13,083mm s = 13.08m s vP = vD = 13.08m s

vD B = 12, 400mm s = 12.4m s

  
vD = vB + vD B

 
w BD = (62.0 rad s )k
19
 Instantaneous Center of Rotation 1

• Angular velocity w is not dependent on the choice of reference point.

• The same translational and rotational velocities at A
àare obtained by allowing the slab to rotate with the same angular velocity
about the point C on a perpendicular to the velocity at A.
• As far as the velocities are concerned, the slab seems to rotate about the
instantaneous center of rotation C.
ü Point C is called the instantaneous center of rotation (IC) and it lies on
the instantaneous axis

ü If the rigid body undergoes general plane motion that is relative to a fixed
reference frame, it must exist at a point where velocity is zero.
• This point is called the instantaneous center of the motion of the rigid body
20
relative to a fixed reference frame.
 Instantaneous Center of Rotation 2

Location of the IC
Velocity of a point on the body is always perpendicular to the
relative-position vector extending from the IC to the point

A
If the v vectors are
parallel

A B

B B
B
A
A

the IC lies at the intersection of the

perpendiculars to the velocity
vectors through A and B

21
A
 Instantaneous Center of Rotation 3

• The particle at the instantaneous center of rotation has zero velocity.

• The trace of the locus of the center of

rotation on the body is the body centrode
and in space is the space centrode.

• The particle coinciding with the instantaneous center of rotation

changes with time and

àthe acceleration of the particle at the instantaneous center of

rotation (IC) is not zero aIC≠0.

• The acceleration of the particles in the slab

cannot be determined as if the slab were simply rotating about C.

22
 Sample Problem 15.9 1

The double gear rolls on the stationary lower rack:

the velocity of its center is 1.2 m/s.
Determine (a) the angular velocity of the gear, and
(b) the velocities of the upper rack R and point D of
the gear.

• Modeling and Analysis:

• The point C is in contact with the stationary lower rack and, instantaneously,
has zero velocity. It must be the location of the instantaneous center of
rotation.
• Determine the angular velocity about C based on the given velocity at A.
#% 1.2 m⁄s
#% = 4% 5 5= =− = 8 rad⁄s
4% 0.15 m

• Evaluate the velocities at B and D based on their rotation about C.

1! = 1" = 2" 3 = 0.25 m 8 rad⁄s #⃗& = 2 m⁄s >⃗

%% = 0.15 m 2 = 0.2121 m #" = 1.697 m⁄s

3% = %% 1 = 0.2121 m 8 rad⁄s #⃗" = 1.2⃗> + 1.2⃗@ m⁄s
23
 Sample Problem 15.10 1

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.
Use method of instantaneous center of rotation

• From Sample Problem 15.3,

!! = 40° + ' = 53.95°
!" = 90° − ' = 76.05°
• The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities through B and D.
B • Determine the angular velocity about the center of
D rotation based on the velocity at B.
40°
A
5!" = 62.0 rad⁄s
&' '2 200 mm
= = • Calculate the velocity at D based on its rotation about
sin 7 6.05° sin 5 3.95° sin50°
the instantaneous center of rotation.
'( = 253.4 mm (0 = 211.1 mm

#2 = #" = 13,080 00/2 = 13.08 0/2 24

 Absolute and Relative Acceleration in
Plane Motion 1

• If the central gear speeds up, a point on one of the

linkages will undergo acceleration because of the
linear motion of the linkage, the angular acceleration
of the point attached to the central gear, and the
normal acceleration of the point attached to the
central gear.

25
• ©Lawrence manning / Corbis RF
 Absolute and Relative Acceleration in
Plane Motion 2

1
L

• Absolute acceleration of a particle of the slab,

"⃗? = "⃗@ + "⃗? ⁄@

• Relative acceleration >⃗"⁄$ associated with rotation about A
includes tangential and normal components.
Vector Magnitude

4⃗:⁄< *
= @2×'⃗:⁄< 4:⁄< *
= '@
"⃗"⁄#
4⃗:⁄< K
= −08'⃗:⁄< 4:⁄< K
= '08
26
 Absolute and Relative Acceleration in
Plane Motion 3

 
• Given a A and v A ,
 
àdetermine B a and a .
  
aB = a A + aB A
= a A + (a B A ) + (a B )
  
n A t

(a)
 (c)
• 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 w. 27
 Absolute and Relative Acceleration in
Plane Motion 4

  
• Write aB = a A + aB A in terms of the two component equations,
+
x components: ® 0 = a A + lw 2 sinq - la cosq
y components: + ­ - a B = -lw 2 cosq - la sin q

à Solve for aB and a.

28
 Analysis of Plane Motion in Terms of a
Parameter
• In some cases, it is advantageous to determine the absolute
velocity and acceleration of a mechanism directly.

x A = l sinq y B = l cosq

v A = x A v B = y B
= lq cosq = -lq sin q
= lw cosq = -lw sin q

a A = xA a B = yB
= -lq 2 sin q + lq cos q = -lq 2 cos q - lqsin q
= -lw 2 sin q + la cos q = -lw 2 cos q - la sin q
29
 Sample Problem 15.13 1

The center of the double gear has a velocity and

acceleration to the right of 1.2 m/s and 3 m/s 5,
respectively. The lower rack is stationary.

Determine (a) the angular acceleration of the gear,

and (b) the acceleration of points B, C, and D.

• The expression of the gear position as a function of q is differentiated twice to define the
relationship between the translational and angular accelerations.
$% 1.2 m⁄s
A$ = −2% C !=− =− = −8 rad⁄s
%6 0.150 m
1$ = −2% Ċ = −2% 3
>$ = −2% C̈ = −2% F 3% 3 m⁄s5
2=− =− '⃗ = ') = − 20 rad⁄s / )
%6 0.150 m
• The acceleration of each point is obtained by adding the acceleration of the gear center and
the relative accelerations with respect to the center.

30
 Sample Problem 15.13 3

"⃗! = "⃗" + "⃗!⁄" = "⃗" + "⃗!⁄" $

+ "⃗!⁄" %
= "⃗" + %&×(⃗!⁄" − *& (⃗!⁄"
= 3 m⁄s& /⃗ + −20 rad⁄s&&× 0.100m 7⃗ − 8 rad⁄s & −0.100m 7⃗
= 3 m⁄s& /⃗ + 2 m⁄s& /⃗ − 6.40 m⁄s& 7⃗

"⃗0 = 5 m⁄' / (⃗ − 6.40 m⁄s / /⃗ "0 = 8.12 m⁄s /

31
Sample Problem 15.13 4

#⃗7 = #⃗# + #⃗7 ⁄# = #⃗# + &'×)⃗7 ⁄# − +9 )⃗7 ⁄#

= 3 m⁄s 9 0⃗ + −20 rad⁄s 9 '× −0.150m 9⃗ − 8 rad⁄s 9
−0.150m 9⃗
= 3 m⁄s 9 0⃗ − 3 m⁄s 9 0⃗ + 9.60 m⁄s 9 9⃗
A⃗7 = 9.60 m⁄s 5 @⃗

#⃗% = #⃗# + #⃗%⁄# = #⃗# + &'×)⃗%⁄# − +9 )⃗%⁄#

= 3 m⁄s 9 0⃗ + −20 rad⁄s 9 '× −0.150m 0⃗ − 8 rad⁄s 9 −0.150m 0⃗
= 3 m⁄s 9 0⃗ + 3 m⁄s 9 9⃗ + 9.60 m⁄s 9 0⃗

A⃗" = 12.6 m⁄2 5 >⃗ + 3 m⁄s 5 @⃗ A" = 12.95 m⁄s 5

32
Sample Problem 15.15 1

Crank AG of the engine system has a constant

clockwise angular velocity of 2000 rpm.
For the crank position shown, determine the angular
acceleration of the connecting rod BD and the
acceleration of point D.

Modeling and Analysis:

• The angular acceleration of the connecting rod BD and
the acceleration of point D will be determined from
#⃗% = #⃗! + #⃗%⁄! = #⃗! + #⃗%⁄! :
+ #⃗%⁄! ;

• The acceleration of B is determined from the given rotation

speed of AB.

33
Sample Problem 15.15 3

B B
G
D D
G

• The directions of the accelerations "⃗1 , "⃗1⁄0 3 , and "⃗1⁄0 4

are
determined from the geometry.
#⃗% = ∓#% 0⃗
From Sample Problem 15.3, wBD = 62.0 rad/s, b = 13.95o.

The direction of (aD/B)t is known but the sense is not known,

34
Sample Problem 15.15 4

• Component equations for acceleration of point D are solved

simultaneously.
>⃗7 = >⃗8 + >⃗7⁄8 = >⃗8 + >⃗7⁄8 9
+ >⃗7⁄8 :
x components:

y components:

Note : if results are + à the assumed direction is correct 35