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KinematicsKinematicsof Planar Motion
Pictures from Dynamics by Benson H. Tongue and Sheri D. Sheppard
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
What is the “program?”
• This chapter deals with kinematics (velocities and accelerations) of extended bodies.
• We compute the velocities and acceleration on moving frame and then we express them on moving frame and then we express them in inertial frame.
– Relative velocity of fixed point
– Absolute velocity
– Relative acceleration of moving points
– Absolute acceleration
Planar motion
TranslationCurvilinear Translation
Rotation about a fixed pointGeneral motion
Curvilinear Translation
Fast and Furious
General planar motion: motion of CM + rotation of body
Vector Mechanics for Engineers: Dynamics
Eig
hth
Ed
ition
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,
ABAB rrr���
+=
© 2007 The McGraw-Hill Companies, Inc. All rights reserved. 15 - 5
• Differentiating with respect to time,
AB
AABAB
vv
rrrr��
��
��
��
��
=
=+=
All particles have the same velocity.
AB
AABAB
aa
rrrr��
���
���
���
���
=
=+=
• Differentiating with respect to time again,
All particles have the same acceleration.
Can I has AMBurger?
Considering the rotation of a rigid body about a fixed axis, the position
of the body is defined by the angle θx
r
P
O
A’
θφ
B
z
of the body is defined by the angle θthat the line BP, drawn from the axis of rotation to a point P of the body,
x
yA
v = = rθ sin φ
forms with a fixed plane. The magnitude of the velocity of P is
ds
dt
.
where θ is the time derivative of θ..
v = = rθ sin φds
dt
.
The velocity of P is expressed as
v = = ωωωω x rdr
dt
where the vectorx
r
P
O
A’
θφ
B
z
where the vector
ωωωω = ωk = θk.
is directed along the fixed axis of rotation and represents the angular velocity of the body.
yA
v = = ωωωω x rdr
dt
Denoting by αααα the derivative dωωωω/dt of
the angular velocity, we express the acceleration of P as
a = αααα x r + ωωωω x (ωωωω x r)
ωωωω = ωk = θk.
x
yA
r
P
O
A’
θφ
B
z
The vector αααα represents the angular acceleration of the body and is directed along the fixed axis of rotation.
differentiating ωωωω and recalling that k is constant in magnitude
and direction, we find that
αααα = αk = ωk = θk. ..
yA
x
y
O
ωωωω = ωk
v = ωk x r
r
Consider the motion of a representative slab located in a plane perpendicualr to the axis of rotation of the body. The angular velocity is perpendicular to the slab, so the velocity of point P of the
slab is
y
v = ωk x r
where v is contained in the plane of the at = αk x r
P
x
y
ωωωω = ωk
where v is contained in the plane of the slab. The acceleration of point P can
be resolved into tangential and normal components respectively equal to
at = αk x r at = rα
an= -ω2 r an = rω2αααα = αk
at = αk x r
O an= -ω2 r
P
The angular velocity and angular acceleration of the slab can be expressed as
ω =dθ
dt
α = =dω
dt
d2θ
dt2
ω = αdω
or
ω = αdω
dθ
Two particular cases of rotation are frequently encountered: uniform rotation and uniformly accelerated rotation. Problems involving either of these motions can be solved by using equations similar to those for uniform rectilinear motion and uniformly accelerated rectilinear motion of a particle, where x,
v, and a are replaced by θ, ω, and α.
A
B
vA
vB
Plane motion = Translation with A + Rotation about A
A
B
vA
vA
The most general plane motion of a rigid slab can be considered
B
y’
x’
vB/A
rB/AA
(fixed)ωk
The most general plane motion of a rigid slab can be considered as the sum of a translation and a rotation. The slab shown can be assumed to translate with point A, while simultaneously rotating about A. It follows that the velocity of any point B of the
slab can be expressed as
vB = vA + vB/A
where vA is the velocity of A and vB/A is the relative velocity of Bwith respect to A.
B
y’
x’
vB/A
rB/AA
(fixed)ωk
A
B
vA
vB
Plane motion = Translation with A + Rotation about A
A
B
vA
vA vA vB
vB/A
vB = vA + vB/AvB = vA + vB/A
Denoting by rB/A the position of B relative to A, we note that
vB/A = ωk x rB/A vB/A = (rB/A )ω = rω
The fundamental equation relating the absolute velocities of points A and B and the relative velocity of B with respect to A
can be expressed in the form of a vector diagram and used to solve problems involving the motion of various types of mechanisms.
C
CAnother approach to the solution of problems involving the velocities of
the points of a rigid slab in plane motion is based on determination of the instantaneous center of rotation
C of the slab.
A
B
vA
vB
vA
vB
Plane motion = Translation with A + Rotation about A
The fact that any plane motion of a rigid slab can be considered the sum of a translation of the slab with reference to point A and
A
B
aA
aBA
B
aA
aA
A
B
y’
x’
(aB/A)n
αkωk
(aB/A)t
aB/A
aB = aA + aB/A
the sum of a translation of the slab with reference to point A and a rotation about A is used to relate the absolute accelerations of any two points A and B of the slab and the relative acceleration of B with respect to A.
where aB/A consists of a normal component (aB/A )n of magnitude
rω2 directed toward A, and a tangential component (aB/A )t of magnitude rα perpendicular to the line AB.
Plane motion = Translation with A + Rotation about A
aB = aA + aB/A
A
B
aA
aBA
B
aA
aA
A
B
y’
x’
(aB/A)n
αkωk
(aB/A)t
aB/A
aB = aA + aB/A
The fundamental equation relating the absolute accelerations of points A and Band the relative acceleration of B with respect to A can be expressed in the form of
a vector diagram and used to determine the accelerations of given points of various mechanisms.
(aB/A)n
(aB/A)t
aA
aB aB/A
Plane motion = Translation with A + Rotation about A
aB = aA + aB/A
A
B
aA
aBA
B
aA
aA
A
B
y’
x’
(aB/A)n
αkωk
(aB/A)t
aB/A
The instantaneous center of rotation C
cannot be used for the determination of accelerations, since point C , in general,
does not have zero acceleration.
aB = aA + aB/A
(aB/A)n
(aB/A)t
aA
aB aB/A
RelativeVelocities
vB/A = ω × rB/A
Velocities
Exercise 6.1.6
Exercise 6.1.16 (p. 294)
Exercise 6.1.29 (p. 296)
Problem 6.1.24
/
/
/
ˆ ˆ1.2 2.4 ft
ˆ ˆGiven: 5.0 rad/s and 1.8 ft
ˆ2.1 ft
Determine: and
B A
C BBC
D C
B C
r i j
k r j
r i
v v
= +
ω = = − =
�
� �
�
� �
Problem 6.1.24
Instantaneous center of rotation
/ /A O A O A Ov v r r= + ω× = ω×� �� � � �
24
ˆ
2 2
ˆˆ ˆCheck: ( ) ( )
ˆ ˆ 0
G
C G CG
v v i
r vt
v t r
v v r
r i k r j
r i r i
= −
π π∆ = → ω = =
∆
= + ω× =
= −ω + ω × − = = −ω + ω =
�
�� � �
Problem 6.2.12
ˆladder is 10 ft long and 4 ft/sAv j= −�
