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Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 1
2 - 1
2.0 INTRODUCTION
There are several practical cases when motion of a particle and a rigid body is
constrained within a plane. Motion of a slider crank, rolling motion of a train wheel,
motion of planar linkages, and etc are among the few examples.
2.1 MOTION OF A POINT ABOUT A FIXED AXIS
2.1.1 Velocity and Acceleration Analysis:
Consider a thin slab which rotates at an angular velocity about a fixed axis through
point O.
Figure 2.1 Rotation about a fixed axis
Attaching a local coordinate Oxyz to the slab (i.e. Oxyz moves or rotates together with
the body at !), we may express the position of point P as
R = Ru (2-1)
The velocity of point P is obtained by differentiating eq(1) to yield
v = R = R u + Ru but u = u (Omega Theorem)
= R u + Ru (2-2)
and similarly, its acceleration by differentiating eq(2) to yield
a = R = R u + R u + Ω R + R
= R u + R u + Ω R + ( R u + R)
= R u + 2 R u + R + (R) (2-3)
However, since the magnitude R is constant, we have R = R = 0. Therefore, the
velocity and acceleration of point P are, respectively, reduced to
v = R (2-4)
a = R + ( R) = at + an (2-5)
as shown in Figure 2-1.
,
O
P
R
Path followed by point P
O
at = R
P
an = (R) x, X
Y, y
v
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 2
2
2.2 GENERAL PLANE MOTION
2.2.1 Velocity and Acceleration Analysis:
A slider-crank mechanism shown in Figure 2-2 is very common in machine design. We
observe that as crank OA rotates about a fixed point O, the connecting rod AB will
undergo the type of motion called general plane motion while slider B having a
rectilinear motion constrained in a horizontal path..
Figure 2-2 General plane motion for connecting AB
The position, velocity and acceleration of point B are determined using the vector
approach as follows. First, we choose a reference coordinate at some fixed point such
as point O. Note in plane motion, the global and local coordinate may be chosen such
that they coincide and each of the axis is parallel to one another.
We may write the position of point B as
OB = OA + AB
Defining RB = OB, R1 = OA, and R2 = AB, we then have the position of point B
expressed as
RB = R1 + R2
O A
B
p
x
y
O A
B
p
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 3
2 - 3
and subsequently the velocity and acceleration are given by
vB = BR = 1R + 2R
aB = BR = 1R + 2R
where the general formula for the position, velocity, and acceleration of the form
R = Ru (2-6)
R = R u + + Ru (2-7)
R = R u + 2 R u + R + (R) (2-8)
may be used to evaluate each of the above expressions.
General Plane motion
Example 2.1: The uniform rod AB rests on the inclined
surface, while end A is attached to a collar which may slide
along the vertical rod as shown in Figure E2.1. Knowing
that the collar A moves downward with a constant speed of
10 m/s when = 30o, determine at this instant the velocity
and acceleration of the roller B.
Solution: We begin our analysis by writing the
position of point B from the reference O as OB = OA + AB (1)
Figure E2.1
Let RB = OB R1 = OA and R2 = AB
Now eq(1) becomes
RB = R1 + R2 and differentiating it to give
vB = BR = 1R + 2R (2)
aB = BR = 1R + 2R (3)
R1 : 1R = – 10J m/s (given)
1R = 0 m/s2 (given)
R2 : Note that for rod AB, the angular velocity (2)and acceleration (2 or 2Ω )are not yet
known at this moment. So, we may express them, respectively as
2 = 2k rad/s (Note k = K and that 2 and Ω 2 are unknowns)
2 = Ω 2 = Ω 2K + 2 K = Ω 2K since K = 0 i.e. no change in its direction!
R2 = R2u2 , u2 = sin i – cosj = 0.5i – 0.866j and R2 = 1.5 m
= 0.75I – 1.3J m (since i = I and j = J)
O
1.5 m
A
B
20o
X
Y
R1 R2
RB
x
y
O
1.5 m
A
B
20o
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 4
4
2R = 2 R2 = (2K) (0.75I – 1.3J) = 1.32I + 0.752J m/s (4)
2R = 2 R2 + 2 (2 R2)
= Ω 2K (0.75I – 1.3J) + (2K) (1.32I + 0.752J)
= 1.3Ω 2I + 0.75Ω 2J + 1.322J – 0.752
2I m/s
2 (5)
Assuming roller B rolls downward, then we also have
vB = vBcos 20o I – vBsin 20
oJ = 0.940vBI – 0.342vBJ (6)
aB = aBcos 20o I – aBsin 20
oJ = 0.940aBI – 0.342aBJ (7)
Solving eqs(2), (4) and (6) yields
vB = 11.31 m/s and 2 = 8.18 rad/s [Ans]
Solving (3), (5), and (7) yields
aB = –131.1 m/s2 and Ω 2 = –56.2 rad/s
2 [Ans]
General Plane motion
Example 2.2: A slider-crank mechanism is
designed and having a motion as shown in Figure E2.2. If crank AB rotates at the constant rate of ω rad/s and is horizontal at the instant
when = tan– 1
(4/3), and roller C is in continuous contact with the horizontal surfaces, determine the velocity and acceleration of roller C at this instant.
Figure E2.2
Solution: We write the position of roller C from a fixed point O or A as
OC = OB + BC (OB = AB)
Let RC = OC R1 = OB and R2 = BC
vC = CR = 1R + 2R (1)
aC = CR = 1R + 2R (2)
R1 : R1 = R1u1 = – bI since i = I ; 1 = K
1R = 1 R1 = K (– bI) = – bJ m/s (3)
1R = 1 R1 + 1 (1 R1) ; 1 = Ω 1 = 0 rad/s2 since is constant
= 0 + K (– bJ) = b2I (4)
A B
C
R1
Y
X
O
R2 RC
A
l
B
C
b
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 5
2 - 5
R2 : For rod BC, the angular velocity (2)and acceleration (2 or 2Ω )are not yet known at
this moment. So, we may express them, respectively as
2 = 2K rad/s (since k = K)
2 = 2Ω = 2Ω K + 2 K = 2Ω
K since K = 0 i.e. no change in direction!
R2 = R2u2 , u2 = – cos i + sinj = – 0.6i + 0.8j and R2 = l m
= – 0.6lI + 0.8lJ m (since i = I and j = J)
2R = 2 R2
= (2K) (– 0.6lI + 0.8lJ) = – 0.6l2J – 0.8l2I m/s (5)
2R = 2 R2 + 2 (2 R2)
= 2Ω K (– 0.6lI + 0.8lJ) + (2K) (– 0.6l2J – 0.8l2I)
= – 0.6l 2Ω J – 0.8l 2Ω
I + 0.6l22I – 0.8l2
2J m/s
2 (6)
Assuming roller C rolls rightward, then we also have
vC = vCi = vCI (7)
aC = aCi = aCI (8)
Solving (1), (3), (5) and (7) yields
vC = 34 b m/s and 2 = –
lb
6.0 rad/s [Ans]
Solving (2), (4), (6) and (8)
aC = b2 +
916
lb
6.0
22 m/s2 and 2Ω
= –34
22
22
6.0 l
b rad/s2 [Ans]
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 6
6
2.2.2 Analysis of Motion With Respect To a Moving Frame
Consider two frames of reference: a fixed frame OXYZ and a rotating frame Oxyz (where Z and
z axis are not shown). We observe that the position vector R of a moving particle B is the same
in both frames; however, its rate of change (i.e. its velocity and acceleration) will be different
depending upon the frame of reference which has been chosen.
Position of Particle B: R = Ru which may be expressed relative to OXYZ or Oxyz
Velocity of Particle B:
Figure 2-3 Velocity analysis of a Particle in Plane Motion
Using a Rotating Frame
vB = ( R )OXYZ = ( R )Oxyz + R = R u + R
= vB/f + vB’ (2-9)
where
vB = ( R )OXYZ is the absolute velocity of particle B observed from OXYZ.
vB’ = R is the velocity of partcle B’ of the moving frame f coinciding with B.
vB/f = R u = but is the velocity of particle B relative to a moving frame f (as observed from
Oxyz).
Note that vB/f or R u may be denoted as R Oxyz . Also observe that if = 0 (i.e. there is no
rotation), then we simply have vB = but which is always tangent to the curved path.
but
R
R
r
X
Y u
B
O
A
x
y B’
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 7
2 - 7
Acceleration of Particle B:
Figure 2-4 Acceleration Analysis of a Particle
in Plane Motion Using a Moving Frame
aB = ( R )OXYZ = R u + 2(but) + Ω R + (R)
= aB/f + acor + aB’ (2-10)
where
aB = ( R )OXYZ is the absolute acceleration of particle B.
aB’ = Ω R + (R) is the acceleration of point B’ of the moving frame f
coinciding with B.
acor = 2(but) is Coriolis, or, complimentary, acceleration.
aB/f = R u = b ut + r
b2
un is the acceleration of B relative to moving frame f.
Note that
(1) aB/f or R u may be denoted as R Oxyz .
(2) If the curved path becomes a straight path, then the term r
b2
un = 0 since r = .
(3) The unit vector un may easily be obtained by performing the cross product of the
direction of rotation of ut (curling toward un) and the vector ut itself. For example, in
the present diagram the direction of rotation of ut curling toward un is clockwise which
may be defined as – k, (and is + k if counterclockwise), therefore, we have
un = – k ut
Again note that if = Ω = 0, then we simply have aB = b ut + r
b2
un which are essentially
the tangential and normal components of acceleration for a motion of B about the center A..
b ut (R)
R
r
X
Y u
B
O
Ω R
2(but)
r
b2
un A
Ω
B’
x
y
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 8
8
Example 2.3
A robot arm AB is rotating counterclockwise
at a rate = 5 rad/s which decreases at 12.5
rad/s2 and at the same time it is being
retracted at a speed b = 1.5 m/s which
increases at 3 m/s2, determine at the instant
shown, the velocity and acceleration of end B
of this robot arm. Use l = 500 mm. Figure E2.3
Solution: The analysis may begin by writing the position vector, and its derivatives as
R = OB or AB = Ru = li = 0.5I (since i = I)
vB = R = R u + R where R = l = – 1.5 m/s; = = k = 5K rad/s (since k = K)
aB = R = R u + 2( R u) + R + (R); where R = l = 3 m/s2;
= Ω = = k + k = –12.5K
vB = R = – 1.5I + (5K)(0.5I)
= – 1.5I + 2.5J m/s [Ans]
aB = R = 3I + 2(5K)(– 1.5I) + (– 12.5K)(0.5I) + (5K)(2.5J)
= 3I – 15J – 7.25J – 12.5I
= – 9.5I – 22.25J m/s2 [Ans]
Each of the components for the velocity and acceleration are shown in Figure E2.3(b).
(i) velocity components (ii) acceleration components
Figure E2.3(b)
B
= R u = – bI
x, X
y, Y
A
R
B
R u = b I
y, Y
A
(R)
acor = 2(– bI)
Ω R
x, X
= Ω
B
b
x, X
y, Y
A
l
O
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 9
2 - 9
Example 2.4
A robot manipulator which consists of a
slider A and a control rod PQ is constrained
to move on horizontal tract as shown in
Figure E2.4. Knowing that at the instant
shown the slider A moves at a constant speed
of 0.5 m/s to the right and that the control rod
PQ retracts at a constant speed of 1.5 m/s
while at the same time rotates at a constant
rate of 10 rad/s about point P, determine the
velocity and acceleration of end Q. Use l =
500 mm.
Figure E2.4
Solution: The velocity and acceleration of point Q may be written, respectively, as
vQ/P = vP + vQ/P (Note that vP = vA = 0.5I m/s) (1)
aQ/P = aP + aQ/P (Note that aP = aA = 0 m/s2 since it slides at the constant speed) (2)
Let RQ/P = PQ = Ru = – lj = – lJ (since j = J)
vQ/P = R Q/P = R u + R (3)
where R = l = 1.5 m/s; = = k = – 10K rad/s (since k = K)
aQ/P = R Q/P = R u + 2( R u) + R + (R); where R = l = 0 ; = Ω = = 0 (4)
vQ/P = R Q/P = 1.5J + (– 10K)(– lJ)
= 1.5J – 10lI m/s relative velocity of end Q to P
aQ/P = R Q/P = 0 + 2(– 10K)(1.5J) + (0)(– lJ) + (– 10K)(– 10lI)
= 0 + 30I + 0 + 100l J
= 30I + 100l J m/s2 relative acceleration of end Q to P
From (1) & (3) vQ = (0.5 – 10l)I + 1.5J m/s [Ans]
From (2) & (4) aQ = 30I + 100l J m/s2 [Ans]
Can you draw the each of the components of the velocity and acceleration of point Q on the
body itself?
Comments: Note that students must be able to handle quantities in both numbers and symbols!
Note also that the term 2(– 10K)(1.5J) should not be 2(– 10K)(pI + 1.5J) since what is
needed here is just vP/ = 1.5J .
P
Q
q
p
l x
y
A
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 10
10
Group work No. 1: Let’s make dynamics fun to learn!
Using the global coordinate attached at a fixed point O, obtain the expressions for the absolute
velocity and absolute acceleration for each of the rollers shown in Figure 2-5 knowning that
each of these rollers moves at a speed e m/s which increases at e m/s2 in the direction as
indicated and that the disk at the same time is rotating anticlockwise at the rate rad/s and is
increasing at rad/s2.
Figure 2-5
Solution: Let R = Ru, note that point P is an arbitrary point. (1)
Then vP = R = R u + R = vP/f + vP’ (2)
aP = R = R u + 2( R u) + R + (R) = aP/f + acor + aP’ (3)
= K rad/s
= Ω = Ω K + K = K rad/s2 [Note Ω = , and K = KK = 0 ]
R : R = Ru = rj = rJ m (since j = J)
vA/f = R u = ei = eI
aA/f = R u = e I + (– r
e
5.0
2
J) [Note aA/f consists of the tangential and normal components.]
acor = 2(vE/f) = 2(K)(eI) = 2eJ
vA’ = R = (K)(rJ) = – rI m/s
aA’ = R + (R) = (K)(rJ) + (K)(– rI) = – rI – 2rJ m/s
2
From (1) vA = R = (e – r)I m/s [Ans]
From (2) aA = R = ( e – r)I + (r
e
5.0
2
+ 2e – 2r)J m/s
2 [Ans]
A
C
B
r
O
e
r r
e
e
e
D
r
0.5r
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 11
2 - 11
Dependent motion
Worksheet 2.1: Arm E is rigidly
welded to a disc of radius 420 mm which
may rotate about the center O. At the instant
shown, the disc has an angular velocity of 5
rad/s clockwise and angular acceleration of
10 rad/s2 counter clockwise. At the same
instant, the rod at C is extruding at a constant
rate of 2 m/s relative to arm E. Determine at
this instant (a) the velocity of point C, and
(b) its acceleration. Figure W2.1
Solution: Position, velocity and acceleration of point C:
OC = R = P + S (1)
vC = R = P + S (2)
aC = R = P + S (3)
where P = PI = 0.74I
P = P I + P I
But I = I and = K = –5K
P = P I + P(I)
P = P I + P = 0 + (–5K) (0.74I) = – 3.7J m/s (4)
P = P I + P I + Ω P + P where Ω = K + K = 10K ( K = 0)
= P I + P (I) + Ω P + ( P I + P)
P = P I + 2 P I + Ω P + (P) (5)
= 0 + 2(–5K)(0) + 10K(0.74I) + (– 5K)(–3.7J)
= –18.5I + 7.4J m/s2
S = S j = 0.38J (since j = J)
S = S j + S j where j = j
= S j + S(j) = S j + S (6)
= 2J + (–5K)(0.38J) = 2J + 1.9I m/s
S = S j + S j + Ω S + S
= S j + S (j) + Ω S + ( S j + S)
S = S j + 2 S j + Ω S + (S) (7)
= 0 + 2(–5K)(2J) + 10K(0.38J) + (–5K)( 1.9I)
= 16.2I – 9.5J m/s2
O
420
C
320
380
380
D
E
X
Y
A
O
C = 10k
= – 5k
E
x, X
Y R
0.74 m
0.38 m
y
A P
S
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 12
12
Finally, we obtain
vC = 1.9I – 1.7J m/s [Ans]
aC = –2.3I – 2.1J m/s2 [Ans]
Alternatively, we may solve this problem as follows. Since P and S have the same angular
motion, we need not to consider them separately. Therefore,
OC = R = OA + AC
= (OA)I + (AC)J = 0.74I + 0.38J m (–5K)(2J) + 10K(0.38J) + (–5K)( 1.9I)
vC = R = ( AO )I + ( CA )J + R
= 0 + 2J + (–5K)( 0.74I + 0.38J)
= 2J + 1.9I – 3.7J
= 1.9I – 1.7J m/s (which is the same as obtained before)
aC = R = ( AO )I + ( CA )J + 2[( AO )I + ( CA )J] + Ω R + (R)
= 0 + 0 + 2(–5K)(2J) + 10K(0.74I + 0.38J) + (–5K)(1.9I – 3.7J)
= –2.3I – 2.1J m/s2 (which is the same as obtained before)
Figure W2.1(b) Graphical depiction of various components of the velocity and acceleration of point C
Comment: We observe that the acceleration for point C and D isn’t the same. This is
expected since point C has another motion relative to a moving frame while point D is
stationary! Suppose that point C is also stationary i.e. no motion relative arm E, will vC = vD
and aC = aD? Explain.
O
C = 10k
= – 5k
E
X
Y
A
aC/ = 0
acor = 20I
aC’,t = –3.8I + 3.7J
aC’, n = –3.8I + 3.7J
O
C = 10k
= – 5k
E
X
Y
A
vC/ = 2J
vC’ = R
= 1.9I – 3.7J
Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 13
2 - 13
Plane Motion of a Rigid Body
Worksheet 2.2: At the instant shown, arm OA has
an angular velocity of 1 rad/s counterclockwise
and is decreasing at a rate of 1 rad/s2. At the
same instant, arm AC is rotating at a rate of 2
rad/s and is increasing at a rate of 2 rad/s2
relative to arm OA. Also that the collar C is
sliding at a rate of u m/s and is increasing at u
m/s2 relative to the arm AC. Determine at the
instant when = tan– 1
(3/4) (a) the velocity of
collar C, and (b) its acceleration.
Figure W2.2
Solution:
Comment:
O
C
l1
1, 1
X
Y
A
2, 2
u, u
l2
Kinematics of Rigid Body In Plane Motion Dynamics and Vibrations 2 - 14
14
Worksheet 2.3
A robot arm is designed to be used in a
welding operation as shown in Figure W2.2
where arm AB at the instant shown is rotating
counterclockwise at a rate = 6 rad/s which
increases at 15 rad/s2 and at the same time it
is being extended at a speed b = 2.5 m/s
which decreases at 8 m/s2, determine at this
instant when = tan – 1
(4/3), the velocity and
acceleration of the welding tip E. Use AB = l1
= 750 mm and BC = l2 = 200 mm. Figure W2.3
Solution:
The analysis may begin by writing the position vector, and its derivatives as
R = OE or AE = Ru = (OB)uOB + (BE)uBE where uOB = cos i – sin j = 0.6I – 0.8J
and uOB = (– k)uOB = – 0.8I – 0.6J
= 0.75(0.6I – 0.8J) + 0.2(– 0.8I – 0.6J)
=
vB = R = R u + R where R = l = – 1.5 m/s; = = k = 5K rad/s (since k = K)
aB = R = R u + 2( R u) + R + (R); where R = l = 3 m/s2;
= Ω = = k + k = –12.5K
vB = R = – 1.5I + (5K)(0.5I)
= – 1.5I + 2.5J m/s [Ans]
aB = R = 3I + 2(5K)(– 1.5I) + (– 12.5K)(0.5I) + (5K)(2.5J)
= 3I – 15J – 7.25J – 12.5I
= – 9.5I – 22.25J m/s2 [Ans]
Each of the components for the velocity and acceleration are shown in Figure E2.3(b).
(i) velocity components (ii) acceleration components
Figure E2.3(b)
B
= R u = – bI
x, X
y, Y
A
R
B
R u = b I
y, Y
A
(R)
acor = 2(– bI)
Ω R
x, X
= Ω
B
b
x, X
y, Y A
O
E
l1
l2