Dynamics and Vibrations Kinematics of Rigid Body In Plane Motion 2 - 1

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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

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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

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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

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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

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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]

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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’

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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

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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

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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

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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

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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

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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

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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

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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