Relative Acceleration Analysis

• Determining the acceleration of various

points in a mechanism is critical for:
– Design of links (e.g stress analysis)
– Prediction of joint wear
– Determination of safe operational parameters
(e.g. max input velocities, loads, etc.)
• Follows similar analysis processes to those
used for velocity analysis

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Recall the 4 cases of referred motion:

Same Point Different Points

Case 2
Same Case 1
DIFFERENCE
Link TRIVIAL CASE
MOTION

Case 4
Different Case 3
DIFFERENCE & RELATIVE
Links RELATIVE MOTION
MOTION

• These can be applied to accel analysis also

MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Consider the simplest non-trivial case first
(different points on the same link, Case 2)
• Clearly, the motion of B relative to A is
B circular (if nonzero) since RB|A has fixed
length
• Differentiate RB|A twice to find AB|A:
|A
RB

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Consider the simplest non-trivial case first
(different points on the same link, Case 2)
• But, since the points are rigidly fixed to
B the same link,
|A
RB

A Tangential component, proportional Normal component, proportional to

to the angular acceleration the angular velocity
(how do we know it is tangential (centripetal acceleration, directed
(perp.) to RB|A?) opposite to RB|A)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Consider the simplest non-trivial case first
(different points on the same link)
α • So, for two points A, B on the same link:
ω
B

• Where:
|A
RB

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Example (Supp Ex A1):
B

3 4

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Solve using a graphical approach:

0A
C

3 4

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Find AB as:

0A
C

3 4

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Expand into tangential and normal
components:

0A
C

3 4

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
C

4 n
3 aA

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
C

aA
4 n
3 aA

t
1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Locates :

0A
C
A AA

aA
4 n
3 aA

t
1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
C
A AA

|A n

aA
4 n
3 aA

t
aB
1 mm = 50 mm/s2

O2 2 O4
A
ω2 α 2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Only direction is known

0A
C
A AA
di
r (a

|A n
3 4 B| t

aB
A
)

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
C
A AA a Bn
di
r (a

|A n
3 4 B

aB
|A t
)

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Only direction is known

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B dir(a B

aB
|A t
)

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Solution for AB comes by intersection

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B| t dir(a B

aB
A
)
AB
1 mm = 50 mm/s2
B
O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X
 Relative Acceleration Analysis
• Intersection also leads to solution for α3

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B| t dir(a B

aB
A
)
AB
t
aB|A 1 mm = 50 mm/s2
B
O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X
 Relative Acceleration Analysis
• Intersection also leads to solution for α3

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B| t dir(a B

aB
A
)
AB
t
aB|A 1 mm = 50 mm/s2
B
O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X X
 Relative Acceleration Analysis
• Similarly, for α4:

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B| t dir(a B

aB
A
) t
AB aB
aB|At 1 mm = 50 mm/s2
B
O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X XX
 Relative Acceleration Analysis
• A similar analysis is used to find AC:

0A
C
A AA a Bn
di t
r (a )

|A n
3 4 B| t dir(a B

aB
A
) t
AB aB
aB|At 1 mm = 50 mm/s2
B
O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X XX
 Relative Acceleration Analysis
• AA is already known:

0A
C
A AA

3 4

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X XX
 Relative Acceleration Analysis
• : Direction and magnitude both known

0A
C
AaC|At AA

3 4

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X XX
 Relative Acceleration Analysis
• : Direction and magnitude both known

0A
C
AaC|At AA

3 4 aC|An

1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
X XX
 Relative Acceleration Analysis
• AC found by measurement:

0A
C
AaC|At AA
AC
3 4 aC|An
C
1 mm = 50 mm/s2

O2 2 O4
A
ω 2 α2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
XXXX
 Relative Acceleration Analysis
• Now consider a more complex case:
(Coincident points on different links, Case 3)
– As in velocity analysis, this is generally seen in the
case of slides
– Slightly more difficult to grasp intuitively, as new
acceleration terms are introduced (slide & Coriolis)
– Unless you’re a large weather system (or a
prismatic slider on a rotating slide) , you’ve likely
not experienced Coriolis acceleration directly
– Caused by change in relative velocity direction due
to rotation of the frame of reference
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Derive the accel. terms for the case below:

Find relative acceleration

4 of C2 with respect to C4

3
C2,C3,C4

2 θ4 = θslide

MECH 335 Lecture Notes

1 © R.Podhorodeski, 2009 1
 Relative Acceleration Analysis
• First, we need the relative displacement
vector, RC2|C3
• By arguments similar to those made for Case 3
velocity analysis, , so:
4

3
C2,C3,C4

2 θ4 = θslide

1 1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Differentiating once more to find
acceleration:

3
C2,C3,C4

2 θ4 = θslide

1 1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Simplifying:

3
C2,C3,C4

2 θ4 = θslide

1 1
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Let’s take a closer look at the terms here:

Sliding acceleration, i.e. instantaneous

rate of change of the slider velocity,
directed along the slide ( )
The Coriolis acceleration ( )
Let’s take a closer look at this one…

Should recognize this as the tangential

acceleration, . (= 0, since rC2|C4 = 0 )
Should recognize this as the normal
acceleration, . (= 0, since rC2|C4 = 0 )

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• The Coriolis acceleration:
Angular velocity Linear velocity of the slider
No particular of the slide along the slide,
qualitative
meaning, just a Equivalent to rotation
consequence of through 90° from θs
differentiation

Just in terms of the direction of , these terms imply rotation through

90°, starting with the direction of and rotating in the sense of (The
magnitude of is easy to determine, just plug in the numbers)
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
Relative Acceleration Analysis
vs vs acC2|C4
ωs

ωs

acC2|C4

acC2|C4
ωs 1 1

ωs

vs vs
acC2|C4

1
MECH 335 Lecture Notes 1
© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Summarizing:

=0 =0

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• NOTE: if the slide is curved (with constant
radius of curvature), a new normal
acceleration term, , is introduced:

• Directed toward the slide’s

ωCC center of curvature
4
|C

• See text sec. 4.5 for further

C2
aN

C
ρC

analysis, including non-

CC constant radius slides

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Example (Supp Ex A3):

3
B

4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Solve using the graphical method:

0A
1 mm = 75 mm/s2

3
B

4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Case 3 acceleration analysis

0A
1 mm = 75 mm/s2

3
B

4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• Expand, remembering the slide and coriolis
terms for the slide relative acceleration:

0A
1 mm = 75 mm/s2

3
=0
B

4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Magnitude & direction both known

0A
1 mm = 75 mm/s2

a
n
B4
3
B

4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Only direction is known

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t )
B4
(a
dir
4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t )
B4
(a B
dir

B2
an
4
2
α2

ω2

O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• : Direction & magnitude both known

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t
B4
)
(a B
dir

B2
an
4
2
α2

ω2

O2 C O4
at
B2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Relative Acceleration Analysis
• The absolute acceleration of B2 is located:

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t )
B4
(a
dir

B2
an
4
2
α2

ω2

AB2
O2 C O4
at
B2

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• Compute the coriolis acceleration:

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t )
B4
(a
dir
4
2
α2

ω2

AB2
O2 C O4

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• But, we wanted the opposite direction:

0A
1 mm = 75 mm/s2

a
n
B4
3
B
t )
B4
(a B
dir
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• : Only the direction is known

dir
(a
s
B4
0A

|B2
1 mm = 75 mm/s2

)
a
n
B4
3
B
t )
B4
(a
dir
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is located by intersection:

dir
(a B
t
4|B
0A

2)
1 mm = 75 mm/s2 AB4
B4

a
n
B4
3
B
t )
B4
(a
dir
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is also located by intersection, giving :

dir
(a B
t
4|B
0A

2)
1 mm = 75 mm/s2 AB4
B4

a
n
t

B4
3 a B4
B
t )
B4
(a
dir
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is now found easily:

dir
(a B
t
4|B
0A

2)
1 mm = 75 mm/s2 AB4
B4

a
n
t

B4
3 a B4
B
t )
B4
(a
dir
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is now found easily:

dir
(a B
t
4|B
0A
a nC

2)
1 mm = 75 mm/s2 AB4
B4

a
n
t

B4
3 a B4
B
t )
B4
(a
dir

a B4
t
|B2
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is now found easily:

dir
(a B
t
a tC

4|B
0A
a nC

2)
1 mm = 75 mm/s2 AB4
B4

a
n
t

B4
3 a B4
B
t )
B4
(a
dir

a B4
t
|B2
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Relative Acceleration Analysis
• is now found easily:

dir
C

(a B
AC

t
a tC

4|B
0A
a nC

2)
1 mm = 75 mm/s2 AB4
B4

a
n
t

B4
3 a B4
B
t )
B4
(a
dir

a B4
t
|B2
4
2
α2

ω2

AB2
O2 C O4 c
B4
|B2
a

X X X MECH 335 Lecture Notes

© R.Podhorodeski, 2009
B2
 Analytical Acceleration Solutions
• As we did for velocity, we can now look at
purely analytical acceleration solutions:
• Illustrate the same 3 approches:
– Analytical treatment of relative accel. equations
– Use of vector cross products
– Time derivatives of loop closure equations

MECH 335 Lecture Notes

© R.Podhorodeski, 2009
 Analytical Acceleration Solutions
• Return to the well-worn example, recalling
our velocity solutions:
A

α2
α3 = ?
R2
ω2 R3

ω3
θ2

θ3
aB = ?
R0 R1 vB
B
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Acceleration Solutions
• Solution 1: Analytical treatment of relative acceleration
– Form the relative acceleration equation: (1)
– Express each term in polar-complex form: (2)
(3)
(4)
– Substitute (2)-(4) back into (1), and substitute :
A

α2
α3 = ?
R2
ω2 R3

ω3
θ2
• Split into real and imaginary equations:
θ3
aB = ?
R0 R1 vB
B
Re:
Im:
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Acceleration Solutions
• Solution 1: Analytical treatment of relative acceleration
Re:
Im:

– Solve the Im equation for :

– Substitute back into the Re equation:

A

α2
α3 = ?
R2
ω2 R3

θ2
ω3
• NOTE: Several simplifying steps have been left out
aB = ?
θ3 of this, and the following solutions. You should
R0 R1 vB attempt these solutions in full yourselves, as it is
B
easy to make errors on exams if you’re not REALLY
familiar with the steps involved
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Acceleration Solutions
• Solution 2: Vector cross-products:
– Start again from the relative velocity eqn: (1)
– Express each term using vector cross-products:

• Where:

α2
R2
α3 = ? • Substitute into (1), expand, and split the
ω2 R3
resulting equation into i and j equations. The
ω3
θ2 solution can be found from this system of 2
aB = ?
θ3 equations in 2 unknowns
R0 R1 vB
B • Again, practice this in full! These solutions
are lengthy, and easy to make errors on.
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
 Analytical Acceleration Solutions
• Solution 3: Time derivatives of loop closure equation:
– Write the loop closure equation:
– Substitute the p-c forms:
– Take the time derivative to get the velocity closure equation:
Note: only the non-zero terms
have been expanded here. Make
sure you can do this in full!
A

α2 • Take the time derivative again to get the

α3 = ?

ω2
R2
R3
acceleration closure equation:
ω3
θ2

θ3
aB = ?
R0 R1 vB
B
• Substitute , form a system
of 2 equations, and solve
MECH 335 Lecture Notes
© R.Podhorodeski, 2009
End of Lecture Pack 5