Problem 4.3
In the linkage shown below, locate all of the instant centers.
3.3"
AB = 1.35"
BD = 3.9"
DE = 0.9"
BC = 0.9"
CF = 2.0"
F
B
6
5
C
2
3
55˚
A
E
4
D\
Solution
1
6
2
5
3
I 36
I13
I1
B I2
4
I15
6
F 6
5
I56
3
2
I25 C I
3
5
3
A
I26
I12
I 24
I14
E
I46
4
I3
D
4
I4
5
Problem 4.4
Find all of the instant centers of velocity for the mechanism shown below.
-4-
I 15
I 45
D
I34
I35
E
5
I 25
4
I 24
I 23
C
2
E
5
D
A
I 12
4
C
5.0 cm
I14
2
28˚
3
A
1
AB = 8.0 cm
AC = 4.5 cm
BD = 13.0 cm
DE = 2.9 cm
3
5
B
I 13
2
B
4
-5-
3
Solution
Problem 4.5
Locate all of the instant centers in the mechanism shown below. If link 2 is turning CW at the
rate of 60 rad/s, determine the linear velocity of points C and E using instant centers.
AD = 3.8"
AB = 1.2"
BC = 3.0"
CD = 2.3"
CE = 1.35"
EB = 2.05"
C
3
4
B
E
2
125˚
A
D
Velocity Analysis
The two points of interest are on link 3. To find the angular velocity of link 3, use I13 and I23.
Then
1vI
23
= 1ω2 × rI23 /I12 = 1ω3 × rI23/I13
Therefore,
1ω3
= 1ω2
rI23/ I12
= 60 1.2 = 17.7 rad / s
4.07
rI23 /I13
Because the instant center I23 lies between I12 and I13, 1ω3 is in the opposite direction of 1ω2.
Therefore, 1ω3 is counterclockwise.
Then,
1vC
= 1ω3 × rC/I13 ⇒ 1vC3 = 1ω3 rC/I13 = 17.7⋅2.11 = 37.3 in / s
1vE
= 1ω3 × rE/I13 ⇒ 1v E3 = 1ω3 rE/I13 = 17.7⋅3.25 = 57.5 in / s
3
and
3
The directions for the velocity vectors are shown in the drawing.
-6-
I 13
1
4
I 34
2
3
C
vC 3
4
3
E
vE 3
vB3
I23
B
2
D
I 24
A
I 14
I12
-7-
Problem 4.6
Locate all of the instant centers in the mechanism shown below. If the cam (link 2) is turning
CW at the rate of 900 rpm, determine the linear velocity of the follower using instant centers.
3
103˚
R
B
AB = 1.5"
R = 0.75"
2
70˚
A
Instant Centers
-8-
Velocity of the Follower
Convert the angular velocity from “rpm” to “rad/s”
ω2 = 900 rpm =
1
900(2π )
= 94.25 rad / s CW
60sec
At the point I 23 the linear velocity of follower and cam is same.
1
v I 23 = 1 v A2 + 1 v I23 / A2 = 0 + 1ω2 × rI 23 / A = (94.25 rad / s )(0.82 in) = 77.285 in / s Down
Problem 4.7
Locate all of the instant centers in the mechanism shown below. If link 2 is turning CW at the
rate of 36 rad/s, determine the linear velocity of point B4 by use of instant centers. Determine the
angular velocity of link 4 in rad/s and indicate the direction. Points C and E have the same
vertical coordinate, and points A and C have the same horizontal coordinate.
D
5
E
4
C
6
100˚
B
3
A
2
AB = 1.1"
AC = 0.9"
CD = 1.5"
DE = 3.25"
Solution:
Find all instant centers and linear velocity of point B2.
1v B
2
= 1ω2 × rB2 /A2 ⇒ 1v B2 = 1ω 2 ⋅ rB2 /A2 = 36⋅1.1 = 39.6 in / s
Using rotating radius method,
-9-
1v B
4
= 32.5 in / s
To calculate the angular velocity of link 4, we can use the relations between related instant
centers.
1ω 2 × rI 24 /I12 =1ω 4 × rI24 /I14
1ω 4
= 1ω 2 ⋅
rI24 /I12
= 36 ⋅ 1.283 = 21.1 rad / s
rI24 /I14
2.186
Therefore,
1ω 4
= 21.1 rad / s CW
- 10 -
I15
I 46
I45
D
4
I14
C
1v
B2
I 56
I 26
I12
I 23
B
I34
2
3
1v 'B
I16
E
6
4
24
I35
I25
1v B
1 v 'I
5
I 36
1
I13
A
2
1v I
6
2
5
3
I 24
24
4
Problem 4.8
Using the instant-center method, find angular velocity of link 6 if link 2 is rotating at 50 rpm
CCW.
- 11 -
Problem 4.12
If ω2 = 5 rad/s CCW, find ω5 using instant centers.
C
B
4
3
5
ω2
62˚
A
E
2
AE = 4.1"
EF = 2.0"
AB = 1.5"
BC = 1.55"
CF = 4.0"
DE = 1.0"
D
F
Solution:
Draw linkage to scale and find necessary instant centers ( I12, I15, and I25 ).
The relationshp between 1ω 2 and 1ω 5 is
1ω2
× rI25/I12 = 1ω5 × rI25 /I15
(1)
Solve Eq. (1) for 1ω5 ,
1 ω5
= 1ω2 ⋅
rI25/ I12
= 5⋅ 1.83 = 4.03 rad / s
rI25 / I15
2.27
So,
1ω5
= 4.03 rad / s CW
- 19 -
C
4
I 34
B
I 45
I 35
3
5
A
E
I 12
I 25
I15
2
I23
1
D
5
2
I13
F
3
4
Problem 4.13
If ω 2 = 1 rad/s CCW, find the velocity of point A on link 6 using the instant center method.
Show vA6 on the drawing.
AC = BC = 1.4"
BE = 3.15"
DF = 1.6"
A
ω2
Y
6
F (3.6", 1.45")
B
2
3
30˚
5
C
35˚
D
E
4
- 20 -
X
Solution:
I 13
I 12
A
I46
B
6
I26
F
I36
2
1 vA 6
1 v'I
I25
D
26
C
5
1 v'
D2
I 35
E
I34
1 vD
I 16
1 vI
3
I 14
4
2
26
1
6
2
5
3
4
I 23
Find necessary instant centers, i.e. I12, I16, and I26 , and the velocity of point D as
- 21 -
Problem 4.19
If the velocity of A2 is 10 in/s to the right, find ω6 using instant centers.
D
AB = 1.75"
BC = 1"
BD = 3"
ED = 2.25"
CE = 1.45"
5
6
B
98˚
3
C
E
4
A
vA2
2
Solution:
Find necessary instant centers as shown in the sketch above, i.e. I12, I16, and I26 . All points in
link 2 have the same velocity; therefore,
1v A
2
= 1v' A2 = 1v I26
Using the rotating radius method,
1vD
6
= 1v D6 / E6 = 13.2 in / s
Now,
1vD / E
6 6
= 1ω 6 × rD6 / E6 ⇒ 1ω6 =
1vD / E
6 6
rD6 / E6
= 13.2 = 5.87rad / s
2.25
Therefore,
1ω6
= 5.87 rad / s CW
- 29 -
1
I56
D
6
2
5
3
I26
vA' 2
5
6
4
I34 , I45
I24
B
C
E
I16
3
4
A
I 46
I14
I23
vA2
2
I12
Problem 4.20
Crank 2 of the push-link mechanism shown in the figure is driven at ω2 =60 rad/s (CW). Find
the velocity of points B and C and the angular velocity of links 3 and 4 using the instant center
method.
B
Y
O2 A = 15 cm
O4 B = 30.1 cm
4
3
D
C
O2
A
B
O4
= 29.5 cm
X
30˚
2
A
Solution:
Find all instant centers and velocity of point A
1v A
2
= 1ω2 × rA2 / O2 ⇒ 1vA2 = 1ω2 × rA2 /O2 = 60⋅0.015 = 0.9 m / s
Using rotating radius method,
1v B
3
= 1.15 m / s
- 30 -
A = 14.75 cm
D
D = 7.5 cm
C
7.5 cm
O
2 O4 =