1. The disk D rolls to the left without slipping and has an angular velocity of 3 rad/s. Determine the angular
velocity of link AB and the velocity of center of disk E. Use theMethod of Instantaneous Center of Zero Velocity.
60o
w=3 rad/s
45o
R
R
0.4 m
0.3 m
3R
A
B
D E
R=0.5 m
O1 O2
Instantaneous Center of Zero Velocity
2. The mechanism used in a marine engine consists of a single crank AB and twoconnecting rods BC and BD. Determine the velocity of the pistons at C and D andangular velocities of rods BC and BD when the crank is in the position shown and hasan angular velocity of 5 rad/s.
s/m.ABv ABB 1205 w
Cv
Dv
P2 (IC of link BD)
P1 (IC of link BC)
Instantaneous Center of Zero Velocity
0.5 m
s/m.v
ABv
B
ABB
1205
w
Cv
Dv
P2 (IC for link BD)
P1 (IC for link BC)
P2
D
B
30o
105o
45o
0.6
83
m
s/m.v
s/rad.
.
v
.
v
m.DP
m.BP
sin
DP
sin
BP
sin
.
D
BD
DBBD
5180
4641
35406830
1
3540
6830
3010545
50
2
2
22
w
w
P1
C
B60o
75o
45o
0.5
46
m
s/m.v
s/rad.
.
v
.
v
m.BP
m.CP
sin
BP
sin
CP
sin
.
D
BC
CBBC
8960
83151
48905460
1
5460
4890
756045
40
1
1
11
w
w
0.5 m
Motion Relative to Rotating Axes
4. The gear has the angular motion shown. Determine the angular velocity and angular
acceleration of the slotted link BC at this instant. The pin at A is fixed to the gear.
w=2 rad/s
a=4 rad/s20.5 m
0.7 m
2 m
O
B
A C
Motion Relative to Rotating Axes
Kinetics of Rigid Bodies
wAB=w1=10 rad/s (cw)wBC=2.5 rad/s (ccw)aBC=21.09 rad/s2 (ccw)ac=21.204 m/s2 (↑)
5. A compressor is shown in the figure. Member AB is rotating at a constant angular
velocity of 10 rad/s. Member BC has a mass of 2 kg and piston C has a mass of 1 kg. The
pressure p on the circular surface of the piston is 10 kPa. At the instant shown, what are the
forces transmitted by pins B and C?
6. The unbalanced 20 kg wheel with the mass center at G has a radius of gyration
about G of 202 mm. The wheel rolls down the 20o incline without slipping. In the
position shown. The wheel has an angular velocity of 3 rad/s. Calculate the friction
force F acting on the wheel at this position.
“General Motion”FBD
N
mg
Ff
KD
=
am
aI
222
8160202020 mkg.).(kmI
aa 25.0 raox x
y
j.i..i.kki.ki.aaa
yx
aa
O/GOG
075067502500750330750250 aaa
604805
67502502020
.F
..FsinmgamF
f
fxefx
a
a
a
a51367184
07502020
..N
.cosmgNamF yefy
aa 81602500750 .).(F).(NIM fefGN.N
N.F
s/rad.
f
971160
6172
59715 2
a❷
❸
❶
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