ENME361: Vibrations, Controls and Optimization I
January 6, 2012
Winter 2012
Solutions for ENME361 HW 1
1.5
In Figure E1.5, a slider of mass
M
r
is located on a bar whose angular displacement in
the plane is described by the coordinate
.
The motion of the slider from the pivot point
is measured by the coordinate
r
1
.
The acceleration due to gravity acts in a direction
normal to the plane of motion.
Assume that the point
O
is fixed in an inertial reference
frame and determine the absolute velocity and absolute acceleration of the slider.
Solution
Choose unit vectors
e
1
and
e
2
fixed to the slider as
shown in the figure.
Then, the position vector from
point
O
to the slider is
11
m
r
re
The absolute velocity is then given by
1
1 1
1 1
1
1 1
1
1
m
dd
r
r
r
r
r
dt
dt
e
V
e
e
e
e
(
t
)
O
2
e
r
1
Noting that
k
, where
k
is the unit vector that points out of the plane, we find that
1 1
1
1
1 1
1
2
m
r
r
r
r
V
e
k
e
e
e
The absolute acceleration is found from
12
1 1
1
2
1 1
1
1
2
1
2
1
1 1
1
1
1
2
1
2
1
2
2
1
1
1
1
1
2
2
m
d
d
d
r
r
r
r
r
r
r
dt
dt
dt
r
r
r
r
r
r
r
r
r
ee
a
e
e
e
e
e
e
e
e
e
e
1.10
Determine the linear momentum for the system shown in Figure E1.5 and discuss if
it is conserved. Assume that the mass of the bar is
bar
M
and the distance from the point
O
to the center of the is
bar
L
.
Solution
The linear momentum of this system is given by
bar
slider
p
p
p
where
2
bar
bar
bar
bar
bar
M
M
L
p
V
e
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View Full Documentwhere
bar
V
is the velocity of the center of mass of the bar. Making use of the slider
velocity determined in the solution to Exercise 1.5, we arrive at
1 1
1
2
slider
r
M
r
r
p
e
e
Thus,
1
1
2
r
bar
bar
r
M r
M
L
M r
p
e
e
Because reaction forces exist at point
O
, we conclude that the linear momentum is not
conserved.
1.14
For the system shown in Figure E1.13, construct the system kinetic energy.
Solution
We make use of Eq. (1.23) to determine
the system kinetic energy.
The position vector
to the center of mass G can be written as:
sin
cos
m
x l
l
r
i
j
And the velocity of the center of mass
G
can be
obtained as,
cos
sin
m
x l
l
V
i
j
Then, from Eq. (1.23), the kinetic energy is
l
G
m
,
J
G
O
O
j
i
2
2
2
2
11
22
2
cos
m
m
G
G
T
m
J
m x
l
xl
J
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 Winter '11
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 Kinetic Energy, Mass, vbar, Lbar

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