Dynamics
487
Problem 4.67
The rope of mass
m
and length
l
is released from rest with a
very
small amount
of it hanging over the edge (i.e.,
s > 0
but it is very close to zero). Determine
the speed of the rope as a function of
s
. Assume the surface is smooth and that
there is no dissipation in the rope as it bends.
Solution
Since the system is conservative the workenergy principle is
T
1
C
V
1
D
T
2
C
V
2
;
(1)
where position
±
is taken to be when
s
±
0
and position
²
occurs at some
point when
s > 0
. With the datum as the table surface, it follows that
V
1
D
0
and
V
2
D
W
v
.
²
s=2/ ;
(2)
where
W
v
refers to the weight of the vertical portion of the rope. The weight
of the horizontal portion of the rope is
W
h
, the constraint force perpendicular
to the table is
N
and the rope’s mass is
m
; it follows that
W
h1
D
mg;
W
h2
D
O.l
²
s/=l
±
mg;
and
W
v2
D
.s=l/ mg:
(3)
Applying Eqs. (3) to Eqs. (2) leads to the following potential energy terms:
V
1
D
0
and
V
2
D
²
s
2
mg=2l:
(4)
The kinetic energies are given by
T
1
D
0;
T
2
D
1
2
O.l
²
±
mv
2
h2
C
1
2
.s=l/ mv
2
D
1
2
mv
2
2
;
(5)
where we note that
v
h2
D
v
D
v
2
. Applying Eqs. (4) and (5) to Eq. (1), the answer for part (a) is
0
D
mv
2
2
=2
²
s
2
mg=2l
)
v
2
D
s
r
g
l
,
(6)
for
s > 0
.
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Solutions Manual
Problems 5.23 through 5.25
Two persons
A
and
B
weighing
150
lb and
180
lb, respectively, jump off a
Foating platform (in the same direction) with a velocity relative to the platform
that is completely horizontal and with magnitude
v
0
D
6
ft
=
s for both
A
and
B
.
The Foating platform weighs
800
lb and assume that
A
,
B
, and the platform are
initially at rest.
Problem 5.23
Neglecting the water resistance to the horizontal motion of the
platform, determine the speed of the platform after
A
and
B
jump at the same
time.
Problem 5.24
Neglecting the water resistance to the horizontal motion of the
platform, and knowing that
B
jumps ±rst, determine the speed of the platform
after both
A
and
B
have jumped.
Problem 5.25
Neglecting the water resistance to the horizontal motion of the
platform, and knowing that
A
jumps ±rst, determine the speed of the platform
after both
A
and
B
have jumped.
WARNING:
the weight of
A
has been changed to
140
lb.
Solution to 5.23
There are no external forces acting on the system. Therefore, the momentum is
conserved in the
x
direction. Let
±
and
²
represent the time instants immediately
before and after the jump.
m
A
.v
Ax
/
1
C
m
B
.v
Bx
/
1
C
m
P
.v
Px
/
1
D
m
A
.v
Ax
/
2
C
m
B
.v
/
2
C
m
P
.v
/
2
:
(1)
Recall that all masses are at rest at
±
and that
.v
Ax
/
2
D
.v
/
2
. Because the velocity
v
0
is a relative velocity
we must write
±
v
0
D
.v
Ax
/
2
±
.v
/
2
)
.v
Ax
/
2
D
.v
/
2
±
v
0
:
Eq. (1) becomes
0
D
±
.v
/
2
±
v
0
²
.m
A
C
m
B
/
C
m
P
.v
/
2
:
Solve for
.v
/
2
.v
/
2
D
.m
A
C
m
B
/ v
0
m
A
C
m
B
C
m
P
D
1:71
ft
=
s
:
Solution to 5.24
There are no external forces acting on the system. Therefore, the momentum
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 Fall '08
 York
 Angular Momentum, MB, .vp, mB C mP

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