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In-Depth Information
=
.
25 kgislaunchedwith the velocity
v
0
=
Aball of mass
m
50 m/s in the direc-
tion shown. If the aerodynamic drag force acting on the ball is
F
D
=
0
C
D
v
3
/
2
, the
differentialequations describing the motionare
C
D
C
D
m
xv
1
/
2
m
yv
1
/
2
x
=−
y
=−
−
g
x
2
where
v
=
+
y
2
.Determine the timeofflight and the range
R
. Use
C
D
=
0
.
03
80665 m/s
2
.
14.
The differentialequationdescribing the angular position
s)
1
/
2
and
g
kg/(m
·
=
9
.
θ
of amechanical
arm is
2
˙
−
θ
−
θ
θ
a
(
b
)
¨
θ
=
2
1
+
θ
and
˙
and
˙
100 s
−
2
and
b
where
a
=
=
15. If
θ
(0)
=
2
π
θ
(0)
=
0, compute
θ
θ
when
t
=
0
.
5 s.
15.
L
=
undeformed length
k
=
stiffness
r
m
The mass
m
issuspended fromanelasticcord with an extensionalstiffness
k
and
undeformed length
L
. If the mass is released fromrest at
θ
=
60
◦
with the cord
unstretched, find the length
r
of the cord when the position
0 is reached for
the first time. The differentialequations describing the motionare
θ
=
k
m
(
r
2
r
˙
r
=
θ
+
g
cos
θ
−
−
L
)
2
r
˙
−
θ
−
g
sin
θ
¨
θ
=
r
80665 m/s
2
,
k
25 kg.
16.
SolveProb. 15 if the pendulum is released from the position
Use
g
=
9
.
=
40 N/m,
L
=
0
.
5 m and
m
=
0
.
θ
=
60
◦
with the
cord stretchedby0
.
075 m.
17.
y
k
m
µ
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