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Exercises 3.4
Using equation (3.11) and noting that
dv/dt
=(
dv/dx
)(
dx/dt
)=(
dv/dx
)
v
, we can determine
the maximum height attained by the shell.
With the above substitution, equation (3.11)
becomes
mv
dv
dx
=
−
(
mg
+0
.
1
v
2
)
,v
(0) = 500
.
Using separation of variables and integration, we get
vdv
10
mg
+
v
2
=
−
dx
10
m
⇒
1
2
ln
(
10
mg
+
v
2
)
=
−
x
10
m
+
C
⇒
10
mg
+
v
2
=
Ke
−
x/
(5
m
)
.
Setting
v
= 500 when
x
= 0, we Fnd
K
=
e
0
(
10(3)(9
.
81) + (500)
2
)
= 250294
.
3
.
Thus the equation of velocity as a function of distance is
v
2
+10
mg
= (250294
.
3)
e
−
x/
(5
m
)
.
The maximum height will occur when the shell’s velocity is zero, therefore
x
max
is
x
max
=
−
5(3) ln
±
0 + 10(3)(9
.
81)
250294
.
3
²
≈
101
.
19 (meters)
.
15.
The total torque exerted on the ﬂywheel is the sum of the torque exerted by the motor and
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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