Chapter 3
17.
Since the motor is turned off, its torque is
T
= 0, and the only torque acting on the ﬂywheel
is the retarding one,
−
5
√
ω
. Then Newton’s second law for rotational motion becomes
I
dω
dt
=
−
5
√
ω
with
ω
(0) =
ω
0
= 225 (rad
/
sec)
and
I
= 50
(
kg
/
m
2
)
.
The general solution to this separable equation is
ω
(
t
) =
−
5
2
I
t
+
C
=
−
0
.
05
t
+
C.
Using the initial condition, we find
ω
(0) =
−
0
.
05
·
0 +
C
⇒
C
=
ω
(0) =
√
225 = 15
.
Thus
t
=
1
0
.
05
15
−
ω
(
t
)
= 20 15
−
ω
(
t
)
.
At the moment
t
=
t
stop
when the ﬂywheel stops rotating we have
ω
(
t
stop
) = 0 and so
t
stop
= 20(15
−
√
0) = 300 (sec)
.
19.
There are three forces acting on the object:
F
1
, the force due to gravity,
F
2
, the air resistance
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Force, Moment Of Inertia, initial condition

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