7.2.
Solve:
Since
(
dd
t
=
)
ωθ
we have
θ
f
=
i
+
area under the
ω
versus
t
graph between
t
i
and
t
f
From
t
=
0 s to
t
=
2 s, the area is
( )( )
1
2
20 rad/s
2 s
20 rad
=
. From
t
=
2 s to
t
=
4 s, the area is
. Thus, the area under the
versus
t
graph during the total time interval of 4 s is 60 rad
or (60 rad)
×
(1 rev/2
π
rad)
=
9.55 revolutions.
()
(
)
20 rad/s
2 s
40 rad
=
7.4.
Model:
The airplane is to be treated as a particle.
Visualize:
Solve:
(a)
The angle you turn through is
21
5000 miles
180
1.25 rad
1.25 rad
71.6
4000 miles
rad
s
r
θθ
°
−==
=
=
×
=
°
(b)
The plane’s angular velocity is
1.25 rad
0.139 rad/hr
9 hr
tt
−
==
=
−
5
rad
1 hr
0.139
3.86
10
rad/s
hr
3600 s
−
=×
=
×
Assess:
An angular displacement of approximately onefifth of a complete rotation is reasonable because the
separation between Kampala and Singapore is approximately onefifth of the earth’s circumference.
7.8.
Solve:
Let
R
E
be the radius of the earth at the equator. This means
R
E
+
300 m is the radius to the top of
the tower. Letting
T
be the period of rotation, we have
( )
E
2
E
top
bottom
2
300 m
2
300 m
26
0
0
m
2.18
10
m/s
24 hrs
24 3600 s
R
R
vv
TT
ππ
−
+
−=
− =
=
7.10.
Solve:
The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the
plane must fly from east to west. The speed is
3
2
r
a
d
km
1
m
i
l
e
6.4
10 km
1680
1680
1040 mph
24 hr
hr
hr
1.609 km
vr
×
=
=
×
=
7.12.
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 Spring '06
 staff
 Acceleration, Circular Motion

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