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C
HAPTER
10
S
OLUTION FOR
P
ROBLEM
21
(a) Use 1 rev = 2
π
rad and 1 min = 60 s to obtain
ω
=
200 rev
1 min
=
(200 rev)(2
π
rad
/
rev)
(1 min)(60 s
/
min)
=20
.
9 rad
/
s
.
(b) The speed of a point on the rim is given by
v
=
ω
r
, where
r
is the radius of the flywheel
and
ω
must be in radians per second. Thus
v
= (20
.
9 rad
/
s)(0
.
60 m = 12
.
5m
/
s.
(c) If
ω
is the angular velocity at time
t
,
ω
0
is the angular velcoty at
t
= 0, and
α
is the angular
acceleration, then since the anagular acceleration is constant
ω
=
ω
0
+
α
t
and
α
=
ω
−
ω
0
t
=
(1000 rev
/
min)
−
(200 rev
/
min)
1
.
0 min
= 800 rev
/
min
2
.
(d) The flywheel turns through the angle
θ
, which is
θ
=
ω
0
t
+
1
2
α
t
2
= (200 rev
/
min)(1
.
0 min) +
1
2
(800 rev
/
min
2
)(1
.
0 min)
2
= 600 rev
.
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View Full Document C
HAPTER
10
S
OLUTION FOR
P
ROBLEM
41
Use the parallelaxis theorem. According to Table 10–2, the rotational inertia of a uniform slab
about an axis through the center and perpendicular to the large faces is given by
I
com
=
M
12
(
a
2
+
b
2
)
.
A parallel axis through a corner is a distance
h
=
0
(
a/
2)
2
+(
b/
2)
2
from the center, so
I
=
I
com
+
Mh
2
=
M
12
(
a
2
+
b
2
)+
M
4
(
a
2
+
b
2
)=
M
3
(
a
2
+
b
2
)
=
0
.
172 kg
3
J
(0
.
035 m)
2
+(0
.
084 m)
2
o
=4
.
7
×
10
−
4
kg
·
m
2
.
C
HAPTER
10
S
OLUTION FOR
P
ROBLEM
55
(a) Use constant acceleration kinematics. If down is taken to be positive and
a
is the acceleration
of the heavier block, then its coordinate is given by
y
=
1
2
at
2
,so
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This note was uploaded on 02/10/2010 for the course BIOL 202 taught by Professor Bloom during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 Bloom

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