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Bodet (ngb299) – ohw 22 optional – turner – (57340)
1
This printout should have 14 questions.
Multiplechoice questions may continue on
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beFore answering.
001 (part 1 of 3) 10.0 points
As a result oF Friction, the angular speed oF a
wheel c hanges with time according to
dθ
dt
=
ω
0
e

σt
,
where
ω
0
and
σ
are constants. The angular
speed changes From an initial angular speed
oF 3
.
87 rad
/
sto3
.
17 rad
/
sin33
.
93 s
.
Determine the magnitude oF the angular
acceleration aFter 2
.
68 s.
Correct answer: 0
.
0224014 rad
/
s
2
.
Explanation:
Let :
ω
0
=3
.
87 rad
/
s
,
t
0
=0
,
ω
2
=3
.
17 rad
/
s
,
t
2
=33
.
93 s
,
and
t
3
=2
.
68 s
.
The equation oF motion is
ω
=
ω
0
e

σt
,
so
ω
2
ω
0
=
e

σt
2
ln
±
ω
2
ω
0
²
=

σt
2
σ
=

ln
±
ω
2
ω
0
²
t
2
=

ln
±
3
.
17 rad
/
s
3
.
87 rad
/
s
²
33
.
93 s
=0
.
00588043 s

1
.
Thus the angular acceleration at
t
3
is
α
(
t
3
)=
dω
dt
=
ω
0
(

σ
)
e

σt
3
=

(3
.
87 rad
/
s) (0
.
00588043 s

1
)
×
e

(0
.
00588043 s

1
)(2
.
68 s)
=

0
.
0224014 rad
/
s
2
±
%α
(
t
3
)
±
=
0
.
0224014 rad
/
s
2
.
002 (part 2 of 3) 10.0 points
How many revolutions does the wheel make
aFter 2
.
48 s ?
Correct answer: 1
.
51642 rev.
Explanation:
Let :
t
f
=2
.
48 s
.
θ
=

³
t
f
0
ω
0
e

σt
dt
=

ω
0
σ
³
t
f
0
e

σt
(

σdt
)
=

ω
0
σ
e

σt
´
´
´
t
f
0
=
ω
0
σ
(
1

e

σt
f
)
=
3
.
87 rad
/
s
0
.
00588043 s

1
×
µ
1

e

(0
.
00588043 s

1
)(2
.
48 s)
¶
=9
.
52794 rad
,
so the number oF revolutions is
n
=
θ
2
π
=
9
.
52794 rad
2
π
=
1
.
51642 rev
.
003 (part 3 of 3) 10.0 points
±ind the number oF revolutions it makes be
Fore coming to rest.
Correct answer: 104
.
742 rev.
Explanation:
θ
=
ω
0
σ
(
1

e

σt
f
)
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View Full DocumentBodet (ngb299) – ohw 22 optional – turner – (57340)
2
For
t
=
∞
,
this reduces to
θ
end
=
ω
0
σ
=
3
.
87 rad
/
s
0
.
00588043 s

1
1rev
2
π
rad
=
104
.
742 rev
.
004 (part 1 of 2) 10.0 points
Consider a 6 kg square which has its mass
concentrated along its perimeter, with each
side of length 4 m.
d
d
What is the moment of inertia of the square
about an axis perpendicular to the plane of
the square at its center of mass?
Use the
parallel axis theorem and divide the square
into parts. The moment of inertia of a rod
rotated about its CM is
I
CM
rod
=
1
12
md
2
.
Correct answer: 32 kg
·
m
2
.
Explanation:
Let :
m
rod
=
6kg
4
and
$
=4m
.
By the parallel axis theorem, the moment
of inertia of each rod about the center of the
square is
I
rod
square
=
1
12
m
rod
d
2
+
m
rod
±
d
2
²
2
=
1
3
m
rod
d
2
.
Since there are four sides to the square,
the moment of inertia of the square about its
center of mass is
I
CM
square
=4
I
rod
square
=4
±
1
3
m
rod
d
2
²
=
4
3
m
rod
$
2
=
4
3
(
6kg
4
)(4m)
2
=
32 kg
·
m
2
.
005 (part 2 of 2) 10.0 points
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 Spring '08
 Turner
 Physics, Friction

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