arellano (aa39398) – Practice Homework Chapter 10 Solutions – weathers – (17101)
1
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printout
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have
53
questions.
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before answering.
001
10.0 points
A girl sitting on a merrygoround moves
counterclockwise through an arc length of
2.57 m.
If the girl’s angular displacement is 2.27
rad, how far is she from the center of the
merrygoround?
Correct answer: 1
.
13216 m.
Explanation:
Let :
Δ
s
= 2
.
57 m
and
Δ
θ
= 2
.
27 rad
.
Δ
s
=
r
Δ
θ
r
=
Δ
s
Δ
θ
=
2
.
57 m
2
.
27 rad
=
1
.
13216 m
.
keywords:
002
10.0 points
A car tire rotates with an average angular
speed of 29 rad/s.
In what time interval will the tire rotate 3
.
9
times?
Correct answer: 0
.
84498 s.
Explanation:
Let :
ω
avg
= 29 rad
/
s
and
Δ
θ
= 3
.
9 rev
.
ω
avg
=
Δ
θ
Δ
t
Δ
t
=
Δ
θ
ω
avg
=
3
.
9 rev
29 rad
/
s
·
2
π
rad
1 rev
=
0
.
84498 s
.
003 (part 1 of 2) 10.0 points
A wheel starts from rest and rotates with
constant angular acceleration to an angular
speed of 15 rad
/
s in 3
.
05 s.
Find the magnitude of the angular acceler
ation of the wheel.
Correct answer: 4
.
91803 rad
/
s
2
.
Explanation:
Let :
Δ
ω
= 15 rad
/
s
and
Δ
t
= 3
.
05 s
.
Angular acceleration is
α
=
Δ
w
Δ
t
=
15 rad
/
s
3
.
05 s
=
4
.
91803 rad
/
s
2
.
004 (part 2 of 2) 10.0 points
Find the angle in radians through which it
rotates in this time.
Correct answer: 22
.
875 rad.
Explanation:
The angle through which the wheel rotates
during this time interval is
θ
=
1
2
α t
2
=
1
2
(4
.
91803 rad
/
s
2
) (3
.
05 s)
2
=
22
.
875 rad
.
keywords:
005 (part 1 of 2) 10.0 points
A grinding wheel,
initially at rest,
is ro
tated with constant angular acceleration of
5
.
4 rad
/
s
2
for 3
.
9 s. The wheel is then brought
to rest with uniform deceleration in 8
.
75 rev.
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arellano (aa39398) – Practice Homework Chapter 10 Solutions – weathers – (17101)
2
Find the angular acceleration required to
bring the wheel to rest.
Note that an in
crease in angular velocity is consistent with a
positive angular acceleration.
Correct answer:

4
.
03365 rad
/
s
2
.
Explanation:
Let :
ω
0
= 0 rad
/
s
,
α
1
= 5
.
4 rad
/
s
2
,
t
1
= 3
.
9 s
,
and
θ
2
= 8
.
75 rev
.
From kinematics
ω
f
=
ω
0
+
α t .
First find the speed attained before the
wheel begins to slow down. Since
ω
0
= 0 rad
/
s
for the acceleration, the final speed
ω
1
=
α
1
t
1
for the acceleration is also the initial speed for
the deceleration.
Considering the wheel as it comes to rest,
ω
2
2
=
ω
1
2
+ 2
α
2
θ
2
= 0
α
2
=

ω
2
1
2
θ
2
=

α
2
1
t
2
1
2
θ
2
=

(5
.
4 rad
/
s
2
)
2
(3
.
9 s)
2
2 (8
.
75 rev)
·
1 rev
2
π
rad
=

4
.
03365 rad
/
s
2
.
006 (part 2 of 2) 10.0 points
Determine the time needed to bring the wheel
to rest.
Correct answer: 5
.
22107 s.
Explanation:
The initial speed for the deceleration is
ω
1
=
α
1
t
1
, so from kinematics,
ω
f
=
ω
1
+
α
2
t
2
= 0
t
2
=

ω
1
α
2
=

α
1
t
1
α
2
=

(5
.
4 rad
/
s
2
) (3
.
9 s)

4
.
03365 rad
/
s
2
=
5
.
22107 s
.
007 (part 1 of 2) 10.0 points
A remotecontrolled car’s wheel accelerates at
22.7 rad/s
2
.
If the wheel begins with an angular speed of
10.3 rad/s, what is the wheel’s angular speed
after exactly seventeen full turns?
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 Fall '10
 Weathers
 Work, Moment Of Inertia, Correct Answer, Arellano

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