zhou (cz3574) – HW8 – mackie – (10611)
1
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printout
should
have
16
questions.
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before answering.
001
10.0 points
A wheel rotating with a constant angular ac
celeration turns through 20 revolutions during
a 4 s time interval. Its angular velocity at the
end of this interval is 17 rad
/
s.
What is the angular acceleration of the
wheel?
Note that the initial angular veloc
ity is
not
zero.
Correct answer:
−
7
.
20796 rad
/
s
2
.
Explanation:
Let :
N
= 20
,
t
= 4 s
,
and
ω
= 17 rad
/
s
.
From kinematics
α t
=
ω
f
−
ω
0
ω
0
=
ω
f
−
α t
and
Δ
θ
=
N
2 (
π
)
,
so
Δ
θ
=
ω
0
t
+
1
2
α t
2
2
N π
= (
ω
f
−
α t
)
t
+
1
2
α t
2
=
ω
f
t
−
1
2
α t
2
α
= 2
ω
f
t
−
2
N π
t
2
= 2
(17 rad
/
s) (4 s)
−
2 (20)
π
(4 s)
2
=
−
7
.
20796 rad
/
s
2
.
keywords:
002
10.0 points
A copper block rests 44.9 cm from the center
of a steel turntable. The coefficient of static
friction between the block and the surface
is 0.60.
The turntable starts from rest and
rotates with a constant angular acceleration
of 0.40 rad/s
2
.
After what time interval will the block start
to slip on the turntable? The acceleration of
gravity is 9
.
81 m
/
s
2
.
Correct answer: 9
.
05163 s.
Explanation:
Let :
r
= 44
.
9 cm
,
μ
s
= 0
.
60
,
α
= 0
.
40 rad
/
s
2
,
and
g
= 9
.
81 m
/
s
2
.
From kinematics
ω
f
=
ω
i
+
α
Δ
t
=
α
Δ
t
since
ω
i
= 0 rad/s,
F
c
=
m a
c
=
m
(
r ω
2
f
)
when the block starts to slip, and
F
s
=
μ
s
F
n
=
μ
s
m g ,
so
F
c
=
F
s
m r ω
2
f
=
μ
s
m g
r
(
α
Δ
t
)
2
=
μ
s
g
Δ
t
=
radicalbigg
μ
s
g
r α
2
=
radicalBigg
0
.
6 (9
.
81 m
/
s
2
)
(0
.
449 m) (0
.
4 rad
/
s
2
)
2
=
9
.
05163 s
.
003
10.0 points
A rotating wheel requires 2
.
86 s to rotate
through 18
.
5 rev.
Its angular speed at the
end of the 2
.
86 s interval is 130
.
2 rad
/
s.
What is its constant angular acceleration?
Assume the angular acceleration has the same
sign as the angular velocity.
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zhou (cz3574) – HW8 – mackie – (10611)
2
Correct answer: 62
.
6273 rad
/
s
2
.
Explanation:
Let :
t
= 2
.
86 s
,
Δ
θ
= 18
.
5 rev
,
and
ω
f
= 130
.
2 rad
/
s
.
From kinematics
ω
f
=
ω
0
+
α t
ω
0
=
ω
f
−
α t ,
so
Δ
θ
=
ω
0
t
+
1
2
α t
2
= (
ω
f
−
α t
)
t
+
1
2
α t
2
=
ω t
−
1
2
α t
2
α
=
2 (
ω t
−
Δ
θ
)
t
2
Since
ω t
−
Δ
θ
= (130
.
2 rad
/
s) (2
.
86 s)
−
(18
.
5 rev)
2
π
1 rev
= 256
.
133 rad
,
α
=
2 (256
.
133 rad)
(2
.
86 s)
2
=
62
.
6273 rad
/
s
2
.
004
10.0 points
A rotating bicycle wheel has an angular
speed of 62
◦
/
s at 4
.
8 s and a constant angular
acceleration of 36
◦
/
s
2
. With the center of the
wheel at the origin, the valve stem is on the
positive
x
axis (horizontal) at
t
0
= 1
.
1 s.
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 Spring '10
 Hiyishi
 Acceleration, Mass, Correct Answer, kg, Zhou

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