boland (kbb544) – HW8 – mackie – (10611)
1
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printout
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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 3 s time interval. Its angular velocity at the
end of this interval is 13 rad
/
s.
What is the angular acceleration of the
wheel?
Note that the initial angular veloc
ity is
not
zero.
Correct answer:
−
19
.
2586 rad
/
s
2
.
Explanation:
Let :
N
= 20
,
t
= 3 s
,
and
ω
= 13 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
(13 rad
/
s) (3 s)
−
2 (20)
π
(3 s)
2
=
−
19
.
2586 rad
/
s
2
.
keywords:
002
10.0 points
A copper block rests 21.0 cm from the center
of a steel turntable. The coefficient of static
friction between the block and the surface
is 0.50.
The turntable starts from rest and
rotates with a constant angular acceleration
of 0.60 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: 8
.
05487 s.
Explanation:
Let :
r
= 21
.
0 cm
,
μ
s
= 0
.
50
,
α
= 0
.
60 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
.
5 (9
.
81 m
/
s
2
)
(0
.
21 m) (0
.
6 rad
/
s
2
)
2
=
8
.
05487 s
.
003
10.0 points
A rotating wheel requires 3
.
9 s to rotate
through 43
.
1 rev.
Its angular speed at the
end of the 3
.
9 s interval is 108
.
8 rad
/
s.
What is its constant angular acceleration?
Assume the angular acceleration has the same
sign as the angular velocity.
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boland (kbb544) – HW8 – mackie – (10611)
2
Correct answer: 20
.
186 rad
/
s
2
.
Explanation:
Let :
t
= 3
.
9 s
,
Δ
θ
= 43
.
1 rev
,
and
ω
f
= 108
.
8 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
= (108
.
8 rad
/
s) (3
.
9 s)
−
(43
.
1 rev)
2
π
1 rev
= 153
.
515 rad
,
α
=
2 (153
.
515 rad)
(3
.
9 s)
2
=
20
.
186 rad
/
s
2
.
004
10.0 points
A rotating bicycle wheel has an angular
speed of 74
◦
/
s at 5 s and a constant angular
acceleration of 35
◦
/
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
.
5 s.
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 Spring '11
 MACKIE
 Physics, Acceleration, Mass, Correct Answer, kg, Boland

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