Bell, Evan – Homework 7 – Due: Mar 2 2005, 11:00 pm – Inst: Furlan
1
This
printout
should
have
34
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
A
record
has
an
angular
speed
of
39
.
4 rev
/
min.
What is its angular speed?
Correct answer: 4
.
12596 rad
/
s.
Explanation:
1 rev = 6
.
28319 rad, and
1 min = 60 s
,
Therefore,
ω
1
= (39
.
4 rev
/
min)
2
π
rad
rev
¶
1 min
60 s
¶
= 4
.
12596 rad
/
s
.
002
(part 2 of 2) 10 points
Through what angle, in radians, does it rotate
in 1
.
42 s?
Correct answer: 5
.
85887 rad.
Explanation:
θ
=
ω t
= (4
.
12596 rad
/
s) (1
.
42 s)
= 5
.
85887 rad
.
003
(part 1 of 3) 10 points
A ball at the end of a string of length 2
.
1 m
rotates at a constant speed in a horizontal
circle. It makes 5
.
7 rev
/
s.
What is the period of the ball’s motion?
Correct answer: 0
.
175439 s.
Explanation:
The period
T
is the time for one complete
revolution and in this case the ball makes
5
.
7 rev
/
s complete revolutions per second, so
T
=
1 rev
5
.
7 rev
/
s
= 0
.
175439 s
.
004
(part 2 of 3) 10 points
What is the frequency of the motion?
Correct answer: 5
.
7 Hz.
Explanation:
f
=
1
0
.
175439 s
= 5
.
7 Hz
.
005
(part 3 of 3) 10 points
What is the ball’s angular velocity?
Correct answer: 35
.
8142 rad
/
s.
Explanation:
ω
=
2
π
rad
T
=
2
π
0
.
175439 s
= 35
.
8142 rad
/
s
.
006
(part 1 of 3) 10 points
A(n) 54
.
5 kg child stands at the rim of a
merrygoround of radius 1
.
51 m, rotating
with an angular speed of 1
.
44 rad
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What is the child’s centripetal accelera
tion?
Correct answer: 3
.
13114 m
/
s
2
.
Explanation:
a
r
=
r ω
2
= (1
.
51 m)(1
.
44 rad
/
s)
2
= 3
.
13114 m
/
s
2
.
007
(part 2 of 3) 10 points
What minimum force between her feet and
the floor of the carousel is required to keep
her in the circular path?
Correct answer: 170
.
647 N.
Explanation:
F
=
m a
r
= (54
.
5 kg)(3
.
13114 m
/
s
2
)
= 170
.
647 N
.
008
(part 3 of 3) 10 points
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Bell, Evan – Homework 7 – Due: Mar 2 2005, 11:00 pm – Inst: Furlan
2
What minimum coefficient of static friction is
required?
Correct answer: 0
.
319504 .
Explanation:
We know the centripetal force is equal to
the force of friction. Therefore,
m a
r
=
μ m g
Also, the normal force,
N
, is equal to
m g
.
Thus,
μ
=
a
r
g
=
3
.
13114 m
/
s
2
9
.
8 m
/
s
2
= 0
.
319504
.
009
(part 1 of 1) 10 points
A small wheel of radius 1
.
8 cm
drives a
large wheel of radius 12
.
9 cm by having their
circumferences pressed together.
If the small wheel turns at 410 rad
/
s, how
fast does the larger one turn?
Correct answer: 57
.
2093 rad
/
s.
Explanation:
Basic Concept:
Linear and angular ve
locity are related by
v
=
r ω ,
where
ω
is in radians per unit time.
Solution:
The rim of the small wheel moves
at
v
1
=
r
1
ω
1
.
Since their circumferences are in contact, the
rim of the larger one is forced to travel at the
same linear speed, and
v
2
=
r
2
ω
2
r
1
ω
1
=
r
2
ω
2
.
Thus
ω
2
=
r
1
ω
1
r
2
=
(1
.
8 cm) (410 rad
/
s)
(12
.
9 cm)
= 57
.
2093 rad
/
s
.
010
(part 1 of 3) 10 points
A wheel has a radius of 4
.
77 m.
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 Spring '07
 GALLIS,MICHAELRO
 Angular Momentum, Work, Moment Of Inertia, Rotation, Correct Answer

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