Le (tl8426) – HW04 – gentle – (56245)
1
This
printout
should
have
33
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A point on the outer rim of a tire on a moving
vehicle exhibits uniform circular motion with
a frequency of 28 Hz. The diameter of the tire
is 39
.
2 cm.
How fast is the car moving?
Correct answer: 124
.
136 km
/
hr.
Explanation:
You first need to find the length
λ
of one
cycle, which is the circumference of the tire.
Thus
λ
=
πd
Thus the velocity is defined by
v
=
fλ
Conversion
for v:
1
s
cm
1
1 m
100 cm
1 km
1000 m
3600 s
1 hr
= km
/
hr
002(part1of2)10.0points
Young
David,
who
slew
Goliath,
experi
mented with slings before tackling the gi
ant.
He found that with a sling of length
0
.
724 m, he could revolve the sling at the
rate of 10
.
9 rev
/
s. If he increased the length
to 0
.
963 m, he could revolve the sling only
6
.
9 rev
/
s.
a) What is the larger of the two linear
speeds?
Correct answer: 49
.
5844 m
/
s.
Explanation:
BasicConcept:
The angular velocity
ω
is
defined as
ω
≡
v
r
,
in units of
rad
s
.
Solution:
v
=
r ω,
so
v
1
=
r
1
ω
1
= (0
.
724 m) (10
.
9 rev
/
s)
parenleftbigg
2
π
rad
rev
parenrightbigg
= 49
.
5844 m
/
s
.
and
v
2
=
r
2
ω
2
= (0
.
963 m) (6
.
9 rev
/
s)
parenleftbigg
2
π
rad
rev
parenrightbigg
= 41
.
7499 m
/
s
.
Therefore
the
greater
linear
speed
is
49
.
5844 m
/
s.
003(part2of2)10.0points
b) Using the sling length 0
.
963 m, what is the
centripetal acceleration at 6
.
9 rev
/
s?
Correct answer: 1810
.
02 m
/
s
2
.
Explanation:
The centripetal acceleration is given by
a
2
=
v
2
2
r
2
=
r
2
ω
2
2
= (0
.
963 m) (6
.
9 rev
/
s)
2
parenleftbigg
2
π
rad
1 rev
parenrightbigg
2
= 1810
.
02 m
/
s
2
.
004(part1of2)10.0points
The orbit of a Moon about its planet is ap
proximately circular, with a mean radius of
3
.
89
×
10
8
m. It takes 13
.
7 days for the Moon
to complete one revolution about the planet.
Find the mean orbital speed of the Moon.
Correct answer: 2064
.
88 m
/
s.
Explanation:
Dividing the length
C
= 2
πr
of the trajectory
of the Moon by the time
T
= 13
.
7 days = 1
.
18368
×
10
6
s
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Le (tl8426) – HW04 – gentle – (56245)
2
of one revolution (in seconds!), we obtain that
the mean orbital speed of the Moon is
v
=
C
T
=
2
π r
T
=
2
π
(3
.
89
×
10
8
m )
1
.
18368
×
10
6
s
= 2064
.
88 m
/
s
.
005(part2of2)10.0points
Find the Moon’s centripetal acceleration.
Correct answer: 0
.
0109608 m
/
s
2
.
Explanation:
Since the magnitude of the velocity is con
stant, the tangential acceleration of the Moon
is zero. The centripetal acceleration is
a
c
=
v
2
r
=
(2064
.
88 m
/
s )
2
3
.
89
×
10
8
m
= 0
.
0109608 m
/
s
2
.
006
10.0points
A ball on the end of a string is whirled around
in a horizontal circle of radius 0
.
308 m. The
plane of the circle is 1
.
77 m above the ground.
The string breaks and the ball lands 2
.
61 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Find the centripetal acceleration of the ball
during its circular motion.
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 Fall '07
 Swinney
 mechanics, Force, Friction, Mass, Correct Answer

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