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001
10.0 points
There is a moon orbiting an Earthlike planet.
The mass of the moon is 5
.
03
×
10
22
kg, the
centertocenter separation of the planet and
the moon is 7
.
48
×
10
5
km, the orbital period
of the moon is 25
.
1 days, and the radius of the
moon is 1520 km.
What is the angular momentum of the
moon about the planet?
Correct answer: 8
.
15387
×
10
34
kg m
2
/
s.
Explanation:
The angular speed
ω
in radians per unit
time, for a complete circle is
ω
=
2
π
T
.
Angular momentum
L
is
L
=
mvr
=
mr
2
ω
=
2
πmr
2
T
= 2
π
(5
.
03
×
10
22
kg)
×
bracketleftbigg
(7
.
48
×
10
5
km)
1000 m
1 km
bracketrightbigg
2
(25
.
1 days)
parenleftbigg
24 hr
1 day
parenrightbigg
= 8
.
15387
×
10
34
kg m
2
/
s
.
002
(part 1 of 3) 10.0 points
A
3
kg
bicycle
wheel
rotating
at
a
2753 rev
/
min angular velocity has its shaft
supported on one side, as shown in the fig
ure. The wheel is a hoop of radius 0
.
6 m, and
its shaft is horizontal. The distance from the
center of the wheel to the pivot point is 0
.
4 m.
When viewing from the left (from the posi
tive
x
axes), one sees that the wheel is rotat
ing in a clockwise manner.
Assume:
All of the mass of the system is
located at the rim of the bicycle wheel. The
acceleration of gravity is 9
.
8 m
/
s
2
.
y
z
x
b
mg
ω
R
The magnitude of the angular momentum
of the wheel is given by
1.
bardbl
vector
L
bardbl
=
mR
2
ω
correct
2.
bardbl
vector
L
bardbl
=
1
4
mR
2
ω
3.
bardbl
vector
L
bardbl
=
mR
2
ω
2
4.
bardbl
vector
L
bardbl
=
1
4
mR
2
ω
2
5.
bardbl
vector
L
bardbl
=
1
2
mR
2
ω
2
6.
bardbl
vector
L
bardbl
=
1
2
mR
2
ω
Explanation:
Basic Concepts:
vector
τ
=
d
vector
L
dt
Solution:
The magnitude of the angular
momentum
I
of the wheel is
L
=
Iω
=
mR
2
ω,
and is along the negative
x
axis.
003
(part 2 of 3) 10.0 points
Let :
m
= 3 kg
,
ω
= 2753 rev
/
min
b
= 0
.
4 m
,
and
R
= 0
.
6 m
.
Find the change in the precession angle
after a 1
.
2 s time interval.
Correct answer: 2
.
59689
◦
.
Explanation:
homework 29 – Turner – (58185)
1
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hinojosa (jlh3938) – homework 29 – Turner – (58185)
2
Let :
ω
= 2753 rev
/
min
=
2
π
(2753 rev
/
min)
(60 s
/
min)
= 288
.
293 rad
/
s
.
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 Spring '08
 Turner
 Angular Momentum, Force, Mass, Work, Correct Answer, Orders of magnitude, Hinojosa

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