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001
10.0 points
An airplane of mass 998 kg flies level to the
ground at a constant speed of 175 m
/
s relative
to the Earth.
An observer on the ground
along the path of the plane sees the plane a
distance 13003
.
7 m away at an angle above
the horizontal of 57
.
77
◦
.
What is the magnitude of the airplane’s
angular momentum relative to a ground ob
server directly below the airplane?
Correct answer: 1
.
92115
×
10
9
kg m
2
/
s.
Explanation:
Basic Concepts:
vector
L
=
vector
r
×
vector
p
L
=
r m v
sin(
θ
)
= (13003
.
7 m)(998 kg)(175 m
/
s) sin(57
.
77
◦
)
= 1
.
92115
×
10
9
kg m
2
/
s
002
(part 1 of 2) 10.0 points
Given:
Use counterclockwise as the positive
angular direction.
The acceleration of gravity is 9
.
8 m
/
s
2
.
A particle of mass
m
is shot with an initial
velocity 2 m
/
s, making an angle 46
◦
, with the
horizontal as shown in figure.
The particle
moves in the gravitational field of the Earth.
x
y
v
R
v
o
2 m
/
s
v
h
46
◦
R
h
O
C
A
Using the origin as the pivot, find the an
gular momentum (along the
z
axis, using a
righthand coordinate system) when the par
ticle is at the highest point of the trajectory.
Correct answer:

0
.
440144 N m.
Explanation:
Basic Concepts
vector
L
=
vectorr
×
vectorp
vector
L
=
I vectorω
We also need the vector cross product

vectora
×
vector
b

=
a b
sin
φ ,
where
φ
is the angle between the vectors
vectora
and
vector
b
.
Let :
v
o
= 2 m
/
s
,
θ
= 46
◦
,
and
g
= 9
.
8 m
/
s
2
.
Solution:
At the highest point (denote
the position vector to that point with
vector
r
h
) we
know that
vector
v
h
=
vector
v
x
=
v
o
cos
θ
ˆ
ı
and
vector
L
=
vector
r
h
×
vector
p
h
=
vector
r
h
×
m vector
v
h
And,

vector
L

=
m

vector
v
h

vector
r
h

sin
θ
h
.
Let us analyze this expression.
We see that
vector
r
h
sin
θ
h
is the maximum height of
y
max
=
h
h
=

vector
r
h

sin
θ
h
=
v
o
2
sin
θ
2
2
g
.
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 Spring '08
 Turner
 Force, Mass, Work, 2 w, 2 m, 2.9 m, 2 2.9 m

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