Hence, the angular momentum vector is
H
G
=[
I
]
ω
=
⎡
⎣
1
2
mr
2
00
0
1
4
mr
2
0
00
1
4
mr
2
⎤
⎦
⎧
⎨
⎩
ω
cos
β
−
ω
sin
β
0
⎫
⎬
⎭
=
⎧
⎨
⎩
1
2
mr
2
ω
cos
β
−
1
4
mr
2
ω
sin
β
0
⎫
⎬
⎭
Thus, in standard vector form, we have
H
G
=
1
2
mr
2
ω
cos
β
i
I
−
1
4
mr
2
ω
sin
β
j
I
In terms of the
xyz
coordinate system, this becomes
H
G
=
1
2
mr
2
ω
cos
β
(cos
β
i
+sin
β
j
)
−
1
4
mr
2
ω
sin
β
(
−
sin
β
i
+cos
β
j
)
=
1
4
mr
2
ω
D
2cos
2
β
+sin
2
β
i
i
+
1
4
mr
2
ω
sin
β
cos
β
j
We can simplify this expression using standard halfangle trigonometric identities, viz.,
cos
2
β
=
1
2
(1 + cos 2
β
)
,
sin
2
β
=
1
2
(1
−
cos 2
β
)
,
sin
β
cos
β
=
1
2
sin 2
β
Substituting into the equation above and combining like terms yields
H
G
=
1
8
mr
2
ω
(3 + cos 2
β
)
i
+
1
8
mr
2
ω
sin 2
β
j
(b)
The angle between
H
G
and the axle is given by
tan
θ
=
(
H
G
)
y
(
H
G
)
x
=
1
8
mr
2
ω
sin 2
β
1
8
mr
2
ω
(3 + cos 2
β
)
Therefore, we conclude that
θ
=tan
−
1
w
sin 2
β
3+cos2
β
W
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

Click to edit the document details