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Focusing on the vertical and horizontal directions, clearly
T
cos
β
−
mg
=0
and
T
sin
β
=
m

a

Therefore, the magnitude of the tension force is
T
=
mg
cos
β
Using the equation for the acceleration of the center of mass above to determine

a

, the horizontal component
of Newton’s Second Law (
T
sin
β
=
m

a

) simplifies to
g
tan
β
=(
f
sin
β
+
c
sin
θ
)
˙
φ
2
Finally, since
T
acts at Point G while gravity acts at the center of mass (and contributes no moment at G),
the moment about the sphere’s center of mass is given by
M
G
=
r
B/G
×
T
Inspection of the geometry shows that
r
B/G
=
c
k
and
T
=
T
sin(
β
−
θ
)
i
+
T
cos(
β
−
θ
)
k
Therefore, the moment is
M
G
=
c
k
×
[
T
sin(
β
−
θ
)
i
+
T
cos(
β
−
θ
)
k
]
Now,
k
×
i
=
j
and
k
×
k
=
0
, wherefore
M
G
=
cT
sin(
β
−
θ
)
j
(c)
From the angularmomentum equation, we know that
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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

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