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2
V
mR
mR
R
φ
∂
=−
∂
±
±±
2
R
mR
mR
Q
−=
±
2
L
mR
∂
=
∂
±
±
,
2
2
dL
mR
mRR
dt
∂
=+
∂
±
±
±
,
LV
∂
∂
∂
∂
2
2
V
mR
mRR
Q
φφ
∂
+=
−
=
∂
±
±
L
mz
z
∂
=
∂
±
,
mz
dt
z
∂
=
∂
±
,
zz
∂
∂
∂
∂
z
V
mz
Q
z
∂
=
∂
G
For
, using the components of
Fm
a
=
G
a
G
from equation 1.12.3:
()
2
R
RR
±
,
( )
2
FmRR
±
±
z
,
z
=
From Section 10.2, since
R
and
are distances,
and
are forces.
However, since
z
R
Q
z
Q
is an angle,
Q
is a torque. Since
F
is coplanar with and perpendicular to
R
,
QR
F
=
… and all equations agree.
(b)
222
22
sin
vrr
r
θ
φθ
+
±
sin
2
m
LTV
r r
r
V
θφ
=−=
+
+
−
±
L
mr
r
∂
=
∂
±
±
,
mr
dt
r
∂
=
∂
±
,
2
sin
mr
mr
rr
∂
∂
−
∂
∂
±
2
sin
r
V
mr
mr
mr
Q
r
∂
−−
=
−
=
∂
2
L
mr
∂
=
∂
±
±
,
2
2
mr
mrr
dt
∂
∂
±
±
±
,
sin
cos
mr
∂
∂
∂
∂
±
2
2s
i
n
c
o
s
V
mr
mrr
mr
Q
θθ
∂
+−
=
−
=
∂
±
±
±
sin
L
mr
∂
=
∂
±
±
,
2
2
sin
2
sin
2
sin
cos
mr
mrr
mr
dt
∂
=+ +
∂
±
± ±
±
±
,
∂∂
2
2
sin
2
sin
2
sin
cos
V
mr
mrr
mr
Q
∂
+
−
∂
±
±
=
is a force
.
r
Q
r
F
Q
and
Q
are torques.
Since
is in the xy plane, the moment arm for
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.
 Fall '11
 JohnAnderson
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