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10.94:
a), g)
b) Using the vector product form for the angular momentum,
so
,
and
2
1
2
1
r
r
v
v

=

=
,
1
1
2
2
v
r
v
r
×
=
×
m
m
so the angular momenta are the same.
c) Let
.
ˆ
j
ω
ϖ
=
Then,
(
29
and
,
ˆ
ˆ
1
1
k
i
v
x
z
r

ϖ
=
×
ϖ
=
(
29
(
29
(
29
(
29
.
ˆ
ˆ
ˆ
2
2
1
1
1
k
j
i
v
L
xR
y
x
xR
mω
r
m
+
+
+

=
×
=
With
,
2
2
2
R
y
x
=
+
the magnitude of
,
and
,
2
is
2
2
1
2
R
m
R
m
ϖ
=
⋅
ω
L
L
and so
.
and
,
cos
6
2
1
)
)(
2
(
2
2
2
π
θ
=
=
=
R
m
R
m
This is true for
2
L
as well, so the total angular
momentum makes an angle of
6
with the +
y
axis.
d) From the intermediate
calculation of part (c),
,
2
1
mvR
R
m
L
y
=
=
so the total
y
component of angular
momentum is
y
y
L
mvR
L
e)
.
2
=
is constant, so the net
y
component of torque is zero. f)
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