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Past exam of the mathematics course of the University of Cambridge
/
2024
/
iii
/
Paper 329
/
1
/
i
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2024
iii
Paper 329
1
2026-09-28
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Solution
i
Solution
0
0
0
i
Axisymmetric
mass conservation
in the spherical gap gives
∂
t
∂
h
+
a
s
i
n
θ
1
∂
θ
∂
(
q
sin
θ
)
=
0.
(1)
Since
h
t
=
−
V
cos
θ
and regularity requires
q
sin
θ
=
0
at
θ
=
0
, integration gives
q
sin
θ
=
aV
∫
0
θ
sin
ϑ
cos
ϑ
d
ϑ
=
2
1
aV
sin
2
θ
,
q
=
2
1
Va
sin
θ
.
(2)
The leading
pressure-driven lubrication flux
is
q
=
−
12
μ
a
h
3
d
θ
d
p
.
(3)
Substitution of
h
and
q
gives
d
θ
d
p
=
−
Δ
3
(
1
−
λ
c
o
s
θ
)
3
6
μ
a
2
V
s
i
n
θ
,
(4)
and therefore
p
(
θ
)
=
λ
Δ
3
(
1
−
λ
cos
θ
)
2
3
μ
a
2
V
+
p
0
.
(5)
Take downward
as
the positive vertical direction. The constant
pressure
contributes no
resultant
, while the
pressure
force
on the inner
sphere
is opposite its outward normal. Thus
F
z
=
−
2
π
a
2
∫
0
π
(
p
−
p
0
)
cos
θ
sin
θ
d
θ
.
(6)
With
t
=
λ
cos
θ
and the supplied
integral
,
F
z
=
−
λ
3
Δ
3
6
π
μ
a
4
V
[
1
−
λ
2
2
λ
+
lo
g
(
1
+
λ
1
−
λ
)
]
.
(7)
The
sign
is upward for
V
>
0
, so this is
a
drag force
.
As
λ
→
0
, the
bracket
is
4
λ
3
/3
+
O
(
λ
5
)
and
F
z
→
−
8
π
μ
a
4
V
/
Δ
3
.
Ancestors
(10)
1
Paper 329
iii
2024
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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