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Past exam of the mathematics course of the University of Cambridge
/
2023
/
ii
/
Paper 1
/
39C
/
b
/
iii
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Past exam of the mathematics course of the University of Cambridge
2023
ii
Paper 1
39C
b
Created
2026-09-24
Updated
2026-09-24
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Solution
iii
Solution
0
0
0
iii
The
polar shear stress in a Newtonian fluid
is
σ
r
θ
=
μ
[
r
∂
r
∂
(
r
u
θ
)
+
r
1
∂
θ
∂
u
r
]
.
(1)
On
r
=
a
, no slip holds for every
θ
, so
u
θ
=
0
and
∂
θ
u
r
=
0
. Therefore
σ
r
θ
(
a
,
θ
)
=
μ
∂
r
u
θ
(
a
,
θ
)
=
−
μ
ψ
rr
(
a
,
θ
)
.
(2)
Differentiating the solution from part (ii) gives
ψ
rr
(
a
,
θ
)
=
γ
−
2
γ
cos
2
θ
,
(3)
and hence
σ
r
θ
(
a
,
θ
)
=
−
μ
γ
+
2
μ
γ
cos
2
θ
.
(4)
The moment arm and line element are both
a
, so the
torque
exerted by the fluid on the disk per unit axial
length
is
T
z
=
a
2
∫
0
2
π
σ
r
θ
(
a
,
θ
)
d
θ
=
−
2
π
μ
γ
a
2
.
(5)
For
γ
>
0
the negative
sign
denotes the
clockwise
sense
of the ambient shear
rotation
. This agrees with the
hydrodynamic torque on a fixed disk in planar shear
.
Ancestors
(11)
b
39C
Paper 1
ii
2023
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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