# Example Problem: Rotating Disk A disk rotates inside an oil bath, as

```AE 160 / ME 111 – Dr. Nikos J. Mourtos – Fluid Properties (Viscosity)
Example Problem:
Rotating Disk
A disk rotates inside an oil bath, as shown in the figure. The clearance is filled with oil.
Derive an expression for the viscous torque on the disk.
Given
Disk angular velocity
Oil viscosity
Clearance between disk and container walls
R
Ω
&micro;
S
Assumptions
•
•
Negligible shear forces on the outer edges of the disk.
Negligible shaft thickness, which implies negligible shear forces on the shaft.
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What is different between this problem and the falling cylinder?
In the falling cylinder case, at any point on the surface of the descending cylinder the
velocity of the oil in contact with the cylinder was the same (equal to the velocity of the
cylinder itself). In this problem the velocity of the oil is again equal to the velocity of the
disk itself, at any point on the surface of the disk (no-slip condition). However, the
velocity at various points of the disk surface is different, depending on their distance
from the vertical axis of rotation (shaft). For example, the center of the disk has zero
velocity because r = 0 there, while the edges of the disk have velocity ΩR. Τhe velocity
of other points on the surface of the disk will fall between these two values and it will be
a function of r:
u=!r
(1)
This complicates things because if u = f(r), then τ = f(r), which means that Fv = f(r), and
finally T=T(r).
AE 160 / ME 111 – Dr. Nikos J. Mourtos – Fluid Properties (Viscosity)
Solution
What we need to do here is look for an infinitesimally thin slice of the disk, of thickness
dr, located at some random distance r from the center, such that at any point on its
surface, the velocity and everything else that depends on it, may be considered constant.
The top view of such a slice is shown in the figure below.
The resisting viscous torque of this slice is:
dT = 2r FV
(2)
The factor “2” is there because we have 2 surfaces on the disk (top + bottom) that
generate viscous resistance.
Now: FV = ! dA
(3)
Where dA is the surface area of the slice in contact with the oil:
dA = 2 ! r (dr)
(4)
The viscous shear stress is given by Newton’s law of viscosity:
τ =&micro;
du
dy
where y is measured perpendicular to the surface of the disk.
(5)
AE 160 / ME 111 – Dr. Nikos J. Mourtos – Fluid Properties (Viscosity)
Referring to the side view of the disk shown below:
the no-slip condition allows us to write the derivative du/dy as:
du ! r &quot; 0
=
dy
S
Thus, eq(2) becomes:
(6)
dT = 2r &micro;
!r
&micro;!! 3
2! rdr = 4
r dr
S
S
(7)
Clearly, T depends on r, so we need to integrate dT from r = 0 (disk center) to r = R (disk
edge). The total resisting torque is then:
4!&micro;! 3
!&micro;! R 4
r
dr
#
T
=
S &quot;0
S
R
T=
(8)
```