Consider a solid cylinder housed inside another cylinder as shown in the figure. Radius of the inner cylinder is 1 m and its height is 2 m. The gap between the cylinders is 5 mm and is filled with a fluid of viscosity \( 10^{-4} \) Pa-s.
The inner cylinder is rotating at a constant angular speed of 5 rad/s while the outer cylinder is stationary. Friction at the bottom surfaces can be ignored. Velocity profile in the vertical gap between the cylinders can be assumed to be linear.
The driving moment required for the rotating motion of the inner cylinder is _________ Nm (rounded off to two decimal places).
Show Hint
For problems involving rotating cylinders and fluid flow, use the velocity profile and shear stress relations to calculate torque and other quantities.
The torque required to rotate the inner cylinder is given by:
\[
T = \int_0^h \left( \frac{r \Delta P}{\ln\left(\frac{r_o}{r_i}\right)} \right) r \, dz,
\]
where \( r \) is the radius, \( \Delta P \) is the pressure difference across the fluid gap, \( h = 2 \, \text{m} \), and \( r_o \) and \( r_i \) are the radii of the outer and inner cylinders respectively.
Using the assumption of a linear velocity profile and the relation for shear stress:
\[
\tau = \mu \frac{du}{dy},
\]
we calculate the moment of the inner cylinder. This yields:
\[
T = 1.24 \, \text{Nm}.
\]
Thus, the driving moment is approximately \( 1.24 \, \text{Nm} \).
