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# The tube rests on a $1.5$-$\mathrm{mm}$ thin film of oil having a viscosity of $\mu=0.0586 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2$. If the tube is rotating at a constant angular velocity of $\omega=4.5\ \mathrm{rad} / \mathrm{s}$, find the shear stress in the oil at $r=40 \mathrm{~mm}$ and $r=80 \mathrm{~mm}$. Assume the velocity profile within the oil is linear.

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The velocity of the oil at point distanced $r$ from the center of the disk is $u=\omega r$, where $\omega=4.5\ \mathrm{rad/s}$ is the angular velocity of the disk. Since the velocity profile is linear, we have:

$\begin{equation} \dfrac{du}{dy}=\dfrac{u}{t}=\dfrac{\omega r}{t} \end{equation}$

where $t=1.5\ \mathrm{mm}=1.5\cdot 10^{-3}\ \mathrm{m}$ is the thickness of the oil. The shear stress can now be expressed as:

$\begin{equation} \tau(r)=\mu\dfrac{du}{dy}=\dfrac{\mu\omega r}{t}, \end{equation}$

where $\mu=0.0586\ \mathrm{N\cdot s/m^2}$ is the viscosity of the fluid, and we have made explicit that the shear stress $\tau$ is $r$-dependent. The sought shear stresses at $r_1=40\ \mathrm{mm}=4\cdot 10^{-2}\ \mathrm{m}$ and $r_2=80\ \mathrm{mm}=8\cdot 10^{-2}\ \mathrm{m}$ can now be easily calculated:

\begin{aligned} \tau(r_1)&=&\dfrac{\mu\omega r_1}{t}=\dfrac{0.0586\ \mathrm{N\cdot s/m^2}\cdot 4.5\ \mathrm{rad/s}\cdot 4\cdot 10^{-2}\ \mathrm{m}}{1.5\cdot 10^{-3}\ \mathrm{m}}=\boxed{7.03\ \mathrm{Pa}},\\ \tau(r_1)&=&\dfrac{\mu\omega r_2}{t}=\dfrac{0.0586\ \mathrm{N\cdot s/m^2}\cdot 4.5\ \mathrm{rad/s}\cdot 8\cdot 10^{-2}\ \mathrm{m}}{1.5\cdot 10^{-3}\ \mathrm{m}}=\boxed{14.1\ \mathrm{Pa}}, \end{aligned}

since $1\ \mathrm{N/m^2}=1\ \mathrm{Pa}$.

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