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The tube rests on a 1.51.5-mm\mathrm{mm} thin film of oil having a viscosity of μ=0.0586 Ns/m2\mu=0.0586 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2. If the tube is rotating at a constant angular velocity of ω=4.5 rad/s\omega=4.5\ \mathrm{rad} / \mathrm{s}, find the shear stress in the oil at r=40 mmr=40 \mathrm{~mm} and r=80 mmr=80 \mathrm{~mm}. Assume the velocity profile within the oil is linear.


Answered 1 year ago
Answered 1 year ago
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The velocity of the oil at point distanced rr from the center of the disk is u=ωru=\omega r, where ω=4.5 rad/s\omega=4.5\ \mathrm{rad/s} is the angular velocity of the disk. Since the velocity profile is linear, we have:

dudy=ut=ωrt\begin{equation} \dfrac{du}{dy}=\dfrac{u}{t}=\dfrac{\omega r}{t} \end{equation}

where t=1.5 mm=1.5103 mt=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:

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

where μ=0.0586 Ns/m2\mu=0.0586\ \mathrm{N\cdot s/m^2} is the viscosity of the fluid, and we have made explicit that the shear stress τ\tau is rr-dependent. The sought shear stresses at r1=40 mm=4102 mr_1=40\ \mathrm{mm}=4\cdot 10^{-2}\ \mathrm{m} and r2=80 mm=8102 mr_2=80\ \mathrm{mm}=8\cdot 10^{-2}\ \mathrm{m} can now be easily calculated:

τ(r1)=μωr1t=0.0586 Ns/m24.5 rad/s4102 m1.5103 m=7.03 Pa,τ(r1)=μωr2t=0.0586 Ns/m24.5 rad/s8102 m1.5103 m=14.1 Pa,\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 N/m2=1 Pa1\ \mathrm{N/m^2}=1\ \mathrm{Pa}.

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