Name: Angle of Contact
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: What is the contact angle? Which factor does the contact angle depend on? Why does the contact angle for the same liquid on different surfaces differ? Let's...

Question

A long cylinder of the radius

R_{1}

is displaced along its axis with a constant velocity

v_{0}

, inside a stationary co-axial cylinder of radius

R_{2}

. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance

r

from the axis of the cylinders. The flow is laminar.

Accepted Answer

Let us take a long cylindrical element of radius $r$ $v_{0}$ and thickness $d r$ , coaxial with the other cylinders.

Assume that, this element is moving with a velocity $v$ . The viscous forces acting on the length $l$ of this element will be,

$F = - 2 π r l η \frac{d v}{d r}$ , where $η$ is the coefficient of viscosity.

The net force on this element and other layers must be zero as the velocity is constant. That means the magnitude of this force does not vary with $r$ .

$\frac{F d r}{r} = - 2 π l η d ν$ .

Integrating,

$F \int_{R_{2}}^{r} \frac{d r}{r} = - 2 π l η \int_{0}^{v} d v$

$F \ln (\frac{r}{R_{2}}) = - 2 π l η v$

Putting $r = R_{1}$ , we get,

$F \ln (\frac{R_{1}}{R_{2}}) = - 2 π l η v_{0}$ .

Dividing by this equation, we get,

$v = v_{0} \frac{\ln (\frac{r}{R_{2}})}{\ln (\frac{R_{1}}{R_{2}})}$

Please note that this force $F$ is originally applied by the agency which pulls the inner cylinder with velocity $v_{0}$ .

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