A long cylinder of radius R1 is displaced along its axis with a constant velocity v0 inside a stationary co-axial cylinder of radius R2. The space between the cylinders is filled with viscous liquid. The velocity of the liquid as a function of the distance r from the axis of the cylinders is v=v0lnR2R1lnR2k. What will be k?

Question

A long cylinder of radius R1​ is displaced along its axis with a constant velocity v0​ inside a stationary co-axial cylinder of radius R2​. The space between the cylinders is filled with viscous liquid. The velocity of the liquid as a function of the distance r from the axis of the cylinders is v=v0​lnR2​R1​​lnR2​k​​. What will be k?

Filo · Accepted Answer

Let us consider a coaxial cylinder of radius

r

and thickness

d r

, then force of friction or viscous force on this elemental layer,

F = 2 π r l η \frac{d v}{d r}

.

This force must be constant from layer to layer so that steady motion may be possible.

or,

\frac{F d r}{r} = 2 π l η d v

........................(1)

Integrating,

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

or,

F ln (\frac{r}{R _{2}}) = 2 π l η v

.......................(2)

Putting

r = R_{1}

, we get

F ln \frac{R _{1}}{R _{2}} = 2 π l η v_{0}

...................(3)

From (2) / (3) we get,

v = v_{0} \frac{ln \frac{r}{R _{2}}}{ln \frac{R _{1}}{R _{2}}}

Therefore,

k = r

Note: The force

F

is supplied by the agency which tries to carry the inner cylinder with velocity

v_{0}

.

Question Text	A long cylinder of 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. The velocity of the liquid as a function of the distance $r$ from the axis of the cylinders is $v = v_{0} \frac{ln \frac{k}{R _{2}}}{ln \frac{R _{1}}{R _{2}}}$ . What will be $k$ ?
Answer Type	Text solution:1
Upvotes	150

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