Irodov 1.333: Constant Torque Through Every Shell – Amit & Deepali Sinha Foundation

Problem Irodov 1.333

Viscous fluid of viscosity $\eta$ fills the annular gap between two long, coaxial cylinders of radii $R_1$ and $R_2$ , with $R_1<R_2$ . The inner cylinder ( $R_1$ ) is held fixed while the outer one ( $R_2$ ) is driven at a constant angular velocity $\omega_2$ , and the resulting flow of the fluid between them is laminar. The friction force per unit area on any cylindrical fluid surface of radius $r$ is given as $\sigma=\eta r\,(\partial\omega/\partial r)$ , where $\omega(r)$ is the fluid’s own local angular velocity. Using this, find (a) how the fluid’s angular velocity $\omega$ depends on $r$ for $R_1\le r\le R_2$ , and (b) the moment, per unit length of the cylinders, of the friction forces acting on the outer cylinder.

Top view down the shared axis. The inner cylinder (radius R_1, ticked rim) is held; the outer (radius R_2) is driven at ω_2. The dashed circle is the variable shell at radius r, carrying tangential velocity v(r). — Top view down the shared axis. The inner cylinder (radius $R_1$ , ticked rim) is held; the outer (radius $R_2$ ) is driven at $\omega_2$ . The dashed circle is the variable shell at radius $r$ , carrying tangential velocity $v(r)$ .

Given

$\eta$
$R_1$ , $R_2$
$\omega_2$
$\sigma = \eta r\,\partial\omega/\partial r$

Find

(a)

\omega(r)

for

R_1\le r\le R_2

; (b)

N

, the moment of friction on the outer cylinder per unit length

Solution

Step 1The torque crossing every shell is the same constant

The torque per unit length crossing a cylindrical shell of radius $r$ is the shear stress $\sigma(r)$ times the shell’s area per unit length ( $2\pi r$ ) times the moment arm ( $r$ again). Setting this equal to a constant $M$ and substituting the given $\sigma$ ,

M = \sigma(r)\cdot 2\pi r \cdot r = 2\pi\eta r^3\frac{d\omega}{dr}. \tag{1}

Step 2Integrate the constant-torque condition

Dividing through by $2\pi\eta r^3$ and integrating once gives $\omega$ up to two constants.

\begin{aligned} \frac{d\omega}{dr} &= \frac{M}{2\pi\eta}\,\frac{1}{r^3} \\ \omega(r) &= -\frac{M}{4\pi\eta}\frac{1}{r^2} + C. \end{aligned} \tag{2}

Step 3Apply the two boundary conditions

The inner cylinder is held, $\omega(R_1)=0$ ; the fluid at the outer wall shares that cylinder’s own speed by no-slip, $\omega(R_2)=\omega_2$ .

\begin{align*}-\frac{M}{4\pi\eta}\frac{1}{R_1^2} + C &= 0 \\ -\frac{M}{4\pi\eta}\frac{1}{R_2^2} + C &= \omega_2\end{align*}

Subtracting the first equation from the second eliminates $C$ and isolates $M$ .

\frac{M}{4\pi\eta}\left(\frac{1}{R_1^2}-\frac{1}{R_2^2}\right) = \omega_2

M = 4\pi\eta\omega_2\dfrac{R_1^2R_2^2}{R_2^2-R_1^2}

(3)

The first boundary equation then gives the remaining constant,

C = \frac{M}{4\pi\eta R_1^2} = \frac{\omega_2 R_2^2}{R_2^2-R_1^2}.

Step 4Assemble

\omega(r)

Substituting $M$ from (3) and $C$ back into the general solution (2),

\begin{align*}\omega(r) &= -\frac{1}{4\pi\eta r^2}\left(4\pi\eta\omega_2\frac{R_1^2R_2^2}{R_2^2-R_1^2}\right) + \frac{\omega_2 R_2^2}{R_2^2-R_1^2} \\ &= \omega_2\frac{R_2^2}{R_2^2-R_1^2}\left(1-\frac{R_1^2}{r^2}\right).\end{align*}

Taking $R_1^2$ out of the bracket gives the profile in its symmetric form.

\omega(r) = \omega_2\dfrac{R_1^2R_2^2}{R_2^2-R_1^2}\left(\dfrac{1}{R_1^2}-\dfrac{1}{r^2}\right)

(4)

Step 5The moment on the outer cylinder

Because the torque transmitted across every shell is the same constant $M$ of (1), the moment the fluid exerts on the outer cylinder, at $r=R_2$ , is that very same $M$ : no further integration is needed.

N = 4\pi\eta\omega_2\dfrac{R_1^2R_2^2}{R_2^2-R_1^2}

(5)

Check

As $R_1\to0$ (the inner cylinder shrinks to the axis), the prefactor $R_1^2R_2^2/(R_2^2-R_1^2)\to R_1^2$ , so $\omega(r)\to\omega_2\bigl(1-R_1^2/r^2\bigr)\to\omega_2$ at every fixed $r$ : with no inner boundary to hold it back, the fluid simply co-rotates rigidly with the outer wall, and the shear $\partial\omega/\partial r\to0$ everywhere, consistent with $N\to4\pi\eta\omega_2 R_1^2\to0$ . In the opposite, thin-gap limit $R_1\to R_2\equiv R$ with gap $d=R_2-R_1\ll R$ , $N\to4\pi\eta\omega_2R^4/(2Rd)=2\pi\eta\omega_2R^3/d$ , exactly the flat-plate estimate (the shear $\eta\omega_2R/d$ acting over the area $2\pi R$ per unit length, and that force then carried on the moment arm $R$ ) for a narrow Couette gap, confirming the curved result reduces correctly to the flat one.

Remark

The whole problem turns on one fact: with no torque source hiding anywhere in the fluid’s interior, the viscous torque crosses every concentric shell unchanged in steady state, even though the stress carrying it thins out as $1/r^2$ . That single conserved quantity is what lets (1) set $r^3\,d\omega/dr=\text{const}$ directly from the given $\sigma$ , without ever writing down the fluid’s equation of motion; the same trick, applied to $r\,dv/dr$ , is exactly what makes 1.332’s straight coaxial sliding problem solvable by inspection too. In both problems the quantity that crosses every shell unchanged is whatever the steady balance conserves, the axial force in that straight film and the torque here, and in both the stress itself still falls off with radius.