Irodov 1.331: Shearing Both Faces of a Disc – Amit & Deepali Sinha Foundation

Problem Irodov 1.331

A thin disc of radius $R = 10$ cm, mounted horizontally, sits inside a cylindrical cavity filled with oil of viscosity $\eta = 0.08$ P, separated from each of the cavity’s horizontal faces above and below it by a clearance $h = 1.0$ mm. Neglecting end effects, find the viscous power dissipated as the disc spins at angular velocity $\omega = 60$ rad/s.

The disc (radius R) turns inside a narrow oil-filled cavity, sheared against both the upper and the lower fixed wall. — The disc (radius $R$ ) turns inside a narrow oil-filled cavity, sheared against both the upper and the lower fixed wall.

Given

$R = 10$ cm
$\eta = 0.08$ P
$h = 1.0$ mm
$\omega = 60$ rad/s

Find

Find

P

, the power dissipated by viscous friction on the disc

Solution

Step 1Shear stress on a ring of the disc

At radius $r$ the disc’s own surface moves at speed $\omega r$ , while the cavity wall a distance $h$ away is at rest; since $h\ll R$ the oil film is thin enough to treat as simple Couette flow, linear across the gap. The shear stress there is

\tau(r) = \eta\frac{\omega r}{h}. \tag{1}

Step 2Torque from a ring of width

dr

A ring at radius $r$ , width $dr$ , has area $2\pi r\,dr$ on each face, and carries the shear stress from both faces at once, so the drag force on the ring is $2\tau(r)\cdot 2\pi r\,dr$ . Multiplying by the moment arm $r$ and substituting (1) gives the ring’s contribution $dN$ to the torque:

\begin{align*}dN &= r\cdot 2\tau(r)\cdot 2\pi r\,dr \\ &= r\cdot 2\left(\eta\frac{\omega r}{h}\right)(2\pi r\,dr) \\ &= \frac{4\pi\eta\omega}{h}r^3\,dr.\end{align*}

Step 3Integrate to the total torque, then the power

Summing every ring’s contribution from the axis to the rim gives the total viscous torque:

N = \int_0^R \frac{4\pi\eta\omega}{h}r^3\,dr = \frac{4\pi\eta\omega}{h}\left[\frac{r^4}{4}\right]_0^R = \frac{4\pi\eta\omega}{h}\cdot\frac{R^4}{4}.

Since the disc turns at constant $\omega$ , the viscous torque is balanced by whatever drives the disc, and the power it delivers against friction is the torque times $\omega$ :

P = N\,\omega = \frac{\pi\eta\omega R^4}{h}\cdot\omega.

P = \dfrac{\pi\eta\omega^2R^4}{h}

(2)

Step 4Evaluate

Converting $\eta=0.08$ P $=8.0\times10^{-3}$ Pa $\cdot$ s, with $R=0.10$ m and $h=1.0\times10^{-3}$ m,

P = \pi(8.0\times10^{-3})(60)^2(0.10)^4/(1.0\times10^{-3})\ \text{W}.

P = \dfrac{\pi\eta\omega^2R^4}{h} \approx 9\ \text{W}

(3)

Check

Doubling the gap $h$ halves $P$ , as it should: a wider film means a gentler velocity gradient and so less shear, for the same disc speed. And $P\to0$ as $\omega\to0$ , since a stationary disc drags nothing.

Remark

The torque grows as $r^3$ under the integral, so the outer rim of the disc dominates the dissipation far out of proportion to its width: the outer tenth of the radius, the annulus from $0.9R$ to $R$ , carries about a third of the total dissipation on its own (since $R^4-(0.9R)^4\approx 0.34R^4$ against a total of $R^4$ , roughly half as much as everything inside $0.9R$ ). Any viscous device built on shear times moment arm, a clutch plate or a rotational damper, is dominated by its outer radius in the same way, which is why such devices are built as wide, thin discs rather than thick, narrow ones.