Question

A circular disc of radius R is kept at small height above a fixed bed by means of a layer of oil of viscosity $\mu$. If the disc is rotated at an angular velocity $\omega$ ,what is the expression for viscous torque on the disc ?

• A. $\frac{\pi \mu \omega }{h}{R}^4$
• B. $\frac{\pi \mu \omega R^4}{4h}$
• C. $\frac{\pi \mu \omega R^4}{2h}$
• D. Zero

Answer 1

The correct option is C $\frac{\pi \mu \omega R^4}{2h}$

Assume a small element dr at radius r

Shear stress on element,
$\tau =\mu \frac{V}{h}=\frac{\mu }{h}\omega \times r$

Force,
$F=\tau .dAF=\frac{\mu }{h}\omega \times r\times 2\pi rdr$

$\text{Viscous torque on the element},dT=F\times r=\frac{\mu }{h}\times \omega \times r\times 2\pi rdr\times r$

$\text{Total torque},T={\int }_0^{R}dT={\int }_0^R\frac{\mu w}{h}2\pi r^{3}dr\Rightarrow T=\frac{\pi \mu \omega R^4}{2h}$