Question

A thin horizontal disc of radius R=10 cmR=10cm is located within a cylindrical cavity filled with oil whose viscosity is $\eta =0.08P$ (fig). The clearance between the disc and the horizontal planes of the cavity is equal to $h=1mm$. Find the power developed by the angular velocity $\omega =60rad/s$. The end effects are to be neglected.

• A. $8W$
• B. $9W$
• C. $10W$
• D. $19W$

Answer 1

The correct option is B $9W$

Viscous force at ($dx$) element at lower surface
$df=\eta A\frac{dv}{dz}=\eta \left(2\pi xdx\right)\frac{xw}{h}$
Net viscous force at ($dx$) element from loth lower and upper surface:
$d{f}_{net}=2df=\frac{4\pi \eta \omega }{h}{x}^{2}dx$
Torque due to this force $dt=xd{f}_{net}=\frac{4\pi \eta \omega x^{3}dx}{h}$
Power due to this torque = $dp=\omega dt=4\pi \eta \frac{{\omega }^2{x}^3}{h}dx$
Power total $=\frac{4\pi \eta {\omega }^2}{h}{\int }_0^R{x}^{3}dx=\frac{\pi \eta {\omega }^2{R}^4}{h}=9W$