Answer to Question #194669 in Mechanical Engineering for Kay

Question

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Answer 1

According to Newtons law of viscosity

$\tau\propto\frac{d\theta}{dt}$

$\tau=\frac{\mu d\theta}{dt}=\frac{\mu dy}{dy}$

here $\mu=$ dynamic viscosity

then,$\tau=\mu\times(\frac{v-0}{0.3\times10^{-3}})$

$=\mu\times(\frac{0.05-0}{0.3\times10^{-3}})=\mu166.6666N/m^2$

$f_s=$ forces due to shear$=\mu166.6666N/m^2\times area\times\mu$

$=166.6666\times1.5\mu$

$=250\mu N$

the tension which is applied on ropes will overcome this shear force hence,

$F_s=T=250\mu N$

Torque $=T\times r$

$24.5\times10^{-3}=250\times \mu\times25\times10^{-3}$

$\mu_{oil}=3.92\times10^{-3}\frac{kg}{ms}$ or$\frac{Ns}{m^2}$

then, for kinematic viscosity,

$V=\frac{\mu_{oil}}{S_{oil}}$

as $S_{oil}=8.9s$ $\frac{kN}{m^3{\times10^3}}$

$912.3334\frac{kg}{m^3}$

then,V $=\frac{\mu_{oil}}{S_{oil}}=\frac{3.92\times10^{-3\frac{kg}{ms}}}{912.334\frac{kg}{m^3}}$

$=4.29667\times10^{-6}\frac{m^2}{s}$

$=4.29667\times10^{-2} strokes$

$10^4 strokes$