Question

In a centrifuge, two tubes containing liquids with suspended particles are attached to a rotor by means of smooth pivots. Both tubes are carefully balanced to have the same weight (Position A). When the rotor is made to spin at a high speed, the tubes swing into an almost horizontal position (B).
A bacterial particle of $1$ $\mu m$ radius and a mass of $5$ x 10${}^{-15}$ kg with a density of $1.1$ times of water is suspended. The resistive force of viscous origin opposing the motion of the particle is given as: $f\left(v\right)=0.56rv$${}_d$
Then, if the drift speed of the particle is given as $v$${}_d$ = $n\times {10}^{-6}m/s$, find $n$. (Given that effective accleeration = $250\times g$, where $g$ is acceleration due to gravity)

Answer 1

In the rotating frame , a centrifugal force $m{\omega }^2\overrightarrow{r}$ acts on a particle of mass m and hence an effective acceleration $\overrightarrow{g{}^{\prime }}={\omega }^2\overrightarrow{r}$ acts in the horizontal direction passing away from the axis of rotation.
The centrifugal force on a suspended particle of radius R and density $\rho$ is $\frac{4}{3}\pi r^3\rho g^{\prime }.$
The surrounding liquid supplies a buoyant force equal to $\frac{4}{3}\pi r^3{\rho }_{\iota }{g}^{\prime }$, where ${\rho }_{\iota }$ is the density of the liquid.
$\Rightarrow$ A net force F = $\frac{4}{3}\pi r^3(\rho -{\rho }_{\iota })g^{\prime }$ $=[1-\left(\frac{p_{\iota }}{\rho }\right)]m{g}^{\prime }$ acts on the particle.
Given $\frac{{\rho }_{\iota }}{\rho }=1.1,m=5\times {10}^{-15}kg$, $g^{\prime }=250g$
$\Rightarrow$ $F=1.225\times 10$${}^{-12}$ $N$
As $f\left(v\right)=F$
$v_d=\frac{F}{0.56r}=\frac{1.225\times {10}^{-12}}{0.56\times {10}^{-6}}=2\times {10}^{-6}$ $m/s$
$\Rightarrow$ $n=2$