The analytical ultracentrifuge olympics

Classical Mechanics Level 5

One way that scientists determine the mass of a biological particle is analytical centrifugation. A preparation (several picomoles) of the particle is layered on top of a viscous fluid such as sucrose solution which is then spun at high speeds. The particle experiences several forces: centripetal force (which pushes it down the gradient), buoyancy (which pushes it up the gradient), and drag (that slows its motion). After a short time, these forces balance and the particle travels down the gradient at a constant rate.

Suppose a particle if found to travel at the rate $v$ through sucrose. Now, we measure the sedimentation rate of a dimer (two of the particles stuck together). This new particle travels down the gradient at the rate $v^\prime$ , what is $v^\prime /v$ ?

Notes and assumptions

Assume the particles are spherical.
Assume that the dimer is formed by melting down two particles and forming a sphere with twice the volume of the monomer.
For simplicity, assume that the force pulling particles down the gradient is provided by a strong, constant gravitational field.
The friction the particle feels is Stokes drag $F \sim rv$ where $r$ is the radius of the particle.

The answer is 1.5874.

1 solution

Chew-Seong Cheong
Jul 31, 2014

It is given that Stokes drag $F\sim rv$ . The drag $F$ is balanced by the 'strong gravitational' force such that $F=ma$ , where $m$ is the mass of the particle and $a$ is the acceleration asserted by the gravitational field. Therefore, the drag $F'$ on the dipole of mass $2m$ is: $F'=2ma=2F$ $\Rightarrow r'v'=2rv$ where $r'$ and $v'$ are the radius and travel rate of the dipole.

Since the dipole has twice the volume of the particle, $\frac{4}{3}\pi r'^3=2\times \frac{4}{3}\pi r^3 \Rightarrow r'=\sqrt[3]{2}r$ . Therefore, $\sqrt[3]{2}rv'=2rv \Rightarrow \frac {v'}{v}=2^{\frac{2}{3}}=\boxed {1.587}$

The analytical ultracentrifuge olympics

The answer is 1.5874.

1 solution

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