Question 5.10: Ultracentrifuge. The rotor of an ultracentrifuge rotates at ......
Ultracentrifuge. The rotor of an ultracentrifuge rotates at 50,000 rpm (revolutions per minute). A particle at the top of a test tube (Fig. 5-13) is 6.00 cm from the rotation axis. Calculate its centripetal acceleration, in “g’s.”
APPROACH We calculate the centripetal acceleration from a_{ R }=v^2 / r.
The test tube makes 5.00 × 10^4 revolutions each minute, or, dividing by 60 s/min, 833 rev/s. The time to make one revolution, the period T, is
T=\frac{1}{(833 rev / s )}=1.20 \times 10^{-3} s / rev.
At the top of the tube, a particle revolves in a circle of circumference 2 \pi r = (2 \pi) (0.0600 m) = 0.377 m per revolution. The speed of the particle is then
v=\frac{2 \pi r}{T}=\left(\frac{0.377 \ m / rev }{1.20 \times 10^{-3} s / rev }\right)=3.14 \times 10^2 m / s.
The centripetal acceleration is
a_{ R }=\frac{v^2}{r}=\frac{\left(3.14 \times 10^2 \ m / s \right)^2}{0.0600 \ m }=1.64 \times 10^6 \ m / s ^2,
which, dividing by g = 9.80 m/s², is 1.67 × 10^5 g’s = 167,000 g’s.