Centrifuge acceleration expressed in terms of gravity

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SUMMARY

The discussion focuses on calculating the centripetal acceleration of an analytic ultra-centrifuge at a radius of 0.084 meters while spinning at 60,000 RPM. The formula used is ac = 4π² * r * f², resulting in a centripetal acceleration of 3,312,825.6 m/s². To express this acceleration in terms of Earth's gravity, the biophysicist applies the conversion factor of 1g = 9.8 m/s², leading to a straightforward multiplication of the calculated acceleration by the appropriate factor.

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A Biophysicist wants to separate sub-cellular particles with an analytic ultra-centrifuge. The biophysicist must determine the magnitude of the centripetal acceleration provided by the centrifuge at various speeds and radii.

Calculate the magnitude of the centripetal acceleration at 8.4 cm from the centre of the centrifuge when it is spinning at 6.0*10^4 rpm. Express your answer in terms of g (acceleration due to Earth's gravity).


Given:

r = 0.084m
f = 60000 / 60s

Required: ac

Analaysis:

ac = 4pie^2 * r * f^2

Solution:

ac = 4(3.14)^2 * 0.084m * (60000 / 60s)^2

= 3312825.6 m / s^2

Now I am a bit unclear as to how to express this in terms of g (Earth's gravity).

I could factor in 9.8 m / s^2 but that only changes the direction of the ac force by .001 m / s2

/F/ = /ac^2/ + /Fg^2/

I think I am approaching this from the wrong angle. Any suggestions?
 
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Use the equation 1g = 9.8m/s^2 (g = accel due to gravity, not gram).

Multiply your answer by 1 ( = 1g/(9.8m/s^2) )...
 
Doh! So easy. Thanks
 

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