Centripetal acceleration centrifuge

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Homework Help Overview

The discussion revolves around the calculation of centripetal acceleration in the context of a centrifuge, specifically determining the revolutions per minute (rpm) of a sample located at a certain radius from the axis of rotation. The original poster provides a scenario involving centripetal acceleration being significantly greater than gravitational acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the period of rotation based on centripetal acceleration and radius, using the formula for centripetal acceleration. Participants question the conversion from period to rpm and discuss the relationship between period and frequency.

Discussion Status

Participants are actively engaged in clarifying the conversion process from period to revolutions per minute. Some guidance has been offered regarding the relationship between period and frequency, but there is no explicit consensus on the calculations presented.

Contextual Notes

There is a focus on ensuring unit consistency during calculations, with participants noting potential errors in unit conversions and the need for careful attention to the desired outcome of revolutions per minute.

pookisantoki
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A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is 5.92 x 10^3 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 6.37 cm from the axis of rotation?

Acceleration due to gravity is 9.80 and I converted 6.37cm to .0637m.
So this is how I solved it: Ac=V^2/r so set it up where V^2=Ac*r=sqrt(9.80*5920)*.0637= 60.79160
Then V=(2pi*r)/T so T=(2pi*r)/V: (2pi*.0637)/60.79160=.006584

Is this right??
 
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I didn't double check your numbers, but you solved for period and the question asks for rpm, just remember to do the appropriate conversion.
 
I'm not sure on how to go from period to RPM
 
Well period is seconds per revolution, while rpm is revolutions per minute. Recall 1/T=F, the frequency, which has units revolutions per second.
 
So would it be .006584*60=.39504??
 
No... check your units and make sure they come out alright. What you have is this:
.0065 \frac{\textrm{seconds}}{\textrm{revolution}} \times 60\frac{\textrm{seconds}}{\textrm{minute}} = .395 \frac{\textrm{seconds}^2}{\textrm{revolution minute}}

Which is obviously not right. Remember where you want to go, which is revolutions/minute.
 

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