Acceleration and rotation of centrifuge

AI Thread Summary
The discussion focuses on calculating the acceleration experienced by test tubes in a centrifuge and comparing it to the acceleration from a drop. For part a, the correct approach involves using the angular velocity formula and ensuring the angular velocity is squared in the acceleration calculation, leading to an acceleration of approximately 21278.87 m/s² for a test tube 11 cm from the axis of rotation. For part b, the method used to calculate the acceleration from a drop was incorrect, as the initial approach did not yield the expected result. Participants emphasized the importance of correctly applying formulas and units in these calculations. Accurate calculations are crucial for understanding the forces at play in centrifuge operations.
aligass2004
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Homework Statement



A typical laboratory centrifuge rotates at 4200 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerators. a.) What is the acceleration at the end of a test tube that is 11 cm from the axis of rotation? b.) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of .8 m and stopped in a 1.0 ms long encounter with a hard floor?

Homework Equations



w=2pi radians (f)
a = w^2 r

The Attempt at a Solution



I tried finding the frequency, then using the frequency to find w, which I got to be 439.823. Then I tried finding a, which I got to be 48.381.
 
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aligass2004 said:

Homework Statement



A typical laboratory centrifuge rotates at 4200 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerators. a.) What is the acceleration at the end of a test tube that is 11 cm from the axis of rotation? b.) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of .8 m and stopped in a 1.0 ms long encounter with a hard floor?

Homework Equations



w=2pi radians (f)
a = w^2 r

The Attempt at a Solution



I tried finding the frequency, then using the frequency to find w, which I got to be 439.823. Then I tried finding a, which I got to be 48.381.
you forgot to square the w term!
 
use T= 2(pi)(r)/v to get the velocity and then use a= v^2/r to get part a
 
And be sure to include your units.
 
I did forget to square the w term. I got 21278.87m/s^2 for part a.
 
For part b, I thought to just use Yf = Yi + Viy(t) + 1/2 (a) (t^2). I got 1.6E6 m/s^2, but it said it was wrong.
 
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