Acceleration and rotation of centrifuge

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SUMMARY

The discussion focuses on calculating the acceleration experienced by test tubes in a laboratory centrifuge rotating at 4200 rpm. For a test tube located 11 cm from the axis of rotation, the correct acceleration is calculated using the formula a = w²r, resulting in 21278.87 m/s² after correcting for the squaring of the angular velocity (w). Additionally, the acceleration from a drop of 0.8 m, calculated using kinematic equations, was incorrectly estimated at 1.6E6 m/s², indicating a need for further review of the applied formulas.

PREREQUISITES
  • Understanding of angular velocity and its calculation (w = 2πf)
  • Familiarity with centripetal acceleration formulas (a = w²r)
  • Knowledge of kinematic equations for free fall
  • Basic unit conversion and dimensional analysis
NEXT STEPS
  • Review the derivation and application of centripetal acceleration formulas
  • Study the kinematic equations for motion under gravity
  • Practice unit conversion in physics calculations
  • Explore the implications of high acceleration in laboratory settings
USEFUL FOR

Physics students, laboratory technicians, and engineers involved in centrifuge operations or high-acceleration environments.

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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