A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations.
1) What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation?
Ac = v2/r
The radius is 10cm or .10m
The circumference is 2*pi*r = 2*pi*.10 = .628m
4000rpm = 4000/60rps = 66rps
v = 66*.628m/s = 41.487m/s
Ac = (41.48)2/.1 = 1717 m/s2
2) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hard floor?
We need to find how fast it would drop from one meter above.
a= acceleration due to gravity : 9.81
Using Vf2=Vi2 + 2ad,
Vf2=0+2*9.8*1
V=4.43 m/s
Set the Velocity to zero for this equation
Vf = Vi + a*t
0=4.43 + a* 0.001
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