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