## What is the acceleration from the axis of rotation?

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 = v^{2}/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/s^{2}

**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** Vf ^{2}=Vi^{2} + 2ad,**

Vf^{2}=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

## Leave a reply