acceleration in the centifruge
w^2r=(4100rpm*1min/60 sec * 2PIrad/sec)^2*.11m = appx 275 g's. Work it out. check my math.
Now dropping it,
1/2 mv^2=mgh
v=sqrt(2gh)=sqrt(2*9.8*1.2)=4.95m/s
a=changevelocity/time=4.95/.001=4950m/s^2=about 500 g's
A typical laboratory centrifuge rotates at 4100 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations.
(a) What is the acceleration at the end of a test tube that is 11 cm from the axis of rotation?
dropped from a height of 1.2 m and stopped in a 1.0 ms long encounter with a hard floor?
1 answer
1 answer