In a microhematocrit centrifuge, small samples of blood are placed in capillary

Question

In a microhematocrit centrifuge, small samples of blood are placed in

capillary tubes with heparin, an anticoagulant. The tubes are rotated at

11,500 rpm, with the bottom of the tube being 9.07 cm from the axis of

rotation.

a) What is the linear speed of the bottom of the tubes?

b) What is the centripetal acceleration at the bottom of the tubes?

c) What angular speed must the centrifuge have if the centripetal

acceleration at the bottom of the tubes is 10,000g, i.e. 10,000 times

greater than the acceleration due to gravity on the Earth’s surface?

Explanation / Answer

w =11500 rpm = (11500*2*3.14)/60 rad/s

w = 1203.67 rad/s

r =9.07 cm

(a) Linear speed v =rw = (0.0907*1203.67)

v =109.173 m/s

(b) a =v^2/r = (109.173*109.173)/(0.0907)

a =131408.4 m/s^2

(c) a =rw^2 = 10000g

w^2 = (10000*9.8)/0.0907 = 1080455

w = 1039.46 rad/s

w = (1039.46*60)/(2*3.14)

w= 9931.2 rpm

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