
#1
Feb110, 06:50 PM

P: 23

For this problem, I tried using the equation:
a_{c}=V^{2}/r and put in 2000= ((2*pi*r)/(1/58))^{2}*r What is the distance of a sample of blood from the rotation axis of a centrifuge rotating at a frequency f = 3480 rpm, if it has an acceleration of 2000g? 



#2
Feb110, 07:10 PM

Mentor
P: 39,589

Use a = omega^2 * r instead (close to the form of your equation anyway), and carefully convert 3480rpm into the omega (show the units in your conversion to check yourself). 



#3
Feb210, 11:38 AM

P: 23

Human blood contains plasma, platelets, and blood cells. To separate the plasma from other components, centrifugation is used. Effective centrifugation requires subjecting blood to an acceleration of 2000g or more. In this situation, assume that blood is contained in test tubes of length L = 14.3 cm that are full of blood. These tubes ride in the centrifuge tilted at an angle of 45.0° above the horizontal (see figure below)
(a) What is the distance of a sample of blood from the rotation axis of a centrifuge rotating at a frequency f = 3480 rpm, if it has an acceleration of 2000g? cm (b) If the blood at the center of the tubes revolves around the rotation axis at the radius calculated in Part (a), calculate the accelerations experienced by the blood at each end of the test tube. Express all accelerations as multiples of g. minimum acceleration g maximum acceleration g This is the entire problem. So for the first part: a=omega^2*r a= (2*pi*r/(1/58))^2*r What am I doing incorrectly? 


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