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A centrifuge, r = 0.06 cm, revolves at 24000 rpm. If its tensile strength is 1200 N,...

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qasenior | Salutatorian

Posted June 24, 2012 at 11:15 PM via web

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A centrifuge, r = 0.06 cm, revolves at 24000 rpm. If its tensile strength is 1200 N, what is the maximum mass that it can hold?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 7, 2013 at 5:20 PM (Answer #1)

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When a mass M is placed in a centrifuge that is rotating at w rad/s and if r is the radius of the centrifuge, the centrifugal force on the mass is equal to F = M*w^2*r. The maximum mass that can be held in the centrifuge is equal to a value M where the centrifugal force acting on the mass Mw^2r is equal to the tensile strength.

In the problem the radius of the centrifuge is 0.06 cm = 0.0006 m and the angular velocity is 24000 rpm = `(24000*2*pi)/60` = 2513.27 rad/s.

The centrifugal force acting on a mass M is M*2513.27^2*0.0006. Equating this to the tensile strength 1200 N gives:

M*2513.27^2*0.0006 = 1200

=> `M = 1200/(2513.27^2*0.0006)`

=> `M ~~ 0.32` kg

The maximum mass that the centrifuge can hold given the tensile strength is 0.32 kg.

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