Asked by HELP
How fast (in rpm) must a centrifuge rotate if a particle 69-cm from the axis of rotation is to experience an acceleration of 200438 g's?
(can you show me how you get the answer) be thorough
(can you show me how you get the answer) be thorough
Answers
Answered by
Damon
a = 200438 g's
=200438 * 9.81 m/s^2
You have got to be kidding
r = 0.69 meters
a = omega^2 r
solve for omega
which is in radians/second
divide by 2 pi to get revolutions per second
multiply by 60 to get revolutions per minute
=200438 * 9.81 m/s^2
You have got to be kidding
r = 0.69 meters
a = omega^2 r
solve for omega
which is in radians/second
divide by 2 pi to get revolutions per second
multiply by 60 to get revolutions per minute
Answered by
Scott
centrifuge means centripetal force
v^2 / r = 200438 g ... v in m/s
rpm = v * 60 / (2 π r)
v^2 / r = 200438 g ... v in m/s
rpm = v * 60 / (2 π r)
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