What angular speed is needed for a centrifuge to produce an acceleration of 650 times the gravitational acceleration 9.8m/s^2 at a radius of 6.68 cm?

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The object in the centrifuge is assumed to be traveling at a constant circular speed. In this case there is no tangential acceleration on the object. It is only accelerated outwards at an acceleration given by v^2/r where v is the constant speed and r is the radius of the circle in which it is moving.

The required acceleration of the object in the centrifuge is 650 times that of the gravitational acceleration. This is equal to 650*9.8 = 6370 m/s^2.

The radius of the centrifuge is 6.68 cm = 6.68*10^-2 m.

To produce the required acceleration the linear velocity is v, where v^2/(6.68*10^-2) = 6370

=> v^2 = 6370*6.68*10^-2

=> v = 20.62 m/s

The angular velocity in terms of revolutions per second is 20.62/(2*pi*6.68*10^-2) = 49.14 revolutions/second.

The centrifuge has to rotate at 49.14 revolutions/second to achieve an acceleration equal to 650 times that of the gravitational acceleration.

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