Question

Find the required angular speed, ω , of an ultracentrifuge for the radial acceleration of a...

Find the required angular speed, ω , of an ultracentrifuge for the radial acceleration of a point 1.40 cm from the axis to equal 3.00×105 g (where g is the acceleration due to gravity). Express your answer numerically in revolutions per minute

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Answer #1

Gravitational acceleration = g = 9.81 m/s2

Angular speed of the ultra centrifuge = (rad/s)

Distance of the point from the axis = R = 1.4 cm = 0.014 m

Radial acceleration of the point 1.4 cm from the axis = a = 3 x 105 g = 3x105 x (9.81) m/s2 = 29.43 x 105 m/s2

a = 2R

29.43x105 = 2(0.014)

= 14498.77 rad/s

Converting to revolutions per minute,

= 1.384 x 105 rpm

Angular speed of the ultra centrifuge = 1.384 x 105 rpm

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