Question

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

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