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Suppose you want to design a medical centrifuge to subject samples to normal accelerations of 1000 g’s. (a) If the distance from the center of the centrifuge to the sample is 300 mm, what speed of rotation in rpm is necessary? (b) If you want the centrifuge to reach its design rpm in 1 min, what constant angular acceleration is necessary?

Respuesta :

Answer:

a) 1725.93 rpm

b) α = 3.012 rad/s²

Explanation:

The centrifugal acceleration is given by

a = v²/r

a = 1000 g = 1000 × 9.8 = 9800 m/s²

v = velocity = ?

r = radius of motion = 300 mm = 0.3 m

9800 = v²/0.3

v² = 2940

v = 54.22 m/s

v = rw

w = angular speed

54.22 = 0.3 × w

w = 54.22/0.3 = 180.74 rad/s

w = 2πf

f = frequency in rev/s

180.74 = 2πf

f = 28.77 rev/s

f = 1725.93 rpm

b) Using the equations of motion,

Since the centrifuge starts from rest,

w₀ = 0 rad/s

w = 180.74 rad/s (equivalent to the f = 1725.93 rpm)

t = 1 min = 60 s

α = ?

w = w₀ + αt

180.74 = 0 + 60α

α = 3.012 rad/s²

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