Biomedical laboratories routinely use ultracentrifuges, some of which are able to spin at 1.00 × 10 5 rev/min about the central axis. The turning rotor in certain models is about 20.0 cm in diameter. At its top spin speed, what force does the rotor exert on a 2.00-g sample that is positioned at the greatest distance from the spin axis? Would the force be appreciably different if the sample were spun in a vertical or a horizontal circle? Why or why not?

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valetta valetta · Answer 1 · 2019-10-24T10:02:25+02:00

Answer:

21.932 kN

Explanation:

Given:

Angular velocity, ω = 1.00 × 10⁵ rev/min

Diameter of the turning rotor = 20.0 cm = 0.2 m

or

Radius, r = [tex]\frac{\textup{0.2}}{\textup{2}}[/tex] = 0.1 m

Mass of the sample = 2.00 g = 0.002 Kg

Now,

Force = mass × ω² × r

Here,

ω = 100000 rev/min

or

ω = [tex]\frac{\textup{100000}}{\textup{60}}[/tex] rev/sec

or

ω = 1667.67 rev/sec

also,

ω = 1667.67 × 2π = 10471.97 radian/s

Therefore,

force = 0.002 × 10471.97 × 0.1

or

Force = 21932.43 N

or

Force = 21.932 kN

b) No the force would not be different as it does not depend on the axis of rotation as it can be observed in the formula

Force = mass × ω² × r