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Andru [333]
3 years ago
13

An object weighing 10 grams is spinning in a centrifuge such that an acceleration of 13.0 g is imposed to it. The arm connecting

the shaft to the object is r = 6.0 inches. If a = acceleration = rω2 where ω = angular speed in rad/s, determine:
Mass of the object in lbm
RPM (revolutions per minute) of the shaft
Force acting on the object in lbf
Chemistry
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

1) 0.022 lbm

2) 276.253 RPM

3) 0.287 lbf

Explanation:

Given data:

mass = 10 kg

acceleration - 13 g = 13\times 9.81 m/s^2 = 12

7.53 m/s^2

r =6 inches = 0.1524 m

1) mass in lbm  = 0.01\times 2.2 = 0.022 lbm

as 1 kg = 2.2 lbm

2) acceleration =  r \omega ^2

127.53 = 0.1524 \times \omega^2

\omega^2 = 836.811

\omega = 28.927 rad/s

1 rad/s  = 9.55 RPM    

[ 1 revolution = 2\pi,    1 rad/s = 1/2\pi RPS = \frac{60}{2\pi} RPM]

SO IN 28.927 rad/s = \frac{60}{2\pi} \times 28.297 = 276.253 RPM

3) Force in N = mass \times a = 0.01\times 127.53 = 1.2753 N

                                                 =  1.2753\times 0.225 lbf = 0.287 lbf

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