Question

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

Answer 1

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 $N = mass \times a = 0.01\times 127.53 = 1.2753 N$

                                                 $= 1.2753\times 0.225 lbf = 0.287 lbf$