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

1) 0.022 lbm

2) 276.253 RPM

3) 0.287 lbf

Step-by-step explanation:

Given data:

mass = 10 kg

acceleration -
`13 g = 13* 9.81 m/s^2 = 12`

7.53 m/s^2

r =6 inches = 0.1524 m

1) mass in lbm
`= 0.01* 2.2 = 0.022 lbm`

as 1 kg = 2.2 lbm

2) acceleration
`=  r \omega ^2`

`127.53 = 0.1524 * \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 = (60)/(2\pi) RPM]`

SO IN
`28.927 rad/s = (60)/(2\pi) * 28.297 = 276.253 RPM`

3) Force in
`N = mass * a = 0.01* 127.53 = 1.2753 N`

`=  1.2753* 0.225 lbf = 0.287 lbf`