Question

How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 113000 g's? if the answer has 4 digits or more, enter it without commas,

e.g. 13500?

Answer 1

The acceleration experienced by the particle is given by

`a=113000 g=113000 \cdot 9.81 m/s^2`
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed
`\omega` of the particle and its distance r from the axis by the relationship

`a= \omega ^2 r`
In our problem,
`r=6 cm=0.06 m`, so we can solve for
`\omega`:

`\omega = \sqrt{ (a)/(r) } = \sqrt{ (113000 \cdot 9.81 m/s^2)/(0.06 m) }=4298 rad/s`
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to
`( (1)/(2 \pi) )` revolutions, while
`1 s = (1)/(60) min`. So we the conversion is
`\omega = 4298 rad/s \cdot ( (1)/(2\pi) rev/rad )( 60 s/min)=41067 rpm`