Question

What angular speed (in revolutions per minute) is needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration 9.8 m/s 2 at a radius of 4.83 cm ?

Answer 1

Angular velocity,
`\omega=3747.33\ rev/min`

Step-by-step explanation:

In this case, we need to find the angular speed needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration.

Radius of the circular path, r = 4.83 cm

The acceleration acting on the particle in circular path is given by :

`a=r\omega^2`

`\omega` is the angular speed in rad/s

`\omega=\sqrt{(a)/(r)}`

`\omega=\sqrt{(759* 9.8)/(4.83* 10^(-2))}`

`\omega=392.42\ rad/s`

or

`\omega=3747.33\ rev/min`

So, there are 3747.33 revolutions per minute that is needed. Hence, this is the required solution.