Question

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 2.00 cm from the axis to equal 6.00×105 g (where g is the acceleration due to gravity).

Answer 1

Angular velocity of an ultra centrifuge is 17146.42 rad/s.

Step-by-step explanation:

Given that,

Acceleration of an ultra centrifuge,
`a=6* 10^5\ g=6* 10^5* 9.8=5.88* 10^6\ rad/s2`

Distance from axis, r = 2 cm = 0.02 m

We need to find the angular speed. We know that the relation between angular speed and angular acceleration is given by :

`a=r\omega^2\\\\\omega=\sqrt{(a)/(r)} \\\\\omega=\sqrt{(5.88* 10^6)/(0.02)} \\\\\omega=17146.42\ rad/s`

So, the angular velocity of an ultra centrifuge is 17146.42 rad/s.