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

Answer:

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

Explanation:

Given that,

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

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 :

$a=r\omega^2\\\\\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5.88\times 10^6}{0.02}} \\\\\omega=17146.42\ rad/s$

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