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 (

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).

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Dulcie 3 years 2021-08-16T02:23:56+00:00 1 Answers 7 views 0

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  1. Dulcie
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    2021-08-16T02:25:43+00:00
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    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.

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