Question

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal 6.00×105 g (where g is the acceleration due to gravity).

Answer 1

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² =  5880000 / 0.018

ω ² =  326666667

ω = 18074 rad/sec

Answer 2

The required angular speed ω of an ultra-centrifuge should be considered as the ω = 18074 rad/sec.

Calculation of the angular speed:

Since

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

So,

6 x 10⁵ g = 1.8 cm x ω ²

Now here

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

So,

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² =  5880000 / 0.018

ω ² =  326666667

ω = 18074 rad/sec

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