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Physics Principles with Applications
Found in: Page 198
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

How fast (in rpm) must a centrifuge rotate if a particle 8.0 cm from the axis of rotation is to experience an acceleration of 100,000 g’s?

The number of revolutions per minute is 33422.5 rpm.

See the step by step solution

Step by Step Solution

Given data

The distance from the axis of rotation is \({\rm{r}} = 8\;{\rm{cm}}\).

The acceleration is \({\rm{a}} = 100000{\rm{g}}\)

 Angular velocity and acceleration

In this problem, to calculate the revolutions per minute of a centrifuge, the relation between radial acceleration and angular velocity will be utilized.

Determine the revolutions per minute

The relation to calculate the revolutions per minute is given by:

\(\omega = \sqrt {\frac{a}{r}} \)

Here, r is the radius of the wheel and \(\omega \) is the angular speed.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\omega &= \sqrt {\frac{{100000 \times 9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{\left( {8\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}}} \\\omega &= \left( {3500\;{\rm{rad/s}} \times \frac{{60\;{\rm{rpm}}}}{{2\pi \;{\rm{rad/s}}}}} \right)\\\omega &= 33422.5\;{\rm{rpm}}\end{aligned}\)

Thus, \(\omega = 33422.5\;{\rm{rpm}}\) is the correct answer.

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