Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 222: 13

Answer

The particle will experience an acceleration of 100,000 g's when the centrifuge rotates at a rate of 34,000 rpm.

Work Step by Step

First we can find the velocity of the particle: $\frac{v^2}{r} = a$ $v = \sqrt{ra}$ $v = \sqrt{(0.080 ~m)(100,000)(9.8 ~m/s^2)}$ $v = 280~m/s$ We can use the velocity to find the number of revolutions each second. Let $z$ be the number of revolutions each second. Then, $z = \frac{v}{2\pi r}$ $z = \frac{280 ~m/s}{(2\pi)(0.080 ~m)}$ $z = 560 ~rev/s$ $(560 ~rev/s)(\frac{60 ~s}{1 ~min}) = 34,000 ~rpm$ The particle will experience an acceleration of 100,000 g's when the centrifuge rotates at a rate of 34,000 rpm.
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