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8-13P

Expert-verifiedFound in: Page 198

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

The number of revolutions per minute is 33422.5 rpm.

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

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

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

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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