Question

How fast (in rpm) must a centrifuge rotate if a particle 7.00 cm from the axis of rotation is to experience an acceleration of 125,000 g’s?

Short answer

The rotation of a centrifuge is $3.99\times {10}^4\mathrm{rpm}$.

Step-by-step solution

Step 1. Understanding the speed of the particle

The speed of the centrifuge varies with the variation in the distance between the particle and centre of rotation, acceleration of the particle. With the help of speed, the rotation of the centrifuge can be evaluated.

Step 2. Identification of given data 

The given data can be listed below as,

• The distance of a particle from the centre of the rotation is, $r=7\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.07\mathrm{m}$.
• The acceleration of particle is, $a_c=125000g\text{'}\mathrm{s}$.

Step 3. Determination of the speed of the particle

The speed of the particle can be expressed as,

$v=\sqrt{a_{\mathrm{c}}r}$

Here, $a_c$is the centripetal acceleration of the particle.

Substitute the values in the above equation.

$\begin{array}{c}v=\sqrt{125000g\left(\frac{9.81\mathrm{m}/{\mathrm{s}}^2}{1g}\right)\times 0.07\mathrm{m}}\\ v=\sqrt{85837.5}\mathrm{m}/\mathrm{s}\\ v=292.98\mathrm{m}/\mathrm{s}\end{array}$

The speed of the particle can be expressed as,

$\begin{array}{c}v=\frac{2\pi r}{T}\\ v=\left(\frac{2\pi r}{\frac{1}{N}}\right)\\ \frac{1}{N}=\left(\frac{2\pi r}{v}\right)\\ N=\left(\frac{v}{2\pi r}\right)\end{array}$

Here, N is the rotation of the centrifuge, T is the time taken by the particle.

Substitute the values in the above equation.

$\begin{array}{c}N=\frac{292.98\mathrm{m}/\mathrm{s}\left(\frac{60\mathrm{s}}{1\min }\right)}{2\pi \times 0.07\mathrm{m}}\\ =3.99\times {10}^4\mathrm{rpm}\end{array}$

Thus, the rotation of a centrifuge is $3.99\times {10}^4\mathrm{rpm}$.