Question

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm

Answer 1

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$, the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$, the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$