# Uniform Circular Motion of centrifuge

1. Apr 19, 2006

### MetalCut

Hi there. I need some help with this question. Can anyone help me......

A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Suppose the centripetal acceleration of the sample is 6.25 X 103 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 5.00cm from the axis of rotation?

Any help would be appreciated.

Thanx

2. Apr 19, 2006

### Hootenanny

Staff Emeritus
Could you please show some work or thoughts?

HINT: What is the equation for centripetal acceleration?

~H

3. Apr 19, 2006

### MetalCut

The equation is a = v2/r. But what do they mean when they say the centripetal acceleration is 6.25x10 3 times as large as the acceleration due to gravity????

4. Apr 19, 2006

### Hootenanny

Staff Emeritus
$$a = \left( 6.25\times 10^{3} \right)g$$

~H

5. Apr 19, 2006

### MetalCut

So then that probably means that v2/r = (6,25x10 3)g

And the circumference of the circle its rotating in is 0,314m or 31,4cm

6. Apr 19, 2006

### Hootenanny

Staff Emeritus
But you want to find revolutions per minute, so your next step would be calculating the angular velcoity ($\omega$). You will need to use;

$$v = \omega r$$

~H

7. Apr 19, 2006

### MetalCut

But i can get (v) also with v=(2)(pie)(r)\T
So i still need T

8. Apr 19, 2006

### Hootenanny

Staff Emeritus
They are effectively the same thing, but you don't need to work out v;

$$a = \frac{v^2}{r}$$

$$v = \omega r = \frac{2\pi r}{T}$$

$$a = \frac{\omega^2 r^2}{r}$$

$$\omega^{2} = \frac{a}{r}$$

$$\frac{2\pi}{T} = \sqrt{\frac{a}{r}}$$

~H

9. Apr 19, 2006

### MetalCut

Thanx i think i've got it.