Uniform Circular Motion.

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.

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