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Uniform Circular Motion.

by MetalCut
Tags: circular, motion, uniform
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MetalCut
#1
Apr19-06, 01:52 AM
P: 21
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
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Hootenanny
#2
Apr19-06, 04:30 AM
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Could you please show some work or thoughts?

HINT: What is the equation for centripetal acceleration?

~H
MetalCut
#3
Apr19-06, 05:33 AM
P: 21
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????

Hootenanny
#4
Apr19-06, 05:39 AM
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Uniform Circular Motion.

[tex]a = \left( 6.25\times 10^{3} \right)g[/tex]

~H
MetalCut
#5
Apr19-06, 06:37 AM
P: 21
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
Hootenanny
#6
Apr19-06, 06:44 AM
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But you want to find revolutions per minute, so your next step would be calculating the angular velcoity ([itex]\omega[/itex]). You will need to use;

[tex]v = \omega r[/tex]

~H
MetalCut
#7
Apr19-06, 07:06 AM
P: 21
But i can get (v) also with v=(2)(pie)(r)\T
So i still need T
Hootenanny
#8
Apr19-06, 07:35 AM
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They are effectively the same thing, but you don't need to work out v;

[tex]a = \frac{v^2}{r}[/tex]

[tex]v = \omega r = \frac{2\pi r}{T}[/tex]

[tex]a = \frac{\omega^2 r^2}{r}[/tex]

[tex]\omega^{2} = \frac{a}{r}[/tex]

[tex]\frac{2\pi}{T} = \sqrt{\frac{a}{r}}[/tex]

~H
MetalCut
#9
Apr19-06, 08:21 AM
P: 21
Thanx i think i've got it.


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