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Uniform Circular Motion. 
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#1
Apr1906, 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 


#2
Apr1906, 04:30 AM

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PF Gold
P: 9,772

Could you please show some work or thoughts?
HINT: What is the equation for centripetal acceleration? ~H 


#3
Apr1906, 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????



#4
Apr1906, 05:39 AM

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PF Gold
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Uniform Circular Motion.
[tex]a = \left( 6.25\times 10^{3} \right)g[/tex]
~H 


#5
Apr1906, 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 


#6
Apr1906, 06:44 AM

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PF Gold
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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 


#7
Apr1906, 07:06 AM

P: 21

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


#8
Apr1906, 07:35 AM

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PF Gold
P: 9,772

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 


#9
Apr1906, 08:21 AM

P: 21

Thanx i think i've got it.



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