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Uniform Circular Motion.

  1. Apr 19, 2006 #1
    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. jcsd
  3. Apr 19, 2006 #2

    Hootenanny

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    Could you please show some work or thoughts?

    HINT: What is the equation for centripetal acceleration?

    ~H
     
  4. Apr 19, 2006 #3
    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????
     
  5. Apr 19, 2006 #4

    Hootenanny

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

    ~H
     
  6. Apr 19, 2006 #5
    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
     
  7. Apr 19, 2006 #6

    Hootenanny

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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
     
  8. Apr 19, 2006 #7
    But i can get (v) also with v=(2)(pie)(r)\T
    So i still need T
     
  9. Apr 19, 2006 #8

    Hootenanny

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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
     
  10. Apr 19, 2006 #9
    Thanx i think i've got it.
     
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