Uniform Circular Motion.

  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. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    Could you please show some work or thoughts?

    HINT: What is the equation for centripetal acceleration?

    ~H
     
  4. 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. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

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

    ~H
     
  6. 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. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

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

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    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. Thanx i think i've got it.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook