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Radial Acceleration

  1. Jan 25, 2006 #1
    Find the required angular velocity of an ultracentrifuge in [itex] \frac{rev}{min} [/itex] for the radial acceleration of a point 1.00 cm from the axis to equal 400,000g (that is, 400,000 times the acceleration of gravity.)
    So [tex] a_{rad} = (\omega)^{2}r [/tex]. [tex] 400,000g = \omega^{2}(0.01 m) [/tex]. Would I just solve for [tex] \omega [/tex]? [tex] \omega [/tex] would be in m/s? Then to convert to rev/min, you use the fact that [tex] 2\pi(0.01 m) [/tex] equals 1 revolution?
    Would this be the correct way to solve this problem?
    :smile: [
  2. jcsd
  3. Jan 25, 2006 #2


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    when you solve for [itex]\omega[/itex] with your formula the units is [itex]rad/s[/itex] (if you worked in SI units), which needs to be converted to [itex]rev/min[/itex] now
    [tex]2\pi\ rad=\ 1\ rev[/tex]
    [tex]1\ rad=\ \frac{1}{2\pi}\ rev[/tex]
    and since
    [tex]60\ s=\ 1\ min[/tex]
    it follows that
    [tex]1\ s=\ \frac{1}{60}\ min[/tex]
    [tex]\omega\ =\ \frac{1\ rad}{1\ s}\ =\ \frac{60\ rev}{2\pi\ min}\ =\ \frac{30}{\pi}\ rev/min[/tex]
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