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Finding the velocity for a given centripetal acceleration

  1. Feb 11, 2007 #1
    THE PROBLEM: (my own words) For a rotating circle with radius 2.10cm, the centripetal acceleration on the rim must be 100g. In revolutions per minute, what is the rate of rotation required?


    1. Centripetal acceleration is given by [tex]a_c = \frac{v^2}{r}[/tex] and so [tex]\frac{v^2}{0.021} = 100g \Leftrightarrow v = \sqrt{0.021 \times 100g} \approx 4.54 \frac{m}{s} [/tex]

    2. The velocity is given by [tex]v = \frac{2 \pi r}{T}[/tex] (where T is the period) and so the period is [tex]T = \frac{2 \pi r}{v}[/tex].

    3. Therefore the frequency is [tex]f = \frac{1}{T} = \frac{v}{2 \pi r} = \frac{4.54\ \frac{m}{s}}{0.132\ \frac{m}{rev}} = 34.4 \frac{rev}{s}[/tex]

    4. This is equal to 34.4/60 RPM = 0.57 RPM.

    WHY I'M PISSED OFF: The answer is actually 2060 RPM and I have NO idea what I did wrong. I've checked my answer about a billion times...Please help me.

    ***EDIT*** OMG! I divided by 60 in step 4 rather than multiplied...This is so stupid I spent over an hour on this problem and spent a really long time writing this post. Well uh...I don't have anything left to say.
    Last edited: Feb 11, 2007
  2. jcsd
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