# Finding the velocity for a given centripetal acceleration

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?

MY SOLUTION:

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

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

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

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

***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.

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