- #1
sh86
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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.
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.
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.
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.
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