- #1
sh86
- 19
- 0
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 [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.
MY SOLUTION:
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: