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DarkBlueChocobo

  • one year ago

Help with application problem

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  1. DarkBlueChocobo
    • one year ago
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    A paddle boat is traveling down a lake at 15 mph. If the paddle wheel is 8 ft in diameter, how many revolutions per min is it rotating

  2. DarkBlueChocobo
    • one year ago
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    |dw:1443836986859:dw|

  3. DarkBlueChocobo
    • one year ago
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    so the radius of the wheel is 4 ft then

  4. DarkBlueChocobo
    • one year ago
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    but first wouldnt we have to change units

  5. DarkBlueChocobo
    • one year ago
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    |dw:1443837060765:dw|

  6. DarkBlueChocobo
    • one year ago
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    |dw:1443837124936:dw|

  7. DarkBlueChocobo
    • one year ago
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    Doesn't this give us the angular speed?

  8. DarkBlueChocobo
    • one year ago
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    @zepdrix

  9. zepdrix
    • one year ago
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    Umm umm umm, I'm getting a little confused by your break down of the units. Lemme show you how I'm thinking about it.\[\large\rm boat~speed=\frac{15\color{orangered}{(miles)}}{1\color{royalblue}{hour}}=\frac{15\color{orangered}{(5280ft)}}{\color{royalblue}{60\min}}\]So we get that the boat travels at a rate of:\[\large\rm boat~speed=\frac{15(5280)}{60}\frac{ft}{\min}\]And the circumference of our wheel is:\[\large\rm C=\pi d\qquad\to\qquad C=8 \pi ft\]The distance around the wheel, the circumference, is `one full revolution`. So our wheel is moving at this rate:\[\large\rm wheel~speed=8\pi\frac{ft}{rev}\] So then ummmmm, we divide I guess? \[\large\rm \frac{15(5280)}{60}\frac{ft}{\min}\div8\pi\frac{ft}{rev}\]Sorry I'm a little rusty on these types of problems :) lol

  10. zepdrix
    • one year ago
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    \[\large\rm =\frac{15(5280)}{60}\frac{\cancel{ft}}{\min}\times\frac{1}{8\pi}\frac{rev}{\cancel{ft}}\]

  11. zepdrix
    • one year ago
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    Mmm I need a brief refresher on this stuff XD What is this type of problem called? I'll Google it. Something with rotational motion?

  12. DarkBlueChocobo
    • one year ago
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    Well the beginning of it was Angular speed and then you converted it to linear speed basically because they are related

  13. zepdrix
    • one year ago
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    I guess I made it a little bit of extra work by writing the angular relation like this. I should have written it like this: \(\large\rm \dfrac{1rev}{8\pi ft}\) And then it's just a multiplication problem, canceling out the units in the long string as you did.\[\large\rm \left(\frac{15miles}{hr}\right)\cdot\left(\frac{5280ft}{mile}\right)\cdot\left(\frac{hr}{60\min}\right)\cdot\left(\frac{1rev}{8\pi ft}\right)\]

  14. zepdrix
    • one year ago
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    I think you just had your last term a little off, ya? :o The wheel spins 8pi every revolution.

  15. DarkBlueChocobo
    • one year ago
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    It seems so :<

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