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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,718

$\displaystyle\omega=\frac{2.4 \text {rev}}{\text {min}}$

and the radius are:

$r_{13}=13 \text{ ft} 11 \text { in}= 167 \text { in}$

$r_{19}=19 \text{ ft} 3 \text { in}= 231 \text { in}$

find:

$\displaystyle\frac{\text {mi}}{\text {hr}}$ of $r_1$ and $r_2$

$\displaystyle v_{r13} =

167\text { in}

\cdot\frac{2.4 \text { rev}}{\text {min}}

\cdot\frac{2 \pi}{\text {rev}}

\cdot\frac{\text {ft}}{12\text { in}}

\cdot\frac{\text {mi}}{5280\text { ft}}

\cdot\frac{60 \text{ min}}{\text {hr}}

\approx

2.4\frac{\text{ mi}}{\text {hr}}

$

thus using the same $\displaystyle v_{r19}=3.3\frac{\text{ mi}}{\text {hr}}$

these ans seem reasonable but my question is on the

$\displaystyle\frac{2 \pi}{\text {rev}}$

isn't $\text {rev}$ really to the circumference of the circle

how ever if used the ans are way to large.

not sure why the $2\pi$ works.