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Trigonometry Miles per hour on a carousel

karush

Well-known member
Jan 31, 2012
2,771
the rides on a carousel are represented by $2$ circles with the same center with

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

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.:cool:
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Re: miles per hour on a carousel

the rides on a carousel are represented by $2$ circles with the same center with

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

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.:cool:
It's all about the units, which you didn't include in your rev - rad conversion. 1 revolution = 2 pi radians. For a unit conversion it becomes the factor
\(\displaystyle \frac{2 \pi ~ \text{rad}}{1 ~\text{rev}}\)

-Dan
 

karush

Well-known member
Jan 31, 2012
2,771
Re: miles per hour on a carousel

so my eq should be this. but ans is them same?

$
\displaystyle v_{r13} = 167\text { in}
\cdot\frac{2.4 \text { rev}}{\text {min}}
\cdot\frac{2 \pi\text{ rad}}{\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}}
$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Re: miles per hour on a carousel

so my eq should be this. but ans is them same?

$
\displaystyle v_{r13} = 167\text { in}
\cdot\frac{2.4 \text { rev}}{\text {min}}
\cdot\frac{2 \pi\text{ rad}}{\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}}
$
Yes, the number will be the same, but now the units line up. Keep the 2 pi rad = 1 rev in mind. You'll see it alot in these kinds of problems.

-Dan