MHB How Fast Do Carousel Rides Travel in Miles per Hour?

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Carousel rides can be calculated for speed in miles per hour using their radius and rotational speed. For a radius of 167 inches, the speed is approximately 2.4 miles per hour, while a radius of 231 inches yields about 3.3 miles per hour. The conversion factor of 2π radians per revolution is essential for aligning units correctly in these calculations. This ensures that the results remain consistent and accurate when converting between different measurement units. Understanding these unit conversions is crucial for solving similar problems effectively.
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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:
 
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Re: miles per hour on a carousel

karush said:
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
[math]\frac{2 \pi ~ \text{rad}}{1 ~\text{rev}}[/math]

-Dan
 
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}}
$
 
Re: miles per hour on a carousel

karush said:
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 a lot in these kinds of problems.

-Dan
 
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