# TrigonometryMiles per hour on a carousel

#### karush

##### Well-known member
the rides on a carousel are represented by $2$ circles with the same center with

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

$r_{13}=13 \text{ ft} 11 \text { in}= 167 \text { in}$
$r_{19}=19 \text{ ft} 3 \text { in}= 231 \text { in}$

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.

#### topsquark

##### Well-known member
MHB Math Helper
Re: miles per hour on a carousel

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

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

$r_{13}=13 \text{ ft} 11 \text { in}= 167 \text { in}$
$r_{19}=19 \text{ ft} 3 \text { in}= 231 \text { in}$

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.
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
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
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