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

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Per
AI Thread Summary
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.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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:
 
Mathematics news on Phys.org
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
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top