How Fast Do Carousel Rides Travel in Miles per Hour?

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SUMMARY

The discussion focuses on calculating the speed of carousel rides in miles per hour using the formula for linear velocity. Two radii are considered: 13 feet 11 inches (167 inches) and 19 feet 3 inches (231 inches), with an angular velocity of 2.4 revolutions per minute. The calculated speeds are approximately 2.4 mph for the first radius and 3.3 mph for the second. The conversion factor of 2π radians per revolution is crucial for ensuring unit consistency in the calculations.

PREREQUISITES
  • Understanding of angular velocity and linear velocity
  • Familiarity with unit conversions (inches to feet, feet to miles)
  • Basic knowledge of trigonometric functions (radians and revolutions)
  • Ability to manipulate equations involving physical quantities
NEXT STEPS
  • Learn about angular velocity and its applications in circular motion
  • Study unit conversion techniques for different measurement systems
  • Explore the relationship between linear and angular velocity in physics
  • Investigate more complex problems involving circular motion and forces
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of circular motion and velocity calculations.

karush
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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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