A gondola on an amusement park ride, similar to the Spin Cycle at Silverwood Theme Park, spins at a speed of 13 revolutions per minute. If the gondola is 25 feet from the ride's center, what is the linear speed of the gondola in miles per hour?

A gondola on an amusement park ride, similar to the Spin Cycle at Silverwood Theme Park, spins at a speed of 13 revolutions per minute. If the gondola is 25 feet from the ride's center, what is the linear speed of the gondola in miles per hour?

Question
Algebra foundations
asked 2021-02-18
A gondola on an amusement park ride, similar to the Spin Cycle at Silverwood Theme Park, spins at a speed of 13 revolutions per minute. If the gondola is 25 feet from the ride's center, what is the linear speed of the gondola in miles per hour?

Answers (1)

2021-02-19
The relationship of the linear speed vv to the angular speed ω (in radians per unit time) and radius r is:
\(\displaystyle{v}=ω{r}\)
Convert the angular speed to radians per hour:
\(\displaystyleω={\left({13}{r}{e}\frac{{v}}{\min}\right)}\cdot{\left({2}π{r}{a}\frac{{d}}{{1}}{r}{e}{v}\right)}\cdot{\left({60}\frac{\min}{{1}}{h}{r}\right)}={1560}π{r}{a}\frac{{d}}{{h}}{r}\)
Convert the radius to miles:
\(\displaystyle{r}={25}{f}{t}\cdot{\left({1}{m}\frac{{i}}{{5280}}{f}{t}\right)}=\frac{{25}}{{5280}}{m}{i}\le{s}\)
So, the linear speet is:
\(\displaystyle{v}={1560}π{r}{a}\frac{{d}}{{h}}{r}\cdot{\left(\frac{{25}}{{5280}}\right)}{m}{i}\le{s}\)
\(\displaystyle{v}≈{23.2}{m}\frac{{i}}{{h}}{r}\)
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