Question

# 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?

Algebra foundations
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?

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:
$$\displaystyle{v}={1560}π{r}{a}\frac{{d}}{{h}}{r}\cdot{\left(\frac{{25}}{{5280}}\right)}{m}{i}\le{s}$$
$$\displaystyle{v}≈{23.2}{m}\frac{{i}}{{h}}{r}$$