MHB Arc length & similar questions

[fluffertoes]

Hello! I really don't understand this concept, and I have an example problem that I am working on that I just CANNOT figure out! Any help? Thanks so much in advance! A group of people get on a pirate ship ride at the fair. This ride is a swinging pendulum with a maximum swing angle of 65 degrees from the center of the ship in either direction. The arm of the pendulum holding the ship has a 40 ft radius and the ship is 22 feet long with the last seats positioned 1 foot from the end of the ship.
What is the arc length traveled by the center of the ship between the two maximum points?
ALSO>>>>
What is the maximum height reached by the center of the ship? I attached a picture for reference! Thanks so much! :)View attachment 6233

 

[Prove It]

Hello! I really don't understand this concept, and I have an example problem that I am working on that I just CANNOT figure out! Any help? Thanks so much in advance! A group of people get on a pirate ship ride at the fair. This ride is a swinging pendulum with a maximum swing angle of 65 degrees from the center of the ship in either direction. The arm of the pendulum holding the ship has a 40 ft radius and the ship is 22 feet long with the last seats positioned 1 foot from the end of the ship.
I found that the arc length traveled by the center of the ship between the two maximum points is 40 feet. Is this correct?
ALSO>>>>
What is the maximum height reached by the center of the ship? I attached a picture for reference! Thanks so much! :)

Between the two maximum points, an angle of $\displaystyle \begin{align*} 130^{\circ} \end{align*}$ is swept out, so the arclength would be

$\displaystyle \begin{align*} \mathcal{l} &= \frac{130}{360} \cdot 2\,\pi \cdot 45\,\textrm{ft} \\ &= \frac{65\,\pi}{2} \,\textrm{ft} \\ &\approx 102.1\,\textrm{ft} \end{align*}$

 

[fluffertoes]

Between the two maximum points, an angle of $\displaystyle \begin{align*} 130^{\circ} \end{align*}$ is swept out, so the arclength would be

$\displaystyle \begin{align*} \mathcal{l} &= \frac{130}{360} \cdot 2\,\pi \cdot 45\,\textrm{ft} \\ &= \frac{65\,\pi}{2} \,\textrm{ft} \\ &\approx 102.1\,\textrm{ft} \end{align*}$

The radius is 40 feet, not 45 feet. So technically wouldn't the arc length end up being (in terms of pi (π)):

l=130/360⋅ (2π⋅40ft) = 260π/9 feet

 

[MarkFL]

Yes:

$$s=r\theta=\left(40\text{ ft}\right)\left(2\cdot65^{\circ}\frac{\pi}{180^{\circ}}\right)=\frac{260\pi}{9}\,\text{ft}\approx90.76\text{ ft}$$

To find the maximum height (above the lowest point) of the middle of the ship, we can observe that the height $h$ of the ship in terms of the angular displacement $\theta$ is given by:

$$h=r(1-\cos(\theta))$$

So, use $r=40\text{ ft}$ and $\theta=65^{\circ}$ in the above formula to get the maximum height of the center of the ship. :D

 

[fluffertoes]

Yes:

$$s=r\theta=\left(40\text{ ft}\right)\left(2\cdot65^{\circ}\frac{\pi}{180^{\circ}}\right)=\frac{260\pi}{9}\,\text{ft}\approx90.76\text{ ft}$$

To find the maximum height (above the lowest point) of the middle of the ship, we can observe that the height $h$ of the ship in terms of the angular displacement $\theta$ is given by:

$$h=r(1-\cos(\theta))$$

So, use $r=40\text{ ft}$ and $\theta=65^{\circ}$ in the above formula to get the maximum height of the center of the ship. :D

So would that leave me with:

(40)Cos(65°) = 16.9
40 - 16.9 = 23.095 feet