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Circle Angle, Time, xy coordinates
- Thread starter Chaim
- Start date
Is the ferris wheel beginning its rotation where you jump on?. The time you're on the ferris wheel is different, but the ferris wheel still rotates at 3 rpm. So, in a whole number of minutes, it comes back where it started. Perhaps I am not getting the gist of the problem.
If so, the Ferris wheel makes 3 revs per minute. So, in 4 minutes it makes 12 revs and comes back where you jumped on.
So, assuming the usual xy axis, \(\displaystyle x=24cos(\frac{17\pi}{90}), \;\ y=24sin(\frac{17\pi}{90})\)
Same with the other problem, only use \(\displaystyle \frac{\pi}{9}\) for 20 degrees. The ferris wheel makes more revs, but still comes back where it started.
I am used to using radians. If you want to use degrees, that's fine.
Ferris wheel models are actually rather interesting.
Let's say the ferris wheel starts where you jump on. Then, it is \(\displaystyle \frac{14\pi}{45}\) rad to the top.
This takes \(\displaystyle 7/135\) minutes.
Thus, the height at any time t is given by \(\displaystyle h(t)=24cos(6\pi(t-\frac{7}{135}))+24\)
You can plug in any time t in minutes and find the height from the ground.
For instance, sub in t=12 and we get 37.42
Subtract 24 and we get the 13.42 from before using
\(\displaystyle 24sin(34)=24sin(\frac{17\pi}{90})=13.42\)
If so, the Ferris wheel makes 3 revs per minute. So, in 4 minutes it makes 12 revs and comes back where you jumped on.
So, assuming the usual xy axis, \(\displaystyle x=24cos(\frac{17\pi}{90}), \;\ y=24sin(\frac{17\pi}{90})\)
Same with the other problem, only use \(\displaystyle \frac{\pi}{9}\) for 20 degrees. The ferris wheel makes more revs, but still comes back where it started.
I am used to using radians. If you want to use degrees, that's fine.
Ferris wheel models are actually rather interesting.
Let's say the ferris wheel starts where you jump on. Then, it is \(\displaystyle \frac{14\pi}{45}\) rad to the top.
This takes \(\displaystyle 7/135\) minutes.
Thus, the height at any time t is given by \(\displaystyle h(t)=24cos(6\pi(t-\frac{7}{135}))+24\)
You can plug in any time t in minutes and find the height from the ground.
For instance, sub in t=12 and we get 37.42
Subtract 24 and we get the 13.42 from before using
\(\displaystyle 24sin(34)=24sin(\frac{17\pi}{90})=13.42\)
Is the ferris wheel beginning its rotation where you jump on?. The time you're on the ferris wheel is different, but the ferris wheel still rotates at 3 rpm. So, in a whole number of minutes, it comes back where it started. Perhaps I am not getting the gist of the problem.
If so, the Ferris wheel makes 3 revs per minute. So, in 4 minutes it makes 12 revs and comes back where you jumped on.
So, assuming the usual xy axis, \(\displaystyle x=24cos(\frac{17\pi}{90}), \;\ y=24sin(\frac{17\pi}{90})\)
Same with the other problem, only use \(\displaystyle \frac{\pi}{9}\) for 20 degrees. The ferris wheel makes more revs, but still comes back where it started.
I am used to using radians. If you want to use degrees, that's fine.
Ferris wheel models are actually rather interesting.
Let's say the ferris wheel starts where you jump on. Then, it is \(\displaystyle \frac{14\pi}{45}\) rad to the top.
This takes \(\displaystyle 7/135\) minutes.
Thus, the height at any time t is given by \(\displaystyle h(t)=24cos(6\pi(t-\frac{7}{135}))+24\)
You can plug in any time t in minutes and find the height from the ground.
For instance, sub in t=12 and we get 37.42
Subtract 24 and we get the 13.42 from before using
\(\displaystyle 24sin(34)=24sin(\frac{17\pi}{90})=13.42\)
Never mind, I found it was just simple, I got the degree and radius, so all I have to do is use rcos(degree) and rain(degree)
Thank you though!
Last edited: