Ferris Wheel Trig Problem

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NeedHelp27

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Hello, all. For homework, we got a problem that reads as follows: A Ferris wheel 50 ft in diameter makes one revolution every 40 sec. If the center of the wheel is 30 ft above the ground, how long after reaching the low point is a rider 50 ft above the ground? Our teacher said to model the situation with an equation.

When I tried to go about doing this, I drew a graph showing the person's height, at it ended up being the cosine graph shifted up 5 and over so that a low point was on the y-axis (0,5). Next, I tried to write the equation in the form f(x)=acos(bx+c)+d. I did (max value - min value)/2 = (55-5)/2=25 to find a. Then, because one revolution takes 40 seconds, I solved 2pi/b=40 for b and got b=pi/20. The graph is shifted 5 up, so d-5. That gave me f(x)=25cos((pix)/20+c)+5. I also have the point (0,5) on the graph, so I can plug it in to get c. f(0)=5=25cos(c)+5. C ended up being pi/2.

Then I went to solve the problem.
f(t)=50=25cos((pix)/20+pi/2)+5.
When I finally isolated x, I got a domain error. Where did I go wrong?
 
NeedHelp27 said:
Hello, all. For homework, we got a problem that reads as follows: A Ferris wheel 50 ft in diameter makes one revolution every 40 sec. If the center of the wheel is 30 ft above the ground, how long after reaching the low point is a rider 50 ft above the ground? Our teacher said to model the situation with an equation.

When I tried to go about doing this, I drew a graph showing the person's height, at it ended up being the cosine graph shifted up 5 and over so that a low point was on the y-axis (0,5). Next, I tried to write the equation in the form f(x)=acos(bx+c)+d. I did (max value - min value)/2 = (55-5)/2=25 to find a. Then, because one revolution takes 40 seconds, I solved 2pi/b=40 for b and got b=pi/20. The graph is shifted 5 up, so d-5. That gave me f(x)=25cos((pix)/20+c)+5. I also have the point (0,5) on the graph, so I can plug it in to get c. f(0)=5=25cos(c)+5. C ended up being pi/2.

Then I went to solve the problem.
f(t)=50=25cos((pix)/20+pi/2)+5.
When I finally isolated x, I got a domain error. Where did I go wrong?
Click to expand...
First, a typo; you are using x & t interchangeably [you used f(x) and f(t)]. Although it doesn't make a difference as long as you are consistent, I'll use t for time and make it f(t) and save x for the horizontal distance and y for vertical distance.. Next, you are shifting around your point of origin which, in part, would also change your cosine to what would normally be my sine function. Note that if you do the usual angle's (positive counter clockwise) as measured from x=0 and y=0 on the unit circle, we have x = cos(t), y= sin(t) where t = zero is the line along the positive x axis. If we wanted to measure the angle from the positive y axis (as is done in some disciplines), then we would have x = sin(t) and y=cos(t). Following what you have, we'll use straight up as an angle of zero.

So, we write
f(t) = a cos( b t + c) + d
Since, in this formulation, the angle is measured from the center of the circle, i.e. at x=0 and y = 25, we have at time zero the angle is straight down or c = \(\displaystyle \pi\). So, we have
f(t) = 25 cos(\(\displaystyle \frac{\pi}{20}(t + 20)) + d\)
and that will change your d to 30.
 
Denis said:
Assume wheel centered at (r,r).
Then we travel clockwise from (r,0) to (x,h).
(if you want the darn thing 5' above ground, draw a horizontal
line crossing (0,-5)! Doesn't affect the problem...:rolleyes:)

P = pi
Givens:
r = radius (25)
h = height (45)
v = wheel velocity = 2rP / 40 : ~3.927 feet per second

Solution:
we have a right triangle, vertices (25,25), (x,45) and (25,45); so:
(25-x)^2 + 20^2 = 25^2
solve: x = 40 or 10 : we want the 10 solution

c = chord joining (x,45) and (25,0)
c is hypotenuse with legs (r-x)=15 and h=45; so:
c = SQRT(45^2 + 15^2) : ~15.903

u = central angle
Law of Sines: (c/2) / SIN(u) = r
solve: u = 2*ASIN[c/(2r)] : ~143.13

a = arc = path travelled from (25,0) to (x,45)
a = Pur / 180 : ~62.452

t = time taken travelling arc a
t = a / v : ~15.903 seconds

Above's the step-by-stepper; arrange 'em as you like...
Click to expand...

But Denis, center at (x0,y0) and
(x-x0) = r sin(a t + b),
(y-y0) = r cos(a t + b),
where r is the radius, a is 'the period divided by the time of the period' and b is the initial condition seems so much simpler to me.
Thus,
t = [\(\displaystyle 2 n \pi + atan2((y-y_0)/r, \pm \sqrt{1-((y-y_0)/r)^2}) - b\)]/a > 0
depending on the quadrant of x-x0 [+ for 1st and 2nd, - for 3rd and 4th for clockwise revolution]


In this case the radius is 25, a = \(\displaystyle 2 \pi / 40 = \pi / 20\), the center is 5 feet above the center of the ferris wheel so x0 = 0 and y0 = 30 and the initial position is straight down from the center so b is \(\displaystyle (2 n + 1) \pi\), take your choice for n (choose your starting Riemann sheet/start time).
 
Hi,
I must admit that the admins of this website are not very nice. I posted a very nice, clear way of doing this problem. Even got the same exact result as Dennis arrived at, yet my posts on this thread has been moved. Why is that? Why have many of my other threads been removed. I even posted in the admin issues section asking why many of my posts have been removed but not one admin had the courage to respond.
I guess that the admins just do not want anyone other than a select few to answer questions. Is that really what a forum is about? Should the forum be about censorship? How many other volunteers have you pushed out of this forum?
I do not have to wonder how long this post will be up for.
icon13.png

Steven
 
Jomo said:
Hi,
I must admit that the admins of this website are not very nice. I posted a very nice, clear way of doing this problem. Even got the same exact result as Dennis arrived at, yet my posts on this thread has been moved. Why is that? Why have many of my other threads been removed. I even posted in the admin issues section asking why many of my posts have been removed but not one admin had the courage to respond.
I guess that the admins just do not want anyone other than a select few to answer questions. Is that really what a forum is about? Should the forum be about censorship? How many other volunteers have you pushed out of this forum?
I do not have to wonder how long this post will be up for.
icon13.png

Steven
Click to expand...

Still here
 
Denis said:
We wouldn't know why...suggest you PM the moderators.
Click to expand...
I did contact a moderator and I was told by the moderator that she/he did not remove any of my posts and does not know who did. I had no further contact from this moderator even though I ask if they could look into this. Seriously what am I to conclude when no one gets back to me with a real answer after three days.
 
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