Sheldon notices the Ferris Wheel radius is 40 feet and the center of the wheel is 50 feet high. Amy times the ride and finds it takes 45 seconds to complete one turn around the wheel.

Question

How many seconds does it take for the platform to go each of these distances?

a. From 3 o'clock to 10 o'clock

b. From 3 o'clock to 7 o'clock

c. From 3 o'clock to 4 o'clock

What is the platform height off the ground at each of these times?

a. 3 seconds after passing 3 o'clock position

b. 12 seconds after passing the 3 o'clock position

c. 24 seconds after passing the 3 o'clock position

d. 35 seconds after passing the 3 o'clock position

Answer 1

Look at the ferris wheel as a big clock that rotates clockwise with 12 o'clock at the top, 3 o'clock at 90⁰ to the right, 6 o'clock at the bottom and 9 o'clock at 90⁰ to the left. Let's call an interval the time it takes for the hour hand to go from 1 hour to the next (example from 12 to 1). There are 12 of these intervals in 1 hour (complete 360⁰ rotation) and it takes 45 seconds for 1 rotation. Therefore, each interval would be 45sec/12int--->3.75sec/int. For the first set of problems, simply multiply the 3.75sec/int by the number of intervals:

a) 3 o'clock to 10 o'clock is 7 intervals--->7(3.75sec/int)=26.25sec.

b) 3 o'clock to 7 o'clock is 4 intervals--->4(3.75sec/int)=15sec.

You can do "c" & "d".

For the second part, 45sec=1rev=360⁰. Therefore, 1sec=360⁰/45--->1sec=8⁰.

a)3 seconds means that the wheel has rotated (3sec)(8⁰/sec)=24⁰. Drawing a vertical line from the line running from the center of the wheel to the 3 o'clock position to the line that is rotated 24⁰clockwise from horizontal line we see that we have a right triangle with the hypotenuse equal to 40ft (the radius of the wheel) and an angle equal to 24⁰. We need to find the vertical line of this triangle which is found by sin24⁰=h/40--->h=40sin24⁰--->h=16.27ft. The center of the wheel is 50ft up from the ground, so the height we are looking for would be 50-16.27=33.73ft.

I won't do the other 3 but I'll help you get started:

b)(12sec)(8⁰/sec)=96⁰. 96⁰ from 3 o'clock would be 6⁰ left of the 6 o'clock position. You would have a right triangle with hyp.=40ft and an angle=6⁰ and need to find the length of the leg from the center of the wheel down to the perpendicular line to the point. Subtract this value from 50 and you should get 10.22ft.

c)(24sec)(8⁰/sec)=192⁰. 192⁰ clockwise from 3 o'clock will be 12⁰ above 9 o'clock. Working it out and you should get h=8.32ft and the height you are looking for to be 50 + h = 58.32ft.

d) (35sec)(8⁰/sec)=280⁰ from 3 o'clock would be 10⁰ clockwise from 12 o'clock. Working it out and you should get 89.39ft.

Lots going on here. Hope this helps!

Answer 2

The Ferris Wheel radius is 40 feet and the center of the wheel is 50 feet high, lowest point of the wheel is at a height of 50 – 40 = 10 feet and the highest point at 50 + 40 = 90 feet.

Since there are 12 hours on a clockface each hour represents 1/12 of a full revolution so each ‘hour’ is rotated through in 45/12 = 3.75 seconds.

Therefore the time to rotate to t o’clock from 3 o’clock , t < 12, is (t – 3)(3.75) seconds.

For Part 1 we have:

a. (10- 3)(3.75) = (7)(3.75) = 26.25 sec

b. (7- 3)(3.75) = (4)(3.75) = 15 sec

c. (4- 3)(3.75) = (1)(3.75) = 3.75 sec

For Part 2, the beginning platform height is at 3 o’clock corresponding to 0^o and a height 50 feet.

Because the speed of revolution is 45 seconds, the ratio of the time t seconds after passing 3 o'clock position and 45 seconds gives the proportional part of the 360^o the platform has turned through, namely an angle of (t/45)( 360^o) = (8t)^o. Then the vertical distance y feet from the platform to a horizontal line through the center at 0^o is 40sin(8t). Recall that lowest point of the wheel is at a height of 10 feet and the center at 0^o and is at 50 feet. Therefore, the platform height is 50 – y = 50 – sin(8t) feet.

What is the platform height off the ground at each of these times?

a. 3 seconds after passing 3 o'clock position. h = 50 – sin(8*3) = 50 – sin(24) = 33.73 feet

b. 12 seconds after passing the 3 o'clock position. h = 50 – sin(8*12) = 50 – sin(96) = 49.01 feet

c. 24 seconds after passing the 3 o'clock position. h = 50 – sin(8*24) = 50 – sin(192) = 53.32 feet

d. 35 seconds after passing the 3 o'clock position. h = 50 – sin(8*35) = 50 – sin(280) = 89.39 feet