circular functions: modelling Bob's height on Ferris wheel

[kmeangel27]

Bob is riding on the Ferris Supreme ride at an amusement park. His seat is attached to a small wheel of radius 8 ft rotating clockwise at a constant rate. This wheel is in turn attached to a larger wheel of radius 20 ft that is rotating counterclockwise at a constant rate different than that of the smaller wheel. When Bob gets on the ride (t = 0), his seat is 5 ft off the ground, and it takes 2 minutes for the ride to return to this starting position. Determine a function that models how far Bob is above the ground at time t.]

so far i have 28sin(t(pie)) + 5. Is that right? because i found the period to be (pie), but i dont know what to do with that. And i know that its 5 ft off the ground, and i also know that the radius is 20 but there is also another wheel attached to that which has a radius of 8 so thats how i got the amplitude to be 28.

I also found that the circumfrences are 40(pie) and 16(pie).
i don't know if i was supposed to even find the circumferences.

I dont know what else there is to do. Is there anything? beucase my teacher siad that there was more than one period. Im really not sure though.

 

[BigGlenntheHeavy]

Let x^2+Y^2=(20)^2. Then the height from the ground = 25+y, y in first quarter.

Then sin(?) = y/20, y = 20sin(?), ergo h+ 25+20sin(?)

Since the ferris wheel makes 2 revolutions per minute, theta =2(2?t) = 4?t, t in minutes.

Then h(t) = 25+20sin[?(4t); however this only applies if at t=0, h =25, but we want when t=0 h =5.

So we have a phase shift of 3?/2, hence H(t) = 25 +20sin[?(4t+3/2)] or H(t) = 25+20sin[?((8t+3)/2)]

Check: H(0) = 25 +20sin(3?/2) = 25-20 =5
H(1/8) = 25 + 20sin(2) =25+0 = 25
H (2/8) =25+20sin(5?/2) = 25+20 = 45
H(3/8) = 25+20sin(3?) = 25+0 = 25
H(4/8) = 25+20sin(7?/2) = 25-20 = 5

Since there are 4? revolutions per minute, every ?/2 revolution = 1/8 minutes.