Question

a large ferris wheel is 100 feet in diameter and rises 10 feet off the ground. each revolution of the wheel takes 30 seconds. A) express the vertical distance "H" of the seat off the ground as a function of time "t" if t=0 corresponds to a time when the seat is at the bottom. B) if the seat is rising, how fast is the distance h changing when h=55 feet?

Answer 1

Take a step back and see what's what. At angle
w=0, H = 10
w = 90 deg, H=10+50
w = 180, H = 10 + 100

Looks like H = 10 + 50(1 -cos(w))

But, angle w = t/30 * 2π = πt/15
That is, one revolution = 2π every 30 sec.

dH = -50sin(w) dw
dH = -50sin(w) * π/15 dt

when H = 55, it's just below the level of the axle, so w will be just under 90 deg

55 = 10 + 50(1 - cos(πt/15))
55 = 10 + 50 - 50cos(0.209t)
0.1 = cos(0.209t)
.209t = arccos(.1) = 1.47
t = 7.033 seconds.

That makes sense, since at t=30/4 = 7.5 the wheel has gone a quarter turn.

As above,

dH/dt = -50sin(0.209t) * .209
= -50sin(1.47) = -50*.995 = 49.74 ft/s