A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0, a certain point is at height 0. Write an equation to describe the height of the point above the ground after t seconds.

Question

The answer is apparently d=-3cos(3pi/2)t +3

But I thought it's supposed to be sine since the height is at 0 when t=0, and why is w (3pi/2) instead of 90pi?

Answer 1

since cos(kt) has a maximum at t=0, -3cos(kt) has a minimum of -3 at t=0.

So, since the bottom of the tire touches the ground, the axle is 3 feet up, and if the point starts at the bottom of the circle, it is indeed given by 3-3cos(3pi/2 t). The height is from the ground, not from the axle.

Naturally, since sin(x) = cos(pi/2 - x), they are really the same thing, just starting at different times (phase shift).

If you use sin(kt), then 3+3sin(kt) starts at axle height, since sin(kt)=0 at t=0.

The reason for the 3pi/2 is that 45 rpm is 3/4 rev/second. That is the frequency, not the period. The period is 4/3 seconds. Since the period of cos(kt) is 2pi/k, we have

2pi/k = 4/3, so k = 3pi/2