حركة الجاذبية الدائرية لعجلة فيريس

Problem

A Ferris wheel has a diameter of $60$ feet, a center $35$ feet above the ground, and a period of $120$ seconds; the rider starts at the bottom at $t=0$, and the task is to model the height with a transformed negative cosine function and find how long the rider is above $50$ feet each rotation.

Step 1: Build the height function

The radius is half the diameter, so the amplitude is $30$ feet. Since one revolution takes $120$ seconds, the cosine coefficient is

$$b = \frac{2\pi}{120} = \frac{\pi}{60}.$$

The vertical shift is the center height, $35$, and because the rider starts at the minimum height, the model uses negative cosine:

$$h(t) = -30\cos\left(\frac{\pi t}{60}\right) + 35.$$

Step 2: Set up the height condition

To find when the rider is above $50$ feet, solve

$$-30\cos\left(\frac{\pi t}{60}\right) + 35 > 50.$$

Subtracting $35$ gives

$$-30\cos\left(\frac{\pi t}{60}\right) > 15.$$

Dividing by $-30$ flips the inequality:

$$\cos\left(\frac{\pi t}{60}\right) < -\frac{1}{2}.$$

Step 3: Find the time interval above $50$ feet

Cosine is less than $-\frac{1}{2}$ when the angle is between $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$, so

$$\frac{2\pi}{3} < \frac{\pi t}{60} < \frac{4\pi}{3}.$$

Multiplying through by $\frac{60}{\pi}$ gives

$$40 < t < 80.$$

So the rider is above $50$ feet for

$$80 - 40 = 40$$

seconds during each rotation.

Answer

The height function is $h(t) = -30\cos\left(\frac{\pi t}{60}\right) + 35$, and the rider stays above $50$ feet for $40$ seconds per rotation.