To determine the passenger's apparent weight at the lowest point of the Ferris wheel, we need to account for the forces acting on them, including the gravitational force and the additional force due to circular motion.

Step 1: Definitions and known values

• Weight of the passenger on the ground $W = 902 \, \text{N}$
• Diameter of the Ferris wheel $d = 100 \, \text{m}$, so the radius $r = \frac{d}{2} = 50 \, \text{m}$
• Time for one complete revolution $T = 60.0 \, \text{s}$
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$

Step 2: Find the passenger's mass

The weight $W$ on the ground is the force due to gravity. We can calculate the passenger's mass $m$ as follows: $W = m \cdot g \quad \Rightarrow \quad m = \frac{W}{g} = \frac{902}{9.8} \approx 92.04 \, \text{kg}$

Step 3: Calculate the velocity of the passenger

The velocity $v$ of the passenger on the Ferris wheel can be determined using the formula for the speed of an object in circular motion: $v = \frac{2\pi r}{T} = \frac{2\pi \times 50}{60} \approx 5.24 \, \text{m/s}$

Step 4: Calculate the centripetal acceleration

Centripetal acceleration $a_c$ is given by: $a_c = \frac{v^2}{r} = \frac{(5.24)^2}{50} \approx 0.55 \, \text{m/s}^2$

Step 5: Calculate the apparent weight at the lowest point

At the lowest point of the Ferris wheel, the apparent weight $W_{\text{apparent}}$ is the sum of the gravitational force and the force due to centripetal acceleration: $W_{\text{apparent}} = m(g + a_c)$ Substitute the values: $W_{\text{apparent}} = 92.04 \times (9.8 + 0.55) = 92.04 \times 10.35 \approx 952.6 \, \text{N}$

Final Answer:

The apparent weight of the passenger at the lowest point of the Ferris wheel is approximately 952.6 N.

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Would you like more details on the calculations or concepts?

Here are some related questions to further explore:

1. What would the passenger's apparent weight be at the highest point of the Ferris wheel?
2. How would the apparent weight change if the Ferris wheel’s speed were doubled?
3. What if the radius of the Ferris wheel were halved? How would it affect the apparent weight?
4. How does centripetal force affect objects in circular motion?
5. Can you derive the formula for centripetal force using Newton's second law?

Tip: Always remember that apparent weight changes in circular motion due to the combination of gravitational and centripetal forces.