The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 mm. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). A passenger weighs 902 NN at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?

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.

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). A passenger weighs 902 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?

Calculate the apparent weight of a 902 N passenger at the lowest point of the Cosmoclock 21 Ferris wheel using concepts of circular motion, centripetal acceleration, and forces. Explore step-by-step calculations, including velocity and acceleration, suitable for Grades 10-12 students.

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