a circular path at a constant speed. To yield net force vectors with the same magnitude, the normal force at the bottom must be ---Select--- v that at the top. Categorize Because the speed of the child is constant, we can categorize this problem as one involving a particle (the child) , complicated by the gravitational force acting at all times on the child. ---Select--- Analyze (Use the following as necessary: or m, and g.) We draw a diagram of forces acting on the child at the bottom of the ride as shown in figure (b). The only forces acting on him are the downward gravitational force F, - mg and the upward force not exerted by the seat. The net upward force on the child that provides his centripetal acceleration has a magnitude Fnet bottom Using the particle in uniform circular motion model, apply Newton's second law to the child in the radial direction when he is at the bottom of the ride (Use the following as necessary: rand g.): Solve for the force exerted by the seat on the child (Use the following as necessary: r, v and g. Do not substitute numerical values; use variables only.): m². - md et- mg + Substitute numerical values given for the speed and radius:

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### Ferris Wheel Physics Problem

**Scenario:**
A child rides on a Ferris wheel with a radius of 12.2 m at a constant speed of 2.75 m/s as shown in the figures. The child moves in a vertical circle.

**Figures:**
- **(a)** Shows the side view of a Ferris wheel with two positions: top and bottom for the child.
- **(b)** Top view indicating the forces acting on the child at different points of the path.
- Vector diagrams illustrating forces on the child at the top and bottom positions of the ride.

---

**Problem (a): Determine the force exerted by the seat on the child at the bottom of the ride. Express in terms of the child’s weight (mg).**

**Solution:**

**Conceptualize:**
- When the child is at the top of the Ferris wheel, gravitational force (weight) and the seat's normal force both act downwards.
- At the bottom, gravity acts downward and the seat’s normal force acts upward against the child’s weight.

**Using Uniform Circular Motion:**
- **Equation:** \(\sum F = m a_c\)
- Acceleration (\(a_c\)) at bottom (\(v^2/r\))

**Calculations:**

1. **Net Force at Bottom** \( N_{bottom} = mg + \frac{mv^2}{r} \)

**Substitute values:**
\[ N_{bottom} = mg + \frac{m (2.75)^2}{12.2} \]

**Magnitude of Normal Force:**
- Greater than the child’s weight at the bottom.

---

**Problem (b): Determine the force exerted by the seat on the child at the top of the ride.**

**Solution:**

**Using Uniform Circular Motion:**
- **Equation:** \(\sum F = m a_c\)
- At the top of the ride, the forces are \(N_{top}\) and weight:

**Calculations:**

1. **Net Force at Top** \( N_{top} = mg - \frac{mv^2}{r} \)

**Substitute values:**
\[ N_{top} = mg - \frac{m (2.75)^2}{12.2} \]

**Magnitude of Normal Force:**
- Less than the child’s weight at the top.

---

**Exercise:**
Calculate the force on
Transcribed Image Text:### Ferris Wheel Physics Problem **Scenario:** A child rides on a Ferris wheel with a radius of 12.2 m at a constant speed of 2.75 m/s as shown in the figures. The child moves in a vertical circle. **Figures:** - **(a)** Shows the side view of a Ferris wheel with two positions: top and bottom for the child. - **(b)** Top view indicating the forces acting on the child at different points of the path. - Vector diagrams illustrating forces on the child at the top and bottom positions of the ride. --- **Problem (a): Determine the force exerted by the seat on the child at the bottom of the ride. Express in terms of the child’s weight (mg).** **Solution:** **Conceptualize:** - When the child is at the top of the Ferris wheel, gravitational force (weight) and the seat's normal force both act downwards. - At the bottom, gravity acts downward and the seat’s normal force acts upward against the child’s weight. **Using Uniform Circular Motion:** - **Equation:** \(\sum F = m a_c\) - Acceleration (\(a_c\)) at bottom (\(v^2/r\)) **Calculations:** 1. **Net Force at Bottom** \( N_{bottom} = mg + \frac{mv^2}{r} \) **Substitute values:** \[ N_{bottom} = mg + \frac{m (2.75)^2}{12.2} \] **Magnitude of Normal Force:** - Greater than the child’s weight at the bottom. --- **Problem (b): Determine the force exerted by the seat on the child at the top of the ride.** **Solution:** **Using Uniform Circular Motion:** - **Equation:** \(\sum F = m a_c\) - At the top of the ride, the forces are \(N_{top}\) and weight: **Calculations:** 1. **Net Force at Top** \( N_{top} = mg - \frac{mv^2}{r} \) **Substitute values:** \[ N_{top} = mg - \frac{m (2.75)^2}{12.2} \] **Magnitude of Normal Force:** - Less than the child’s weight at the top. --- **Exercise:** Calculate the force on
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