An amusement park ride consists of a car mvoing an a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.0 kN, the circle's radius is 10m. At the top of the circle, what are the (a) magnitude `F_(B)` and (b) direction (up or down) of the force on the car from the boom if the car's speed is v=5.0m/s? What are (c) `F_(B)` and (d) the direction if v=12m/s?

Answer

To solve the problem, we will analyze the forces acting on the car at the top of the vertical circle. The forces involved are the gravitational force (weight of the car and riders) and the force exerted by the boom (denoted as \( F_B \)).

### Given Data:
- Weight of the car and riders, \( W = 6.0 \, \text{kN} = 6000 \, \text{N} \)
- Radius of the circle, \( r = 10 \, \text{m} \)
- Speed at the top of the circle for part (a) and (b), \( v = 5.0 \, \text{m/s} \)
- Speed at the top of the circle for part (c) and (d), \( v = 12.0 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step-by-Step Solution:

#### Part (a): Magnitude of \( F_B \) when \( v = 5.0 \, \text{m/s} \)

1. **Calculate the mass of the car**:
\[ m = \frac{W}{g} = \frac{6000 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 612.24 \, \text{kg} \]

2. **Calculate the centripetal force required**:
The centripetal force \( F_c \) needed to keep the car moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
Substituting the values:
\[ F_c = \frac{612.24 \times (5.0)^2}{10} = \frac{612.24 \times 25}{10} = 1530.6 \, \text{N} \]

3. **Apply the force balance at the top of the circle**:
At the top of the circle, the net centripetal force is provided by the weight of the car minus the force from the boom:
\[ F_c = W - F_B \]
Rearranging gives:
\[ F_B = W - F_c = 6000 \, \text{N} - 1530.6 \, \text{N} = 4469.4 \, \text{N} \]

4. **Convert to kN**:
\[ F_B \approx 4.47 \, \text{kN} \]

#### Part (b): Direction of \( F_B \) when \( v = 5.0 \, \text{m/s} \)

- Since \( F_B \) is positive, the direction of the force exerted by the boom on the car is **upward**.

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#### Part (c): Magnitude of \( F_B \) when \( v = 12.0 \, \text{m/s} \)

1. **Calculate the centripetal force required**:
\[ F_c = \frac{mv^2}{r} = \frac{612.24 \times (12.0)^2}{10} = \frac{612.24 \times 144}{10} = 8805.6 \, \text{N} \]

2. **Apply the force balance at the top of the circle**:
\[ F_c = W - F_B \]
Rearranging gives:
\[ F_B = W - F_c = 6000 \, \text{N} - 8805.6 \, \text{N} = -2805.6 \, \text{N} \]

3. **Convert to kN**:
\[ F_B \approx -2.81 \, \text{kN} \]

#### Part (d): Direction of \( F_B \) when \( v = 12.0 \, \text{m/s} \)

- Since \( F_B \) is negative, the direction of the force exerted by the boom on the car is **downward**.

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### Summary of Answers:
- (a) \( F_B \approx 4.47 \, \text{kN} \)
- (b) Direction: Upward
- (c) \( F_B \approx -2.81 \, \text{kN} \)
- (d) Direction: Downward