A 40,000 kg stunt plane flies in a 260-m-diameter vertical circle. At the point where the plane is flying straight down, its speed is 55 m/s and it is speeding up at a rate of 12 m/s per second. What angle does the net force make with the horizontal? Let an angle above horizontal be positive and an angle below horizontal be negative.

To solve this problem, we need to find the angle between the net force on the plane and the horizontal axis when the plane is at the lowest point of the vertical circle, flying straight down.

### Step 1: Identify forces acting on the plane at the lowest point
At the lowest point in the vertical circle, two forces act on the plane:
- **Gravitational Force (Weight)**, \( F_g \), which points downward (toward the center of the Earth) and has a magnitude:
  \[
  F_g = m \cdot g = 40,000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 392,000 \, \text{N}
  \]
- **Centripetal Force (Inward toward the center of the circle)**, which provides the inward acceleration. This centripetal force is directed upward (toward the center of the circular path) at the lowest point.

The acceleration at this point is the result of both centripetal and tangential acceleration. The net force has components due to both accelerations.

### Step 2: Determine components of acceleration
- **Tangential Acceleration**, \( a_t = 12 \, \text{m/s}^2 \), which points in the direction of the plane's motion (straight down) and is responsible for speeding up the plane.
- **Centripetal Acceleration**, \( a_c \), which points inward (upward at the lowest point in the circle).

The formula for centripetal acceleration is:
\[
a_c = \frac{v^2}{r}
\]
Where:
- \( v = 55 \, \text{m/s} \) (speed at the lowest point)
- \( r = \frac{260}{2} = 130 \, \text{m} \) (radius of the circle)

Now, calculating \( a_c \):
\[
a_c = \frac{55^2}{130} = 23.27 \, \text{m/s}^2
\]

### Step 3: Find the net acceleration
The total acceleration is a combination of tangential and centripetal accelerations. Since the tangential acceleration points downward and the centripetal acceleration points upward, we combine them as vector components.

- The **tangential component** of the acceleration is \( a_t = 12 \, \text{m/s}^2 \) (downward).
- The **centripetal component** of the acceleration is \( a_c = 23.27 \, \text{m/s}^2 \) (upward).

To find the net acceleration, we subtract these components since they are in opposite directions:
\[
a_{\text{net}} = a_c - a_t = 23.27 \, \text{m/s}^2 - 12 \, \text{m/s}^2 = 11.27 \, \text{m/s}^2 \, (\text{upward})
\]

### Step 4: Calculate the angle of the net force
The net force has two components:
- The vertical component is the difference between the centripetal and tangential forces.
- The horizontal component is zero because no forces are acting horizontally at the lowest point.

Thus, the net force is purely vertical, and since the net acceleration is upward (above the horizontal), the angle with the horizontal is:

\[
\theta = 90^\circ \quad \text{(above the horizontal)}
\]

Therefore, the net force makes an angle of **90° above the horizontal**.

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### 5 Relative Questions:
1. How would the angle change if the plane were at the top of the vertical circle?
2. What would happen to the net force if the tangential acceleration increased?
3. How does the centripetal force vary with speed in a circular motion?
4. What is the relationship between net force and acceleration in circular motion?
5. How does gravity affect the motion of the plane in vertical circular motion?

### Tip:
In circular motion, always break forces and accelerations into centripetal (radial) and tangential components for clarity when solving problems.

Discover how to calculate the angle of the net force acting on a 40,000 kg stunt plane flying in a 260-m diameter vertical circle. The solution involves key physics concepts like centripetal acceleration, gravitational force, and Newton's laws. Suitable for high school physics students.

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