A(n) 40,000 kg stunt plane performs a loop-the-loop, flying 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.

We need to find the **angle between the net force and the horizontal** when the plane is flying straight down in a loop-the-loop. 

Here’s a step-by-step breakdown to solve this:

---

### 1. Known Values:
- Mass of the plane: \( m = 40,000 \, \text{kg} \)
- Diameter of the circle: \( d = 260 \, \text{m} \), so the **radius** \( r = \frac{260}{2} = 130 \, \text{m} \)
- Speed at the point flying straight down: \( v = 55 \, \text{m/s} \)
- Tangential acceleration: \( a_{\text{t}} = 12 \, \text{m/s}^2 \)

---

### 2. Forces Acting on the Plane:
- **Centripetal Force**: Points toward the center of the circle (upward in this case). The magnitude of the centripetal force is found from:

  \[
  F_{\text{c}} = \frac{m v^2}{r}
  \]

  Substituting values:

  \[
  F_{\text{c}} = \frac{40,000 \times (55)^2}{130}
  = \frac{40,000 \times 3025}{130}
  = 930,769.2 \, \text{N}
  \]

- **Tangential Force**: Causes the plane to speed up. This force points **downward** (in the same direction as the velocity). Its magnitude is:

  \[
  F_{\text{t}} = m a_{\text{t}} = 40,000 \times 12 = 480,000 \, \text{N}
  \]

---

### 3. Net Force and the Angle with Horizontal:
The **net force** is the vector sum of the centripetal and tangential forces. Since the centripetal force points **upward** (toward the center of the loop) and the tangential force points **downward** along the direction of motion, we can set up the components:

- **Vertical component of the net force**:  
  \[
  F_y = F_{\text{t}} - F_{\text{c}} = 480,000 - 930,769.2 = -450,769.2 \, \text{N}
  \]

  (A negative value indicates that the net force points downward overall in the vertical direction.)

- **Horizontal component of the net force**:  
  There is no horizontal force in this problem (since all the forces act vertically).  
  \[
  F_x = 0
  \]

---

### 4. Calculating the Angle with the Horizontal:
Since the net force only has a vertical component, the angle \( \theta \) with the horizontal (measured above or below the horizontal) is:

\[
\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{-450,769.2}{0}\right)
\]

Since the denominator is zero, the angle must be either \( \theta = -90^\circ \) or \( \theta = 90^\circ \). Because the net force points **downward**, the angle is:

\[
\theta = -90^\circ
\]

---

### 5. Final Answer:
The angle the net force makes with the horizontal is:

\[
\boxed{-90^\circ}
\]

---

Would you like me to explore how this net force changes at other points in the loop? Or need help understanding the steps?  

Here are 5 follow-up questions for further exploration:  
1. How would the problem change if the plane’s tangential acceleration were negative (slowing down)?  
2. What would happen if the plane had a greater radius for its loop?  
3. How do centripetal force and tangential force combine at other points in the loop?  
4. How would air resistance affect this problem?  
5. What is the maximum force experienced by the plane throughout the loop?  

**Tip:** In circular motion, centripetal force always points toward the center, regardless of the plane's speed or acceleration direction.

A 40,000 kg stunt plane performs a loop-the-loop, flying 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.

Explore how to calculate the angle of the net force acting on a 40,000 kg stunt plane performing a loop-the-loop in a 260-m vertical circle. With a speed of 55 m/s and an acceleration of 12 m/s², this problem involves circular motion, centripetal force, and vector addition. Suitable for Grades 11-12 physics students.

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