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.

To solve this problem, we will calculate the angle the net force makes with the horizontal when the plane is at the lowest point in the loop-the-loop, flying straight down.

### Given:
- Mass of the stunt plane, \( m = 40,000 \, \text{kg} \)
- Diameter of the vertical circle, \( d = 260 \, \text{m} \) 
  (so the radius \( r = \frac{260}{2} = 130 \, \text{m} \))
- Speed of the plane at the lowest point, \( v = 55 \, \text{m/s} \)
- Tangential acceleration (speeding up), \( a_{\text{tangential}} = 12 \, \text{m/s}^2 \)

### Step 1: Calculate the centripetal acceleration
Centripetal acceleration is given by the formula:
\[
a_{\text{centripetal}} = \frac{v^2}{r}
\]
Substitute the known values:
\[
a_{\text{centripetal}} = \frac{55^2}{130} = \frac{3025}{130} \approx 23.27 \, \text{m/s}^2
\]

### Step 2: Net force components
The net force acting on the plane is the combination of two components: 
1. **Centripetal force** (directed towards the center of the circle, which is vertically upwards when the plane is at the lowest point).
2. **Tangential force** (due to tangential acceleration, directed along the plane’s path, which is horizontally to the right at this point since the plane is moving downward).

- **Centripetal force**:
\[
F_{\text{centripetal}} = m \times a_{\text{centripetal}} = 40,000 \times 23.27 = 930,800 \, \text{N}
\]
This force is directed vertically upward.

- **Tangential force**:
\[
F_{\text{tangential}} = m \times a_{\text{tangential}} = 40,000 \times 12 = 480,000 \, \text{N}
\]
This force is directed horizontally to the right.

### Step 3: Find the angle of the net force
To find the angle that the net force makes with the horizontal, we can use the tangent of the angle \( \theta \), which is the ratio of the centripetal force to the tangential force:
\[
\tan \theta = \frac{F_{\text{centripetal}}}{F_{\text{tangential}}}
\]
Substitute the values:
\[
\tan \theta = \frac{930,800}{480,000} \approx 1.9392
\]
Now, take the arctangent to find the angle:
\[
\theta = \tan^{-1}(1.9392) \approx 62.6^\circ
\]

Since the centripetal force is vertical and upwards, and the tangential force is horizontal, the net force is above the horizontal.

### Final Answer:
The angle the net force makes with the horizontal is approximately **+62.6°** (above the horizontal).

Would you like any further details or explanations?

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Here are some related questions:
1. How does centripetal force change as the plane speeds up or slows down in the loop?
2. What is the effect of increasing the radius of the loop on the centripetal acceleration?
3. How would the angle change if the plane were decelerating instead of accelerating tangentially?
4. What would happen if the tangential acceleration were zero—how would the forces balance?
5. How does the weight of the plane contribute to the forces involved at different points in the loop?

**Tip:** Centripetal force always points towards the center of the circular path, while tangential acceleration affects speed but does not contribute to the centripetal force.

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.

Learn how to calculate the angle of the net force in a stunt plane performing a loop-the-loop with a speed of 55 m/s and a 260-meter-diameter vertical circle. The plane experiences both centripetal and tangential acceleration, which combine to form the net force at a 62.6-degree angle above the horizontal. Suitable for high school physics students in Grades 11-12.

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