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 accelerating at a rate of 12 m/s squared. 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 for the angle that the net force makes with the horizontal, we need to analyze the forces acting on the stunt plane at the point where it is flying straight down. We'll use the following information:

- Mass of the stunt plane: \( m = 40,000 \, \text{kg} \)
- Speed of the stunt plane at the bottom of the loop: \( v = 55 \, \text{m/s} \)
- Acceleration of the plane: \( a = 12 \, \text{m/s}^2 \)
- Diameter of the circle: \( D = 260 \, \text{m} \), so the radius is \( r = 130 \, \text{m} \).

### Step 1: Break down the forces

At the point where the plane is flying straight down, two main forces act on it:
- **The gravitational force** \( F_g \), acting downward.
- **The net force** \( F_{\text{net}} \), responsible for the centripetal acceleration and any tangential forces.

#### Gravitational force:
The gravitational force is calculated as:
\[
F_g = m \cdot g = 40,000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 392,000 \, \text{N}
\]

This force points vertically downward.

### Step 2: Centripetal force

The centripetal force is responsible for keeping the plane in a circular path and is directed towards the center of the loop. This force is calculated as:
\[
F_{\text{centripetal}} = m \cdot \frac{v^2}{r} = 40,000 \, \text{kg} \cdot \frac{(55 \, \text{m/s})^2}{130 \, \text{m}} = 933,846.15 \, \text{N}
\]

This force points horizontally to the left when the plane is at the bottom of the loop.

### Step 3: Tangential force

The tangential force comes from the plane's acceleration, which is given as \( 12 \, \text{m/s}^2 \). This force is calculated as:
\[
F_{\text{tangential}} = m \cdot a = 40,000 \, \text{kg} \cdot 12 \, \text{m/s}^2 = 480,000 \, \text{N}
\]

Since the plane is flying straight down, this tangential force acts downward.

### Step 4: Calculate the net force components

Now we can break the net force into its vertical and horizontal components:

1. **Vertical component of the net force** \( F_{\text{net, vertical}} \):
\[
F_{\text{net, vertical}} = F_g + F_{\text{tangential}} = 392,000 \, \text{N} + 480,000 \, \text{N} = 872,000 \, \text{N}
\]
This points downward.

2. **Horizontal component of the net force** \( F_{\text{net, horizontal}} \):
\[
F_{\text{net, horizontal}} = F_{\text{centripetal}} = 933,846.15 \, \text{N}
\]
This points to the left (negative horizontal direction).

### Step 5: Find the angle of the net force with the horizontal

The angle \( \theta \) of the net force with respect to the horizontal can be found using the tangent of the angle:
\[
\tan(\theta) = \frac{F_{\text{net, vertical}}}{F_{\text{net, horizontal}}} = \frac{872,000}{933,846.15}
\]
\[
\theta = \tan^{-1}\left( \frac{872,000}{933,846.15} \right) \approx \tan^{-1}(0.9339)
\]
\[
\theta \approx 43.4^\circ
\]

Since this angle is measured below the horizontal, we consider it negative. Thus, the angle is:
\[
\theta = -43.4^\circ
\]

### Conclusion:
The net force makes an angle of **\( -43.4^\circ \)** with the horizontal, meaning it is directed 43.4 degrees below the horizontal.

---

Do you need further explanation on any step, or would you like to see a breakdown of specific forces involved?

Here are five related questions you might find useful:
1. How does the net force change at different points in the loop?
2. How would the angle change if the speed at the bottom were different?
3. What is the role of centripetal force in circular motion?
4. How does gravitational force affect objects in circular motion?
5. What is the difference between tangential and centripetal acceleration?

**Tip**: In circular motion, the net force often results from a combination of centripetal (radial) and tangential forces, which affect speed and 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 accelerating at a rate of 12 m/s². 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 the physics behind a 40,000 kg stunt plane performing a loop-the-loop in a 260-m diameter vertical circle. Learn how to calculate the angle of the net force with respect to the horizontal when the plane is flying straight down at 55 m/s, using principles of centripetal force, tangential force, and Newton's laws. Suitable for high school students in Grades 11-12.

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