Question

A 50.0-kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is 95.0 m/s, what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00$g$? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Short answer

(a) Minimum radius is 230.0 m. (b) Apparent weight is 2452.5 N.

Step-by-step solution

Understand the Forces Involved

At the lowest point of the circular path, the pilot experiences centripetal acceleration, which is directed towards the center of the circle. The forces acting on the pilot are the gravitational force ($F_g=mg$, downward) and the normal force ($F_n$, upward) exerted by the seat. The apparent weight is equal to the normal force, which is what we perceive when we feel heavier or lighter in situations like these. At the lowest point, for the centripetal acceleration not to exceed 4.00g, we can set the acceleration to $a=4.00g=4.00\times 9.81{\text{m/s}}^2=39.24{\text{m/s}}^2$.

Calculate Minimum Radius of the Circle (Part a)

Using the formula for centripetal acceleration ($a_c=\frac{v^2}{r}$), we can solve for the radius $r$. Rearranging this formula gives $r=\frac{v^2}{a_c}$. Substitute the given $v=95.0\text{m/s}$ and $a_c=39.24{\text{m/s}}^2$:$$r=\frac{(95.0\text{m/s}{)}^2}{39.24{\text{m/s}}^2}\approx 230.0\text{m}$$

Calculate Apparent Weight at the Lowest Point (Part b)

The apparent weight is the normal force at the lowest point, which is given by the equation $F_n=m{a}_c+mg$, where $F_n$ is the apparent weight, $m$ is mass, $a_c$ is centripetal acceleration, and $g$ is gravitational acceleration. We have:$$F_n=50.0\text{kg}\times 39.24{\text{m/s}}^2+50.0\text{kg}\times 9.81{\text{m/s}}^2$$$$F_n=1962\text{N}+490.5\text{N}=2452.5\text{N}$$

Key concepts

Apparent Weight

When you're sitting in a plane or on any ride that involves going in a circle, you might feel either heavier or lighter. This feeling is related to what we call "apparent weight." It's important to understand that your actual weight doesn't change. Your real weight is always given by the force of gravity acting on you, which is your mass multiplied by gravitational acceleration, or simply, gravity. In the exercise about the stunt pilot, the apparent weight is the force you feel because of the airplane's acceleration, on top of gravity pulling you downwards. When the plane reaches the bottom of the circle, the pilot experiences this apparent weight the most. Here, the apparent force, or the normal force, is at its greatest because it's both fighting gravity and providing the extra force needed to keep the plane turning in a circle. To calculate this apparent weight, we use the formula:

• $$F_n=m{a}_c+mg$$Where:
  • $F_n$ is the apparent weight or normal force.
  • $m$ is mass (in Kg).
  • $a_c$ is the centripetal acceleration (in m/s²).
  • $g$ is the standard gravity (9.81 m/s²).

The combination of forces makes the pilot feel heavier than when stationary.

Radius of Curvature

The radius of curvature is essential when talking about circular motion, especially with a plane like the stunt pilot's. In simpler terms, it's the radius of the "circle" the plane makes as it moves. If the curve is tighter, the radius is smaller. If the path is more open or broader, then the radius is bigger.A crucial point here is that the radius of curvature affects how much acceleration is needed for the airplane to stay on its circular path. In our scenario, we want the pilot not to experience more than a 4g-force (where g is gravity). A larger radius means smaller centripetal acceleration, and a smaller radius means bigger centripetal acceleration.To find the definite radius where the acceleration does not exceed a certain level, you use the formula:

• $$r=\frac{v^2}{a_c}$$
• Where:
  • $r$ is the radius of curvature.
  • $v$ is the speed (in m/s).
  • $a_c$ is the maximum centripetal acceleration allowed (in m/s²).

This formula shows us how the radius, speed, and acceleration join together, ensuring the pilot's safety while performing breathtaking stunts.

Centripetal Acceleration

Centripetal acceleration is what makes the stunt pilot's plane stay in its circular path. It's a special kind of acceleration that's always directed toward the center of the circle. It's like an invisible leash, pulling the object into a continuous loop rather than flying straight out into the air.As the speed of the plane increases or as the radius of the circular path decreases, the centripetal acceleration goes up. This means the forces acting on the plane and the pilot increase. That's why keeping centripetal acceleration within safe limits is critical, especially in high-speed and high-risk maneuvers like those performed by stunt pilots.The formula we use to calculate centripetal acceleration is:

• $$a_c=\frac{v^2}{r}$$Where:
  • $a_c$ is the centripetal acceleration.
  • $v$ is the velocity (in m/s).
  • $r$ is the radius (in m).

This gives a clear picture of how velocity and radius are crucial for determining how much centripetal force is needed. Always remember, the greater the speed and the tighter the radius, the stronger the centripetal acceleration.