Problem 52 - University Physics with Modern Physics

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 \(95.0 \mathrm{m} / \mathrm{s},\) what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00\(g ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Step-by-Step Solution

Answer

Minimum radius is \(230.2 \, \text{m}\); apparent weight \(= 2452.5 \, \text{N}\).

Step 1: Understand the problem

We are given a scenario where a stunt pilot pulls out of a dive in a vertical circular path. The speed of the plane at the lowest point is \(95.0 \, \text{m/s}\) and the maximum allowed acceleration is \(4.00 \, g\). We need to find the minimum radius of the circle and the apparent weight of the pilot at the lowest point.

Step 2: Identify given values

We have the following given values:- Mass of the pilot \(m = 50.0 \, \text{kg}\)- Speed at the lowest point \(v = 95.0 \, \text{m/s}\)- Maximum acceleration \(a = 4.00 \, g = 4.00 \, \times \, 9.81 \, \text{m/s}^2 = 39.24 \, \text{m/s}^2\).

Step 3: Apply centripetal acceleration formula

The centripetal acceleration \(a_c\) at the lowest point of the circular path is given by \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius of the circle. We set \(a_c\) equal to the maximum allowed acceleration: \[\frac{v^2}{r} = 39.24 \]

Step 4: Solve for the radius

Rearrange the formula to solve for \(r\):\[r = \frac{v^2}{39.24}\]Substituting the given speed:\[r = \frac{(95.0)^2}{39.24} \approx 230.2 \, \text{m}\]

Step 5: Calculate total force at the lowest point

At the lowest point, the net force is the centripetal force, which is the sum of gravitational force \(mg\) and the normal force \(N\):\[m \cdot \left( \frac{v^2}{r} + g \right) = N\]

Step 6: Solve for the apparent weight

The apparent weight is equal to the normal force \(N\). Using \(r = 230.2 \, \text{m}\) in the centripetal acceleration formula:\[N = m \cdot (39.24 + 9.81) = 50.0 \cdot 49.05 = 2452.5 \, \text{N}\]

Key Concepts

Centripetal Acceleration

In the realm of circular motion, centripetal acceleration plays a vital role. It is the acceleration that keeps an object moving in a circle, constantly changing the direction of the velocity. This acceleration points towards the center of the circle, hence the name "centripetal," which means center-seeking.
For a stunt pilot performing a dive, centripetal acceleration ensures her path curves smoothly instead of continuing in a straight line. Its magnitude depends on both the speed of the object and the radius of the circle it follows. The formula to calculate centripetal acceleration is given by:

• \(a_c = \frac{v^2}{r} \)

where \(v\) is the speed and \(r\) is the radius of the circle.
In this pilot scenario, the challenge is to ensure that the acceleration doesn't exceed a safe threshold—in this case, 4 times the gravitational acceleration \(g\). Balancing speed and radius using the centripetal acceleration formula is crucial to designing safe and thrilling circular maneuvers.

Apparent Weight

When moving in a circle, such as in a vertical loop, the forces a person experiences change due to centripetal force. At the lowest point in a circle, the pilot feels heavier than usual—a sensation known as "apparent weight."
Apparent weight differs from actual weight, which is the gravitational force pulling down. It is the force perceived due to acceleration, in this case, the sum of gravitational force and the centripetal force. The apparent weight is calculated using:

• \( N = m \left( g + \frac{v^2}{r} \right) \)

where \( N \) is the normal force or apparent weight, \( m \) is mass, \( g \) is gravitational acceleration, and \( \frac{v^2}{r} \) is the centripetal acceleration.
In the stunt pilot scenario, calculating apparent weight helps determine the load the pilot endures, ensuring it's within safe limits.

Radius of a Circle

The radius of a circle is fundamental to understanding circular motion as it directly influences the centripetal acceleration an object experiences. In this context, the radius is the distance from the center of the circular path to any point on the edge.
The challenge in physics problems like the stunt pilot scenario is to determine the minimum radius for safe maneuvering. By rearranging the centripetal acceleration formula, the radius can be solved as:

• \(r = \frac{v^2}{a_c}\)

where \(r\) is the radius, \(v\) is the speed, and \(a_c\) is the centripetal acceleration.
This formula emphasizes the inverse relationship between radius and acceleration: a larger radius allows for higher speeds at the same acceleration, crucial for ensuring safety in dynamic activities like piloting.

Physics Problem Solving

Physics problems often seem daunting but breaking them down step-by-step can simplify even the toughest puzzles. The stunt pilot problem is a perfect example of applying systematic approaches to solve for unknowns.
Start by understanding the scenario and identifying what's given. Here, we know the mass, speed, and maximum acceleration allowed. Next, use relevant formulas to connect the dots. For circular motion, the key formula involves centripetal acceleration.

• Identify \( a_c = \frac{v^2}{r} \), rearrange it to find unknowns like radius.
• Use related concepts—such as apparent weight or normal force—to explore effects like forces experienced at specific points.
• Always keep units consistent and double-check calculations for accuracy.

By following these steps, physics problems become less about complexity and more about logical thinking, enabling you to navigate from knowns to unknowns with confidence.