Problem 12 - College Physics

Question

A \(50.0 \mathrm{~kg}\) 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 \(9.50 \mathrm{~m} / \mathrm{s}\) what should the minimum radius of the circle be in order for the cen- tripetal acceleration at this point not to exceed \(4.00 \mathrm{~g} ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Step-by-Step Solution

Answer

Minimum radius: \(2.30 \, \text{m}\); apparent weight: \(2467 \, \text{N}\).

Step 1: Understand the Problem

We have a stunt pilot who pulls out of a dive by moving in a circular path in a vertical plane. We need to find the minimum radius of this circular path such that the centripetal acceleration does not exceed \(4.00 \; g\), where \(g = 9.81 \; \text{m/s}^2\). Additionally, we need to find the apparent weight of the pilot at the lowest point of the pullout, where the speed is \(9.50 \; \text{m/s}\).

Step 2: Convert Maximum Centripetal Acceleration

The maximum allowable centripetal acceleration is given as \(4.00 \, g\). Convert this to meters per second squared: \[ a_c = 4.00 \, g = 4.00 \times 9.81 \, \text{m/s}^2 = 39.24 \, \text{m/s}^2 \]

Step 3: Apply Centripetal Acceleration Formula

Using the formula for centripetal acceleration \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius, set \(a_c = 39.24 \; \text{m/s}^2\) and solve for \(r\):\[ 39.24 = \frac{(9.50)^2}{r} \]Rearrange to find \(r\):\[ r = \frac{(9.50)^2}{39.24} \approx 2.30 \, \text{m} \]

Step 4: Find Net Force at Lowest Point

At the lowest point, the centripetal force is provided by the net force, which is the difference between gravitational force \( (mg) \) and the normal force \( (N) \):\[ N - mg = \, m \cdot \frac{v^2}{r} \]

Step 5: Calculate Apparent Weight

The apparent weight of the pilot is the normal force \(N\) which can be expressed as:\[ N = m \cdot \frac{v^2}{r} + mg \]Substituting the given values \((m = 50.0 \, \text{kg}, \; v = 9.50 \, \text{m/s}, \; r = 2.30 \, \text{m})\):\[ N = 50.0 \cdot \frac{(9.50)^2}{2.30} + 50.0 \cdot 9.81 \]Calculate \(N\):\[ N \approx 1976.74 + 490.5 \approx 2467.24 \, \text{N} \]

Step 6: Conclusion

The minimum radius of the circle is approximately \(2.30 \, \text{m}\) and the apparent weight of the pilot at the lowest point is approximately \(2467 \, \text{N}\).

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a crucial concept when understanding motion in a circular path. This term refers to the acceleration that keeps an object moving along the circular path rather than in a straight line. It's constantly perpendicular to the object's velocity and points toward the center of the circle.
The formula for centripetal acceleration is given by:

• \(a_c = \frac{v^2}{r}\)

Here, \(v\) represents the velocity of the object, and \(r\) is the radius of the circle. Hence, for a given velocity, a smaller radius results in larger centripetal acceleration.
In practical terms, centripetal acceleration needs to be limited within safe thresholds to prevent excessive stress on physical structures or, as in this problem, the pilot's body. The problem exemplifies a real-life scenario where excessive centripetal acceleration could exert dangerous G-forces on a pilot during a stunt.

Apparent Weight

Apparent weight is an interesting concept, especially when discussing objects in motion. It refers to the force exerted by an object, perceived as 'weight,' in non-standard conditions, like being in a moving elevator or during circular motion.In circular motion, the apparent weight can be affected by the direction of acceleration relative to gravity. At the bottom of the circular path, additional force needs to counteract gravity and provide the necessary centripetal force.
In this exercise, the apparent weight corresponds to the normal force acting on the stunt pilot as she transitions through the lowest point of the circle. It's calculated as:

• \(N = m \cdot \frac{v^2}{r} + mg\)

Here, \(m\) is mass, \(v\) the velocity, \(r\) the radius of the circle, and \(g\) the gravitational force. This apparent weight is significantly higher than the regular weight of the pilot due to the added centripetal force required to sustain the circular motion.

Circular Motion

Circular motion is a foundational topic in physics that describes the movement of an object along the circumference of a circle. This is not just limited to objects moving in perfect circles but can also include any object following a curved path.
Key aspects include:

• Uniform Circular Motion: The object's speed remains constant, but its velocity changes due to direction change.
• Non-Uniform Circular Motion: Both speed and direction change; involves tangential acceleration in addition to centripetal.

In circular motion, several forces come into play, most notably the centripetal force that points toward the center of the circle, constantly changing the direction of the object's velocity without altering its speed. This concept is pivotal in understanding how forces act on an object in rotational paths, ensuring the object maintains its circular trajectory.

Physics Problem Solving

Physics problem-solving involves logical steps to analyze and solve a problem, allowing us to apply concepts and formulas effectively. Engaging with physics problems requires breaking down each task into manageable steps.
Here’s how it applies to our exercise:

• Identifying what's given and what's asked: Distinguishing known quantities like speed and weight from unknowns like radius and apparent weight.
• Utilizing appropriate formulas: Knowing the formula for centripetal acceleration and rearranging it to solve for the desired variable.
• Accounting for units: Ensuring unit consistency, especially when working with gravitational constants and acceleration values.
• Logical sequencing: Tackle problems in a step-by-step manner, ensuring thorough understanding of each component before moving to the next.

By developing these problem-solving skills, students can better tackle complex physics challenges, making the learning process both effective and rewarding.