Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A \(80.0-\mathrm{kg}\) pilot in an aircraft moving at a constant speed of \(500 . \mathrm{m} / \mathrm{s}\) pulls out of a vertical dive along an arc of a circle of radius \(4000 . \mathrm{m}\).

Short Answer

Expert verified
Question: Calculate the centripetal force acting on a pilot with a mass of 80 kg when his aircraft is moving at a speed of 500 m/s and following a circular path with a radius of 4000 m as he pulls out of a dive. Answer: The centripetal force acting on the pilot when pulling out of the dive is 5000 N.

Step by step solution

01

Identify the given values

We are given the following information from the exercise: 1. Mass of the pilot (m) = 80 kg 2. Speed of the aircraft (v) = 500 m/s 3. Radius of the circular path (r) = 4000 m
02

Apply the centripetal force formula

We need to apply the formula for centripetal force, which is \(Fc = m*(v^2/r)\), in order to calculate the centripetal force acting on the pilot during the circular motion. We have all the required values, so we can plug them into the formula: \(Fc = 80 * (500^2/4000)\)
03

Calculate the centripetal force

Now we perform the calculations to find the centripetal force: \(Fc = 80 * (250000/4000)\) \(Fc = 80 * 62.5\) \(Fc = 5000 \mathrm{N}\) The centripetal force acting on the pilot when pulling out of the dive is 5000 N.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Circular Motion
Uniform Circular Motion involves movement along a circular path at a constant speed. Though the speed remains constant, the direction of the object's velocity is continuously changing. **This change in direction** implies that the object is constantly accelerating, even if its speed does not change. The acceleration in question is called centripetal acceleration, and it is always directed towards the center of the circle.

Some everyday examples include the motion of planets around the sun, a car navigating a roundabout, and a pilot making a loop in the air. **The concept of uniform circular motion** helps in understanding not only everyday phenomena but also complex systems in physics and engineering. The principles involved, such as the need for a net inward force to maintain circular motion, are vital for designing roller coasters, spinning fairground rides, and even satellite orbits.

In the scenario given, an aircraft maneuvering through a vertical loop at a consistent speed represents a typical uniform circular motion.
Centripetal Acceleration
Centripetal Acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle itself. This central-seeking nature ensures the object continues in a curved path rather than shooting off in a straight line, which would be its natural motion due to inertia.

Mathematically, centripetal acceleration is expressed as \( a_c = \frac{v^2}{r} \), where \( v \) is the linear speed, and \( r \) is the radius of the circle. **This formula illustrates** how acceleration increases with a higher speed or a smaller radius, indicating tighter curves require greater acceleration.

In our context, the pilot's aircraft is experiencing centripetal acceleration to sustain its circular path out of a dive. The level of centripetal acceleration impacts both the forces exerted on the pilot and the structural demands on the aircraft.
  • Higher speeds necessitate greater centripetal acceleration for the same radius, demanding more from the pilot and craft.
  • Larger radii at constant speed reduce acceleration needs.
Understanding these dynamics is key in aviation and automotive industries, ensuring safety and performance.
Newton's Second Law
Newton's Second Law provides the framework for understanding how forces affect motion. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, expressed as \( F = ma \). This principle straightforwardly links to the centripetal force necessary for uniform circular motion.

When an object like an aircraft is moving in a circular path, the force maintaining this motion—called the centripetal force—is calculated using Newton's Second Law adapted for circular motion: \( F_c = m\cdot\frac{v^2}{r} \). This equation shows how various factors like the object's mass, speed, and the radius of the circle affect the centripetal force needed.

In the example of the pilot and aircraft, applying Newton's Second Law helped us determine the centripetal force was \( 5000 \text{ N} \). This force is crucial for safety and control during the maneuver. By relating forces to the aircraft's mass and the pilot's, Newton's insights help ensure the aircraft can withstand the stresses experienced during flight maneuvers, balancing speed, and path curvature to avoid excessive strain or failure.
  • Adjusting speed alters the force requirements drastically, important for flight dynamics.
  • Understanding mass and force links helps engineers design systems from planes to amusement parks safely.
Newton's framework aids not only in calculating these forces but also in predicting and controlling them in real-world applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A CD starts from rest and speeds up to the operating angular frequency of the CD player. Compare the angular velocity and acceleration of a point on the edge of the CD to those of a point halfway between the center and the edge of the CD. Do the same for the linear velocity and acceleration.

Is it possible to swing a mass attached to a string in a perfectly horizontal circle (with the mass and the string parallel to the ground)?

Suppose you are riding on a roller coaster, which moves through a vertical circular loop. Show that your apparent weight at the bottom of the loop is six times your weight when you experience weightlessness at the top, independent of the size of the loop. Assume that friction is negligible.

Two masses hang from two strings of equal length that are attached to the ceiling of a car. One mass is over the driver's seat; the other is over the passenger's seat. As the car makes a sharp turn, both masses swing away from the center of the turn. In their resulting positions, will they be farther apart, closer together, or the same distance apart as they were when the car wasn't turning?

A car of weight \(W=\) \(10.0 \mathrm{kN}\) makes a turn on a track that is banked at an angle of \(\theta=20.0^{\circ} .\) Inside the car, hanging from a short string tied to the rear-view mirror, is an ornament. As the car turns, the ornament swings out at an angle of \(\varphi=30.0^{\circ}\) measured from the vertical inside the car. What is the force of static friction between the car and the road?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free