Flying in a Loop: Plane Speed & Pilot Weight

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SUMMARY

The discussion focuses on the dynamics of an airplane flying in a vertical loop with a radius of 190 meters. At the top of the loop, the pilot experiences weightlessness, which requires calculating the speed of the airplane at that point. The speed at the bottom of the loop is given as 300 km/h, leading to the need to determine the apparent weight of the pilot, whose true weight is 700 N. The forum participants emphasize the importance of correctly applying the concepts of centripetal acceleration and the forces acting on the pilot during the loop.

PREREQUISITES
  • Centripetal acceleration and its formula: accel = v^2/r
  • Understanding of forces: gravitational force and normal force
  • Basic physics of circular motion
  • Knowledge of converting units, specifically km/h to m/s
NEXT STEPS
  • Calculate the speed of the airplane at the top of the loop using centripetal acceleration principles.
  • Learn how to analyze forces in circular motion, focusing on weightlessness scenarios.
  • Study the conversion of speed units from km/h to m/s for accurate calculations.
  • Explore the implications of varying speeds in vertical loops on pilot experience and aircraft design.
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in the mechanics of flight and circular motion dynamics.

PhysicStorm
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An airplane flies in a loop (a circular path in a vertical plane) of radius 190 m. The pilot's head always points toward the center of the loop. The speed of the airplane is not constant. The airplane goes slowest at the top of the loop and fastest at the bottom.


(a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?

(b) At the bottom of the loop, the speed of the airplane is 300 km/h. What is the apparent weight of the pilot at this point? His true weight is 700 N.


I have tried this problem multiple times and i used the formula of
accel = v^2/r
i know that accel is also equal to net force of the pilot divided by his mass so and that at the top of the loop it is equal to force of gravity down and force of normal force from seat

the masses do cancel out using
(9.8(m)+Fn)/m = v^2/190
i might be solving for normal force wrong can someone clear it up for me please?
w/o part a correct part b is always going to be incorrect. ty for your time and effort
 
Physics news on Phys.org
Why do you include the normal force coming from the chair? In normal flight this will just cancelthe gravitational force, but at the top, if the pilot is weightless there will be no force coming from the seat.
 

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