Vertical loop dynamics (aviation mechanics): A pilot flies a small aircraft in a vertical loop of radius r. If he experiences weightlessness at the top of the loop, what speed v must the plane have at that top position (take gravitational acceleration = g)?

Introduction / Context:Looping maneuvers are analyzed with circular motion dynamics. At the top of a vertical loop, a pilot feels weightless when the normal reaction from the seat becomes zero. That condition sets a critical relationship between speed, loop radius, and gravitational acceleration.

Given Data / Assumptions:

• Radius of loop = r.
• Acceleration due to gravity = g.
• Point-mass aircraft model at the loop apex; no lift contribution at the instant (worst case for weightlessness criterion).

Concept / Approach:At the top, centripetal acceleration a_c must be provided entirely by gravity when apparent weight (normal reaction N) is zero. Using the circular motion condition a_c = v^2 / r and setting N = 0 leads directly to the required speed for weightlessness.

Step-by-Step Solution:

Verification / Alternative check:Any speed less than √(g r) makes v^2 / r < g, so N must act downward (cannot), meaning the pilot would not stay in contact without restraints. Speeds greater than √(g r) create positive normal forces (apparent weight).

Why Other Options Are Wrong:

• A: Zero speed cannot maintain the loop and ignores centripetal requirements.
• C: g r is acceleration units times length; missing the square root makes dimensions incorrect.
• D: √(2 g r) corresponds to a larger speed than necessary for N = 0.
• E: Infinite speed is physically unnecessary and impossible.

Common Pitfalls:Forgetting to set N = 0 for weightlessness; neglecting direction of centripetal acceleration at the apex; dimensional mistakes such as choosing g r instead of √(g r).

Final Answer:v = √(g r)