An aircraft initiates a symmetric pull-up maneuver from steady level flight at a true airspeed $V = 200\text{ m/s}$. If the pilot pulls a normal load factor $n = 5.0$ at the bottom of the pull-up, taking $g = 9.81\text{ m/s}^2$, the instantaneous radius of curvature $R$ of the vertical flight path (in meters) is closest to:

(A) 1,019 m

(B) 815 m

(C) 2,039 m

(D) 408 m

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles At the lowest point (bottom) of a vertical pull-up maneuver, the aircraft lift $L$ must support the weight $W$ and provide the vertical centripetal acceleration: $$L - W = \frac{W}{g} \frac{V^2}{R}$$ Dividing by weight $W$: $$n - 1 = \frac{V^2}{g R} \implies R = \frac{V^2}{g(n - 1)}$$ where $n = L/W$ is the normal load factor. ### 2. Analytical Formulation & Calculation Given: - $V = 200\text{ m/s} \implies V^2 = 40{,}000\text{ m}^2/\text{s}^2$ - $n = 5.0 \implies n - 1 = 4.0$ - $g = 9.81\text{ m/s}^2$ ### 3. Step-by-Step Calculation $$R = \frac{40{,}000}{9.81 \times 4.0} = \frac{40{,}000}{39.24} \approx 1{,}019.37\text{ m} \approx 1{,}019\text{ m}$$ ### 4. Distractor Forensics - **Option A (1,019 m):** Correct pull-up radius using $(n - 1)$. - **Option B (815 m):** Divided by $n = 5.0$ instead of $(n - 1) = 4.0$, forgetting that weight already acts opposite to centripetal acceleration at the bottom ($40000 / (9.81 \times 5) = 815.5\text{ m}$). - **Option C (2,039 m):** Used $(n + 1) = 6.0$ or omitted factor of 2. ### 5. Exam Trap & Standard Reference - **Trap Alert:** In a vertical pull-up at the bottom, centripetal load factor is $(n - 1)$, NOT $n$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6.

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