GATE AE 2020 • Q25 NAT • 1 Mark

Flight Mechanics & Space Dynamics

Maneuvers & Load Factors

Subtopic: Vertical Loop Maneuver & Structural Load

Question Statement:

An airplane of mass 5000 kg is flying at a constant speed of 360 km/h at the bottom of a vertical circular loop of radius 400 m. Taking $g = 9.8$ m/s$^2$, the lift force (in kN) generated by the wings at the bottom of the loop is _________ (round off to the nearest integer).

Options:

  • (A) The governing parameter increases monotonically with Space Dynamics & Orbital Mechanics
  • (B) The governing parameter remains invariant across the boundary
  • (C) The parameter satisfies the conservation equation for Space Dynamics & Orbital Mechanics
  • (D) The parameter depends exclusively on viscous dissipation

Official Answer Key:

174.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2020 Q.25)** **1. Intuitive Physical Setup & Given Parameters**: - Airplane mass: $m = 5000\text{ kg}$ - Speed: $V = 360\text{ km/h} = \frac{360 \times 1000}{3600} = 100\text{ m/s}$ - Loop radius: $R = 400\text{ m}$ - Gravitational acceleration: $g = 9.8\text{ m/s}^2$ **2. Step-by-Step Flight Dynamics Derivation**: 1. **Equations of Motion at Bottom of Vertical Loop**: At the lowest point of the circular vertical loop, lift $L$ acts vertically upwards, and weight $W = mg$ acts vertically downwards. Centripetal acceleration $\frac{V^2}{R}$ is directed upwards toward the center of the loop: $L - mg = m \frac{V^2}{R}$ $L = mg + m \frac{V^2}{R} = m \left( g + \frac{V^2}{R} \right)$ 2. **Substituting Numerical Values**: $\frac{V^2}{R} = \frac{100^2}{400} = \frac{10000}{400} = 25\text{ m/s}^2$ $g + \frac{V^2}{R} = 9.8 + 25 = 34.8\text{ m/s}^2$ $L = 5000\text{ kg} \times 34.8\text{ m/s}^2 = 174000\text{ N} = 174\text{ kN}$ The lift generated at the bottom of the loop is exactly **174 kN** (load factor $n = 34.8/9.8 \approx 3.55$). **3. Comprehensive Options & Distractor Analysis**: - **Question Type**: Numerical Answer Type (NAT). - **Exact Analytical Evaluation**: **174.0 kN** (Official Range: [173, 175]). **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **174.0** - **Key Formula**: $L = m\left(g + \frac{V^2}{R}\right)$ at the bottom of a vertical loop.

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