Rocket Propulsion Questions and Answers – Flight Stability

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This set of Rocket Propulsion Multiple Choice Questions & Answers (MCQs) focuses on “Flight Stability”.

1. Stability of a flying vehicle cannot be ensured by __________
a) control surfaces
b) reaction control system
c) anti-sloshing setup
d) hinged multiple rocket nozzles
View Answer

Answer: c
Explanation: Sloshing of liquid propellants in the tank can be prevented by anti-sloshing setup. But when the vehicle moves into an unstable state, it cannot control the vehicle and help it return back to the stable regime.
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2. Suppose an aircraft trimmed at a 10° angle of attack faces a gust that changes the angle of attack to 15°. What is the nature of the moment required to bring it back to the trimmed condition?
a) Positive pitching moment
b) Positive yawing moment
c) Negative pitching moment
d) Negative yawing moment
View Answer

Answer: c
Explanation: Negative pitching moment is a nose-down pitching moment which will act as the restoring moment in this situation. Yawing moment happens about the vehicle’s z-axis.

3. Which of the following statements is true for the center of gravity (cog) and the center of pressure (cop) location along the axis of a fuel powered aircraft?
a) cog and cop both changes
b) cog remains fixed but cop changes
c) cog and cop both remain fixed
d) cog changes but cop remains fixed
View Answer

Answer: a
Explanation: Center of gravity changes due to fuel consumption of the aircraft. Since aerodynamics moments vary with Mach number, center of pressure location is also not fixed.

4. Which of the following can be spin-stabilized?
a) Unguided rocket projectiles
b) Commercial aircrafts
c) Artificial satellites
d) Bullet trains
View Answer

Answer: a
Explanation: Unguided rocket projectiles can be given a rotation about its longitudinal axis by means of inclined fins or smaller rocket nozzles to stabilize it. This is an example of spin stabilization.

5. Determine the period of revolution of an artificial satellite in a circular orbit of altitude 350 km.
a) 1.738 hrs
b) 1.266 hrs
c) 1.395 hrs
d) 1.525 hrs
View Answer

Answer: d
Explanation: T=2π(R0+h)3/2/√μ
= 2×3.14x(6728000)3/2/(3.986×1014)
= 1.525 hrs.
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6. For a satellite revolving around the earth in a circular orbit at an altitude of 350 km, determine the orbital velocity.
a) 7691.3 m/s
b) 7606.9 m/s
c) 7532.3 m/s
d) 7797.2 m/s
View Answer

Answer: a
Explanation: V = \(\sqrt{(\mu/R)}\), where R = Ro + h.
V = \(\sqrt{(3.98 * 10^{14}/6728000)}\)
= 7691.3 m/s.

7. For some type of flying vehicle applications, a low-speed roll is given to avoid thrust vector deviations.
a) True
b) False
View Answer

Answer: a
Explanation: In some cases, a low-speed roll is applied on the rocket to minimize or cancel out the thrust vector deviations due to aerodynamic shape misalignment or due to some other reason. This roll is different from the roll used for spin stability.

8. For an airplane undergoing a steady pull-up maneuver along a curved flight path of the constant angle of attack, constant pitch rate of 20°/s, and having a radius of curvature of 150 m, determine the flight velocity.
a) 47.66 m/s
b) 52.36 m/s
c) 38.33 m/s
d) 69.42 m/s
View Answer

Answer: b
Explanation: Pitch rate q = V/R
Then, V = qR = 20 x π/180 x 150 = 52.36 m/s.

9. The point on the rocket where all the aerodynamic forces balance is called as the ______________
a) center of pressure
b) neutral point
c) maneuvering point
d) center of gravity
View Answer

Answer: a
Explanation: Center of pressure is the point where all the aerodynamic forces balance. Lift and drag are the two aerodynamic forces acting on the rocket.

10. Determine the normal force coefficient of a rocket having a maximum frontal area of 0.2 m2, flying at sea level with a velocity of 100 m/s. Consider the center of gravity to be ahead of the center of pressure by 0.5 m and the corrective moment coefficient to be 2.5.
a) 0.016
b) 0.012
c) 0.008
d) 0.004
View Answer

Answer: d
Explanation: C1 = 1/2 ρ V2 Ar Cna [Z – W], where C1 is the corrective moment coefficient, ρ is the ambient density, V is the velocity, Cna is the normal force coefficient, Z-W is the location of center of gravity ahead of center of pressure, and Ar is the reference area.
Then Cna = 2.5/(0.5×1.225×1002x0.2×0.5) = 0.004.
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Sanfoundry Global Education & Learning Series – Rocket Propulsion.

To practice all areas of Rocket Propulsion, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn