Question:
The rotor of an aeroplane engine has a mass moment of inertia 1.0 kg m\(^2\). The engine rotates at a speed of 500 RPM in the clockwise direction if viewed from the front of the aeroplane. If the aeroplane while flying at 1200 km/hr turns with a radius of 2 km at same elevation, then the magnitude of the gyroscopic moment exerted by the rotor on the aeroplane structure in N m is
The rotor of an aeroplane engine has a mass moment of inertia 1.0 kg m\(^2\). The engine rotates at a speed of 500 RPM in the clockwise direction if viewed from the front of the aeroplane. If the aeroplane while flying at 1200 km/hr turns with a radius of 2 km at same elevation, then the magnitude of the gyroscopic moment exerted by the rotor on the aeroplane structure in N m is
Show Hint
Find the spin angular velocity and the turn rate (precession) separately, then multiply with the inertia.
Updated On: Aug 14, 2026
- 8.73
- 17.46
- 4.37
- 26.19
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Convert the spin speed to rad/s.
Spin speed \(N = 500\) RPM gives angular velocity \(\omega_s = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 500}{60} = 52.36\) rad/s.
Step 2: Convert the aeroplane speed and find the rate of turn.
Flight speed \(V = 1200\) km/hr \(= \dfrac{1200 \times 1000}{3600} = 333.33\) m/s. Turning at radius \(R = 2\) km \(= 2000\) m gives the rate of precession \(\omega_p = \dfrac{V}{R} = \dfrac{333.33}{2000} = 0.1667\) rad/s.
Step 3: Apply the gyroscopic couple formula.
The gyroscopic moment magnitude is \(C = I \omega_s \omega_p\), where \(I\) is the mass moment of inertia of the rotor about its spin axis.
Step 4: Substitute the values.
\(C = 1.0 \times 52.36 \times 0.1667 = 8.73\) N m.
Final Answer:
The gyroscopic moment exerted on the aeroplane structure is close to 8.73 N m. \[ \boxed{C \approx 8.73 \text{ N m}} \]
Spin speed \(N = 500\) RPM gives angular velocity \(\omega_s = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 500}{60} = 52.36\) rad/s.
Step 2: Convert the aeroplane speed and find the rate of turn.
Flight speed \(V = 1200\) km/hr \(= \dfrac{1200 \times 1000}{3600} = 333.33\) m/s. Turning at radius \(R = 2\) km \(= 2000\) m gives the rate of precession \(\omega_p = \dfrac{V}{R} = \dfrac{333.33}{2000} = 0.1667\) rad/s.
Step 3: Apply the gyroscopic couple formula.
The gyroscopic moment magnitude is \(C = I \omega_s \omega_p\), where \(I\) is the mass moment of inertia of the rotor about its spin axis.
Step 4: Substitute the values.
\(C = 1.0 \times 52.36 \times 0.1667 = 8.73\) N m.
Final Answer:
The gyroscopic moment exerted on the aeroplane structure is close to 8.73 N m. \[ \boxed{C \approx 8.73 \text{ N m}} \]
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