Question Details

The rotor of an aeroplane engine has mass moment of inertia 1.0 kg-m2. The engine rotates at speed of 500 RPM in clockwise direction if viewed from front of aeroplane. If aeroplane flying at 1200 km/hr turns with a radius of 2 km at same elevation. Then magnitude of gyroscopic moment by rotor on aeroplane structure is

Options

A

26.19 N-m

B

17.46 N-m

C

4.37 N-m

D

8.73 N-m

Show Answer

Correct Answer :

Option D

8.73 N-m

Solution :

The correct option is 8.73 N-m.

To find the magnitude of the gyroscopic couple (or gyroscopic moment) exerted by the rotor on the aeroplane structure, we use the standard formula for the gyroscopic couple:
C = I ω ω p
where:
I is the mass moment of inertia of the rotor.
ω is the angular velocity of the rotor.
ωp is the angular velocity of precession (associated with the turning of the aeroplane).

Let's calculate each of these values step-by-step from the given data:
1. Mass moment of inertia (I):
Given directly as:
I = 1.0 kg-m 2

2. Angular velocity of the rotor (ω):
The rotational speed of the engine is given as N=500 RPM.
The relation between angular velocity and RPM is:
ω = 2 π N 60
Substituting N=500:
ω = 2 π 500 60 = 50 π 3 52.36 rad/s

3. Angular velocity of precession (ωp):
The linear speed of the aeroplane is v=1200 km/hr. First, convert this speed into meters per second (m/s):
v = 1200 1000 3600 = 1200 5 18 = 1000 3 333.33 m/s
The turn radius is given as R=2 km=2000 m.
The angular velocity of precession is:
ω p = v R = 1000 / 3 2000 = 1 6 0.1667 rad/s

4. Magnitude of the gyroscopic couple (C):
Now substitute these values into the couple formula:
C = 1.0 52.36 0.1667
C = 1.0 50 π 3 1 6 = 50 π 18 = 25 π 9
Using π3.14159:
C 78.54 9 8.73 N-m
Thus, the magnitude of the gyroscopic moment acting on the aeroplane structure is indeed 8.73 N-m.

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