Engineering Questions with Answers - Multiple Choice Questions

Machine Dynamics MCQ – Effect of Gyroscopic Couple on an Aeroplane

1 - Question

A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what speed will it precess about the vertical axis?
a) 14.7 rad/s
b) 15.7 rad/s
c) 16.7 rad/s
d) 17.7 rad/s

View Answer

Answer: c
Explanation: Given: d = 300 mm or r = 150 mm = 0.15 m ; m = 5 kg ; l = 600 mm = 0.6 m ;
N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s
We know that the mass moment of inertia of the disc, about an axis through its centre of gravity and perpendicular to the plane of disc,
I = m.r2/2 = 5(0.15)2/2 = 0.056 kg-m2
and couple due to mass of disc, C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m
Let ωP = Speed of precession
We know that couple (C),
29.43 = I.ω.ωP = 0.056 × 31.42 × ωP = 1.76 ωP
ωP = 29.43/1.76 = 16.7 rad/s




2 - Question

An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft.
a) 10.046 kN-m
b) 11.046 kN-m
c) 12.046 kN-m
d) 13.046 kN-m

View Answer

Answer: a
Explanation: Given : R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ;
N = 2400 r.p.m. or ω = 2π × 2400/60 = 251 rad/s
We know that mass moment of inertia of the engine and the propeller,
I = mk2 = 36 kg-m2
and angular velocity of precession,
ωP = v/R = 55.6/50 = 1.11 rad/s
We know that gyroscopic couple acting on the aircraft,
C = I. ω. ωP = 36 × 251.4 × 1.11 = 100 46 N-m
= 10.046 kN-m




3 - Question

The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius.
a) 100.866 kN-m
b) 200.866 kN-m
c) 300.866 kN-m
d) 400.866 kN-m

View Answer

Answer: b
Explanation: Given: m = 8 t = 8000 kg ; k = 0.6 m ; N = 1800 r.p.m. or ω = 2π × 1800/60
= 188.5 rad/s ; v = 100 km/h = 27.8 m/s ; R = 75 m
We know that mass moment of inertia of the rotor,
I = mk2 = 2880 kg-m2
and angular velocity of precession,
ωP = v / R = 27.8 / 75 = 0.37 rad/s
We know that gyroscopic couple,
C = I.ω.ωP = 2880 × 188.5 × 0.37 = 200 866 N-m
= 200.866 kN-m




4 - Question

The heavy turbine rotor of a sea vessel rotates at 1500 r.p.m. clockwise looking from the stern, its mass being 750 kg. The vessel pitches with an angular velocity of 1 rad/s. Determine the gyroscopic couple transmitted to the hull when bow is rising, if the radius of gyration for the rotor is 250 mm.
a) 4.364 kN-m
b) 5.364 kN-m
c) 6.364 kN-m
d) 7.364 kN-m

View Answer

Answer: d
Explanation: Given: N = 1500 r.p.m. or ω = 2π × 1500/60 = 157.1 rad/s; m = 750 kg;
ωP = 1 rad/s; k = 250 mm = 0.25 m
We know that mass moment of inertia of the rotor,
I = mk2 = 46.875 kg-m2
Gyroscopic couple transmitted to the hull (i.e. body of the sea vessel),
C = I.ω.ωP = 46.875 × 157.1 × 1= 7364 N-m = 7.364 kN-m




5 - Question

The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.
a) 11.27 kN-m
b) 22.27 kN-m
c) 33.27 kN-m
d) 44.27 kN-m

View Answer

Answer: b
Explanation: Given : m = 3500 kg ; k = 0.45 m; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s
When the ship is steering to the left
Given: R =100 m ; v = km/h = 10 m/s
We know that mass moment of inertia of the rotor,
I = mk2 = 708.75 kg-m2
and angular velocity of precession,
ωp = v/R = 10/100 = 0.1 rad/s
Gyroscopic couple,
C = I.ω.ωp = 708.75 × 314.2 × 0.1 = 22 270 N-m
= 22.27 kN-m




6 - Question

The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity. The period of pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching is 12 degrees.
a) 3.675 kN-m
b) 4.675 kN-m
c) 5.675 kN-m
d) 6.675 kN-m

View Answer

Answer: a
Explanation: Given: tp = 40 s
Since the total angular displacement between the two extreme positions of pitching is 12° (i.e. 2φ = 12°), therefore amplitude of swing,
φ = 12 / 2 = 6° = 6 × π/180 = 0.105 rad
and angular velocity of the simple harmonic motion,
ω1 = 2π / tp = 2π / 40 = 0.157 rad/s
We know that maximum angular velocity of precession,
ωp = φ.ω1 = 0.105 × 0.157 = 0.0165 rad/s
Gyroscopic couple,
C = I.ω.ωp
= 708.75 × 314.2 × 0.0165 = 3675 N-m
= 3.675 kN-m




7 - Question

The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum gyroscopic couple.
a) 11.185 kN-m
b) 22.185 kN-m
c) 33.185 kN-m
d) 44.185 kN-m

View Answer

Answer: c
Explanation: Given : m = 20 t = 20 000 kg ; k = 0.6 m ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad ; tp = 30 s
We know that mass moment of inertia of the rotor,
I = m.k2 = 20 000 (0.6)2 = 7200 kg-m2
and angular velocity of the simple harmonic motion,
ω1 = 2π / tp = 2π/30 = 0.21 rad/s
Maximum angular velocity of precession,
ωPmax = φ.ω1 = 0.105 × 0.21 = 0.022 rad/s
We know that maximum gyroscopic couple,
Cmax = I.ω.ωPmax = 7200 × 209.5 × 0.022 = 33 185 N-m
= 33.185 kN-m




8 - Question

The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum angular acceleration of the ship during pitching.
a) 0.0016 rad/s2
b) 0.0026 rad/s2
c) 0.0036 rad/s2
d) 0.0046 rad/s2

View Answer

Answer: d
Explanation: We know that maximum angular acceleration during pitching
= φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2




9 - Question

A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effect when the ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.
a) 38.5 kN-m
b) 48.5 kN-m
c) 58.5 kN-m
d) 68.5 kN-m

View Answer

Answer: a
Explanation: Given : m = 5 t = 5000 kg ; N = 2100 r.p.m. or ω = 2π × 2100/60 = 220 rad/s ; k = 0.5 m
v = 30 km / h = 8.33 m / s ; R = 60 m
We know that angular velocity of precession,
ωp = v/R = 8.33/60 = 0.14 rad/s
and mass moment of inertia of the rotor,
I = m.k2 = 5000(0.5)2 = 1250 kg-m2
Gyroscopic couple,
C = I.ω.ωp = 1250 × 220 × 0.14 = 38 500 N-m = 38.5 kN-m




10 - Question

A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effect when the ship pitches 6 degree above and 6 degree below the horizontal position. The bow is descending with its maximum velocity. The motion due to pitching is simple harmonic and the periodic time is 20 seconds.
a) 6075 N-m
b) 7075 N-m
c) 8075 N-m
d) 9075 N-m

View Answer

Answer: d
Explanation: Given: φ = 6° = 6 × π/180 = 0.105 rad/s ; tp = 20 s
We know that angular velocity of simple harmonic motion,
ω1 = 2π / tp = 2π / 20 = 0.3142 rad/s
and maximum angular velocity of precession,
ωPmax = φ.ω1 = 0.105 × 0.3142 = 0.033 rad/s
Maximum gyroscopic couple,
Cmax = I.ω.ωPmax = 1250 × 220 × 0.033 = 9075 N-m

Get weekly updates about new MCQs and other posts by joining 18000+ community of active learners