1. 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
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
2. 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
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
3. 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
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
4.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
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
5. 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
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
6. 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
Explanation: We know that maximum angular acceleration during pitching
= φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2
7. 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
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
8. 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
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
9. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required to drive the pump.
a) 0.102 kW
b) 0.202 kW
c) 0.302 kW
d) 0.402 kW
Explanation: Work to be done = wQHg
w = 1 Kg/litre
Q = 50 litres per min
H = 20 + 5 + 5 = 30m
Work output/min = 1 x 50 x 30 x 9.81 Nm/min
Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60
= 0.302 kW.
10. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?
a) 141.36 W
b) 241.36 W
c) 341.36 W
d) 441.36 W
Explanation: Work per revolution = Tϴ
= 15 x 2п
= 94.24 Nm per revolution
P = Work/sec
= 94.24 x 1.5 = 141.36 W.