Question: Determine gyroscopic couple and it's effect.
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## The turbine rotor of ship has a mass of 900 kg and radius of gyration 600 mm. It rotates at 1800 rpm clockwise looking from the stern. Determine gyroscopic couple and it’s effect when; (i) the ship is travelling at 40 km/hr and steers to left in a curve of 100m radius ; (ii) the ship is pitching and the bow is descending with maximum velocity. The pitching is simple harmonic, the periodic time being 30 seconds and total angular moment between the extreme positions is 120.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: Dec 2015

 modified 2.8 years ago  • written 2.8 years ago by Sayali Bagwe • 2.3k
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m = 900 kg

N= 1800 rpm

k = 0.6 mω .˙. mass of moment of Interia of the rotar I $= m k² \\ = 900* (0.6)^2$

I=324 kg- m ²

R= 100m

V $= 40 km / hr \\ = 40 *5/18 = 11.11 m /sec$

Tp = 30 sec

Amplitude of angular displacement $θ= 12/2 = 6° = 0.1047$ rad

Case 1:

Ship steering towards left

$V = R ωp \\ 11.11= 100 * ωp \\ ωp = 0.111 rad /sec$

Magnitude of gyroscopic couple

T $= I * ω* ωp \\ = 324 *(2*П*1800/60)* 0.111 \\ =6785 N-m$

Calculate the magnitude of gyroscopic couple

$ω_{p \max} = (2*П/ Tp )2 * Ø \\ =(2*П/30 )2 * 0.1047 \\ ωpmax =0.0219 rad / sec$

Magnitude of gyroscopic couple

$T= I * ω* ωp \\ = 324 *(2*П*1800/60)* 0.219 \\ =1339 N-m$

Calculating the max during pitching

α max = θ * ω1

$=(2*П/ Tp )2 * θ \\ =(2*П/30 )2 * 0.1047 * 0.1047 \\ =0.0219 * 0.1047 \\ α= 4.592 * 10 – 3 rad / sec 2$

 written 2.8 years ago by Sayali Bagwe • 2.3k