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Gyroscope effect on ship.
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Gyroscope effect on ship:

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PART A: Gyroscope effect on ship during steering:

  1. Rotor rotates clockwise and ship takes right turn:

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  1. Rotor rotates clockwise and ship takes left turn:

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Note:

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Diagram

  • Initial angular momentum = $\overrightarrow {OX} \ = \ Iw$

  • Final angular momentum = $\overrightarrow {OX'} \ = \ Iw$

$\therefore$ Change of angular momentum = $\overrightarrow {OX} \ – \ \overrightarrow {OX}$

= $\overrightarrow {xx}$

= $\overrightarrow {ox} . \ \delta \ \theta$

= $I.w. \delta \ \theta$

$\therefore$ rate of change angular momentum = $I.w. \frac{\delta \ \theta}{\delta \ t}$

$$= \lim_{\delta \to\ 0} I.w. \frac{\delta \ \theta}{\delta t}$$

= $I. \ w. \ \frac{d \ \theta}{dt} \ = \ I.w.w_p.$

Where, $I =$ M.M.I of rotor

$w =$ Angular speed of rotor.

$Wp =$ angular speed of precession.

  1. Rotor rotates anticlockwise and ship takes left turn:

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  1. Rotor rotates anticlockwise and ship takes right turn.

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Part B: Gyroscopic effect on ship during pitching:

  1. Rotor rotates anticlockwise and ship pitches up with SHM.

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Ship moves or turns towards starboard.

  1. Rotor rotates anticlockwise and ship pitches down with SHM.

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Ship moves or turns towards portside.

  1. Rotor rotates clockwise and ship pitches up.

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Ship moves towards port side.

  1. Rotor rotates clockwise and ship pitches down.

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Ship moves towards star board.

PART C: Gyroscopic effect of ship during rolling:

In case of rolling action, the axis of rotation of rotor and axis of rolling of ship are same. Hence there is no gyroscopic effect of ship during rolling.

NOTE:

1. PART A: Steering:

$C \ = \ I.\ w. \ w_p$

2. PART B: Pitching (with SHM)

Diagram

  • $\theta \ = \ \phi \ sin \ w_ot$

{ where, $W_o \ = \ \frac{2 \pi}{Tp}$

  • $\frac{d \ \theta}{d \ t} = \phi \ W_o \ cos \ w_ot$

If $cos w_ot = 1$

$\therefore$ $w_{p_{max}} \ = \ W_0$

$C \ = \ I.w.w_{p_{max}}$

$I \ = \ mk^2$

$w_p = \frac{v}{R}$

  • $\alpha = \frac{d^2 \ \theta}{dt^2} \ = \ -\phi \ w_o^2 \ sin \ w_ot$

If $sin w_ot = 1$

$\therefore$ $\alpha \ max = - \phi \ w_o^2$

. . . . .(Negative sign indicates retardation)


Numericals:

$m = 6000 \ kg$

$N = 2400 \ rpm$

$\therefore$ $w = \frac{2 \ \pi \ \times \ 2400}{60}$

$w = 251.32 \ rad/sec$

$k = 450 \ mm = 0.450 \ m$

Case 1: Rotor rotates anti clockwise when viewed from bow end and steers left:

$\therefore v = 18 \ knots = 18 \times 1860 = 33480 \ m/hr$

$= 9.3 \ m/s$

$\therefore R = 60 \ m$

$I = mk^2$

$= 6000 \times 0.450^2$

$I = 1215 \ kgm^2$

  • $wp = \frac{v}{R} = \frac{9.3}{60} = 0.155 \ rad/sec$

$\therefore$ C = I w. Wp

= 1215 x 251.32 x 0.155

$C_{max} = 47329 \times 10^3 \ N.m$

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Case 2: Bow moves down and rotor rotates

$Tp = 18 \ sec, \phi = 7.5° = 7.5 \times \frac{\pi }{180} = 0.1309^c$

$w_o = \frac{2 \pi }{Tp} = \frac{2 \pi}{18} = 0.349$ rad/sec

$w_{p_{max}} = \phi . \ w_0 \ = \ 0.1309 \times 0.349 = 0.0456 \ rad/sec$

$C_{max} \ = \ I. w. \ wp_{max} = 1215 \times 251.32 \times 0.0456$

$C_{max} = 13.92 \times 10^3 \ N.m$

Case 3: There is no gyroscopic effect during rolling.

Case 4:

$\alpha_{max} \ = - \phi . \ w_0^2$

$= - 0.1309 \times 0.349^2$

$\alpha _{max} = -0.01594 \ rad/sec^2$

(Negative sign indicates retardation of ship)

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