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Bow turns portside.
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$m = 900 \ kg$ $k = 0.6 \ m$ $N = 1800 \ rpm$ $\therefore$ $w = \frac{2 \pi \times 1800}{60}$

$\therefore w = 188.496 \ rad/sec$

Case 1:

$v = 40 \ km/hr$

$v = \frac{40 \times 10^3}{3600}$

$I = mk^2 = 324 \ kg m^2$

$w_p = \frac{V}{R} = \frac{40 \times \frac{5}{18}}{100} = 0.111 \ rad/sec$

$C = I . w . W_p$

$= 324 \times 188.496 \times 0.111 = 6.78 \times 10^3 \ N.m$

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Case 2:

$\phi = \frac{12}{2} \times \frac{\pi}{180} = \frac{0.209^c}{2} = 0.1047^c$

(Bow is descending)

enter image description here

Bow turns portside.

$w_0 = \frac{2 \pi }{Tp} = \frac{2 \pi}{30} = 0.2094$

$w_{p_{max}} \ = \ \theta . \ w_o \ = \ 0.0219$

$C = I.w. W_{p_{max}}$

$= 324 \times 188.496 \times 0.0219$

$C = 1337.49$

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