Question

The rotor of the turbine of a ship has a mass of 2500 kg and rotates at a speed of 3200 rpm counter-clockwise when viewed from stern (rear end). The rotor has radius of gyration of 0.4 m. The ship pitches 5 degrees above and 5 degrees below the normal position and the bow is descending with its maximum velocity. The pitching motion is simple harmonic with a periodic time of 40 seconds. Then the of gyroscopic couple (N-m) when view from the stern is

• A. 1837.5 towards left
• B. 1837.5 towards right
• C. 1935.8 towards left
• D. 1387.5 towards right

Answer 1

The correct option is B 1837.5 towards right

$I={\text{mk}}^2=2500\times (0.4{)}^2=400{\text{kg.m}}^2$

$\omega =\frac{2\pi \times 3200}{60}=335\text{rad/s}$

$\phi =5^{\circ }=5\times \frac{\pi }{180}=0.0873\text{rad}$

$T=40s$

$\therefore {\omega }_0=\frac{2\pi }{40}=0.157\text{rad/s}$

${\omega }_p=\phi {\omega }_0=0.0873\times 0.157=0.0137\text{rad/s}$

$C=I\omega {\omega }_p=400\times 335\times 0.0137$

$=1837.5\text{Nm}$

$\overrightarrow{\omega }\text{is acting towards stern (rear end)}$

${\overrightarrow{\omega }}_p\text{is into the plane of paper}$

$\overrightarrow{\omega }\times {\overrightarrow{\omega }}_p\text{is acting towards right when viewed from rear end.}$