Turbine rotor of a ship has a mass of 2.2 tonnes and rotate at 1800 rpm clockwise, when viewed from aft. The radius of gyration of rotor is 320 mm. Determine the gyroscopic couple and its effects, when

1. the ship turns at radius of 250 mm, with speed of 25km/h.
2. The ship pitches with the bow rising at an angular velocity of 0.8 rad/sec.
3. Ship rolls at an angular velocity of 0.1 rad/sec.

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

Marks: 10 Marks

Year: May 2016

M= 2200 kg

R= 250 m

K= 0.32 m

N= 1800 rpm

V$= 25 km/h \\ = 25×1000/3600 \\ = 6.94 m/s$

1. I$= mk^2 \\ = 2200×(0.32)^2 \\ = 225.3 kg-m^2$

   $W= \dfrac{2πN}{60}= \dfrac{2×π×1800}{60}= 188.5 rad/sec. \\ W_p= \dfrac{V}{R} = \dfrac{6.94}{250}= 0.0278 rad/sec.$

2. C$= I_{wwp} \\ = 225.3×188.5×0.0278$

   C= 1180 N.M – $\text{The effect is to lower the bow and raise the \\ Stern; when the ship turns right.}$

   Wp= 0.8 rad/sec

   C $= Iw × Wp \\ = 225.03×188.5×0.8$

   C = 33972 N.M

   The effect of the reaction couple when the bow is raising is to turn the ship towards the sight or toward the starboard.

3. $W_p= 0.1 rad/sec.$

   C $= 225.3×188.5×0.1 \\ = 4246.5 N.M$

   As the axis of spin is always parallel to the axis of precession for all gyrn, there is no gyroscopic effect on the ship.