Irodov 1.288: A Turning Ship Pitches Its Flywheel – Amit & Deepali Sinha Foundation

Problem Irodov 1.288

A ship travels at $v = 36$ km/h along an arc of a circle of radius $R = 200$ m. Its flywheel has moment of inertia $I = 3.8\times10^{3}$ kg $\cdot$ m $^2$ about its own rotation axis and spins at $n = 300$ rpm, with that axis lying along the length of the ship. Find the moment of the gyroscopic forces that the flywheel’s shaft exerts on its bearings.

The ship seen from above, running along its circular arc of radius R at speed v. Going round the arc is a rotation Ω about the vertical through the center O, and the whole ship, flywheel shaft included, shares it. — The ship seen from above, running along its circular arc of radius $R$ at speed $v$ . Going round the arc is a rotation $\vect\Omega$ about the vertical through the center $O$ , and the whole ship, flywheel shaft included, shares it.

Given

$v = 36$ km/h
$R = 200$ m
$I = 3.8\times10^{3}$ kg $\cdot$ m $^{2}$
$n = 300$ rpm
axis along the ship

Find

Find the moment

N

of the gyroscopic forces on the bearings

Solution

Step 1The turn is itself a rotation

Following a circular arc of radius $R$ at speed $v$ , the ship turns about the vertical through the center of that arc at the rate

\Omega = \frac{v}{R} .

This rotation is shared by everything rigidly attached to the ship, and in particular by the flywheel’s shaft. So the flywheel’s angular momentum, which lies along the shaft, is being carried round with it.

Step 2The angular momentum of the flywheel

The flywheel spins at $n$ revolutions per second about its own axis, so its angular velocity is $\omega = 2\pi n$ and

L = I\omega = 2\pi n I

(1)

Step 3How fast that vector is being turned

The shaft lies along the ship, which is horizontal, while $\vect\Omega$ is vertical. The two are perpendicular, which is the case that gives the largest possible effect: the whole of $\vect L$ is swung sideways and none of it merely spins about its own direction.

The angular momentum after a further short time. It has the same length but has turned through dφ = Ω dt, so the change d L is perpendicular to L and has magnitude L dφ. — The angular momentum after a further short time. It has the same length but has turned through $d\varphi = \Omega\,dt$ , so the change $d\vect L$ is perpendicular to $\vect L$ and has magnitude $L\,d\varphi$ .

In a time $dt$ the shaft turns through $d\varphi = \Omega\,dt$ , so $\vect L$ turns through the same angle without changing length, and the chord joining the two tips has the magnitude

|d\vect L| = L\,d\varphi = L\,\Omega\,dt .

By the equation of moments, that rate of change is the moment applied to the flywheel, and by the third law the shaft presses back on its bearings with a moment of the same magnitude:

N = \left|\frac{d\vect L}{dt}\right| = \Omega L .

Substituting $\Omega = v/R$ and (1),

N = \frac{2\pi n I v}{R}

(2)

Step 4Numbers

The data must first be put in consistent units: $v = 36$ km/h $= 10$ m/s and $n = 300$ rpm $= 5.0$ rev/s. Then

N = \frac{2\pi \times 5.0 \times 3.8\times10^{3} \times 10}{200} \approx 6\times10^{3}\ \mathrm{N\cdot m} = 6\ \mathrm{kN\cdot m}.

Check

Test the geometry the answer depends on. If the flywheel’s axis were vertical, along $\vect\Omega$ itself, the turn would merely spin $\vect L$ about its own direction, $\vect L$ would not change at all, and the gyroscopic moment would vanish. The formula gives the other extreme, the perpendicular case, and in between the answer would carry a factor $\sin$ of the angle between the two axes. The units also check: $2\pi n I$ is an angular momentum, $\mathrm{kg\,m^{2}s^{-1}}$ , and $v/R$ is $\mathrm{s^{-1}}$ , so the product is $\mathrm{kg\,m^{2}s^{-2}} = \mathrm{N\cdot m}$ .

Remark

The moment is about the horizontal axis across the ship, since it is perpendicular to both the shaft, which runs fore and aft, and the vertical. So the gyroscopic couple tends to pitch the ship, pressing the bow down in one direction of turn and lifting it in the other, and it is carried entirely by the two bearings of the shaft. At 6 kN $\cdot$ m on bearings a few meters apart, that is a couple of kilonewtons on each of them, a few hundred kilograms of force, which is why large marine flywheels and turbines are mounted with this effect in mind.