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Oscillatory motion of floating bodies, including the angle of heel and period of oscillations.

Introduction

When a body is displaced from its equilibrium position by an external unbalanced force, it will accelerate to its new position with a certain finite velocity. However, the stability of the object will create a net restoring force, which will cause it to accelerate back to the original position. The restoring motion provides momentum for the object to travel past the equilibrium position, and this back and forth motion continues until the velocity of oscillation disappears.

A similar motion occurs if a single hulled ship is subjected to any disturbing force, such as waves on the high seas; it will heel or roll. This page investigates the degree of roll and the periodic time of the movement.

The Period of Oscillation of a Floating Body.

NOTE:

1) Torque = Moment of Inertia $X$ Angular Acceleration.
2) The Solution of
$\frac{d^2y}{dx^2} + a^2y = 0\;is\; given\; by\;y = A\,sin\,a\,x + B\,cos\,a\,x$
(1)

For small angles of heel, the Body can be regarded as Oscillating about it's Metacentre in a manner similar to a Pendulum about it's point of suspension.

W = The weight of the Ship
M = The Metacentric Height
k = The Radius of Gyration of the Ship about a horizontal Axis through the C.of.G.
$\theta$ = The Rotation after a time t.

$Angular\;Acceleration = - \frac{d^2\theta }{dt^2}\;( Direction \;towards\; equilibrium\; position)$

(2)

The Moment of Inertia of the ship about the C of G is $\frac{W}{g}k^2$ .

The Moment of Inertia of the ship about M is $\frac{W}{g}\left(k^2+m^2 \right)$ .

Provided that m is small compared with k, these two can be regarded as equal

$\therefore\;\;\;\;\;I_G\approx I_M$

(3)

For small angles of heel , the Righting Couple = $W\;m\;\theta$

$\therefore\;\;\;\;\;W\;m\;\theta = I_m\times -\frac{d^2\theta }{dt^2}\;\approx I_G\times- \frac{d^2\theta }{dt^2}$

(4)

$= -\frac{W}{g}\;k^2 \frac{d^2\theta }{dt^2}$

(5)

$\therefore\;\;\;\;\;\frac{d^2\theta }{dt^2} + \frac{gm}{k^2\theta } = 0$

(6)

The Solution of this Equation is given by:

$\theta = A\;sin\sqrt{\frac{gm}{k^2}}\;t + b\;cos\sqrt{\frac{gm}{k^2}}\;t$

(7)

When t = 0 $\theta$ = 0 and therefore B = 0

When $\theta$ = o and t = T/2 i.e. T = the time of a complete oscillation.

$A\;sin\sqrt{\frac{gm}{k^2}}\times\frac{T}{2} = 0$

(8)

$Since\;A\;\neq 0\;\;\;\;sin\sqrt{\frac{gm}{k^2}}\times\frac{T}{2} = 0$

(9)

The simplest solution:

$\sqrt{\frac{gm}{k^2}}\times\frac{T}{2} = \pi$

(10)

$\therefore\;\;\;\;\;The\;Periodic\;Time\;T = 2\pi \sqrt{\frac{k^2}{mg}}$

(11)

Example 1

A Solid Cylinder of Uniform material and with height equal to the Diameter is to float in water with it's axis vertical. Calculate the Metacentric hight and the specific gravity of the Cylinder so that it may have a Rolling Period of six seconds when the Diameter is 4 ft.

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Worked Examples

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Example 2

A model ship weighs 80 lb. and floats in a tank of water. A ball of weight 1.6 oz. traveling horizontally at 15 ft./sec is arranged to strike the model broadside at a point 2 ft. vertically above its centre of gravity. The ball after striking, moves with the model and the amplitude of the resulting roll is $2^0$ . The natural period of roll of the system is 1.5 sec. Find the metacentric height of the system assuming it to be constant over this amplitude. ( Assume that the vessel oscillates about its centre of gravity. (B.Sc. Part 2)