The oscillation of Floating Bodies

Oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states.

Torque (called moment or moment of force) is the tendency of a force to rotate an object about an axis,fulcrum, or pivot.

Weight is the name given to the force on an object due to gravity in science and engineering contexts.

The metacentric height is the distance between the centre of gravity of a ship and its metacentre.

The oscillation of floating bodies including the angle of heel and the period of oscillations

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Introduction

A single hulled ship subjected to any disturbing force 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.

Torque formula

The most used formula to define torque is : $\boldsymbol \tau = I\boldsymbol{\alpha}$
where
$\tau$ is the torque
$I$ is the moment of inertia
$\alpha$ is the angular acceleration of the body

Oscillation equation

Let $\frac{d^2y}{dx^2} + a^2y = 0$ General solution is given by: $y = A\,sin\,a\,x + B\,cos\,a\,x$

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.

If:

$W$ = 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 $\alpha = - \frac{d^2\theta }{dt^2}$ (Direction towards equilibrium position)

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

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

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

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

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

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

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

The Solution of this Equation is given by:

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

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

When $\theta$ = 0 and $t =T/2$ i.e. $T$ = the time of a complete oscillation.

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

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

The simplest solution:

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

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

Example 1 [imperial]

Problem

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.

Workings

The Rolling Period is given by \f $T\;=\;2\pi \sqrt{\displaystyle\frac{k^2}{hg}}\;Secs.\f$ .

Where $h$ is the Metacentric Height and $k$ is the Radius of Gyration of the Body about it's $C$ of $G$ .

$T\;=\;6Secs.\;=\;2\pi \sqrt{\frac{k^2}{gh}}$

To find $h$ : $k^2\;=\;\frac{d^2}{16}\;+\;\frac{l^2}{12}\;=\;16\left(\frac{1}{16}\;+\;\frac{1}{12} \right)\;=\;\frac{7}{3}\;ft^2$ $h\;=\;\frac{4\pi ^2k^2}{gT^2}\;=\;\frac{4\pi ^2\times 7/3}{32.2\times 36}\;=\;0.0795\;ft.$

To Find The Specific Gravity: $\frac{\pi }{4}D^2\times 4\times 62.4\times S\;=\;\frac{\pi }{4}D^2x\times 62.4$ $\therefore\;\;\;\;\;\;\;s\;=\;\frac{x}{4}$ $OD\;=\;\frac{x}{2}$ And: $OG\;=\;2\;ft.$ $BM\;=\;\frac{I}{V}\;=\;\frac{\pi \times 4^4}{64}\div \frac{\pi \times h^2x}{4}\;=\;\frac{1}{x}$

$MG(h)\ = BM + BO - BO - OG =$ \f $\displaystyle\frac{1}{x}\;+\;\displaystyle\frac{x}{2}\;-\;2\;=\;0.0795\;ft.\f$ $\therefore\;\;\;\;\;\;2\;+\;x^2\;-\;4.159\;=\;0$