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# The oscillation of Floating Bodies

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

## 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.

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

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.
• $\inline&space;\theta$ = The Rotation after a time t.

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

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

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

$\therefore\;\;\;\;\;I_G\approx&space;I_M$

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

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

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

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

The Solution of this Equation is given by:

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

When t = 0 $\inline&space;\theta$= 0 and therefore B = 0

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

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

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

The simplest solution;_

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

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

### 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.

The Rolling Period is given by $\inline&space;T\;=\;2\pi&space;\sqrt{\frac{k^2}{hg}}\;Secs.$ Where h is the Metacentric Height and k is the Radius of Gyration of the Body about it's C of G.

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

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

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

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

Solving the quadratic X = 3.604 ft.

From which the Specific Gravity S = 0.901

## Worked Examples

The workings associated with these examples have been hidden. They can be seen by clicking on the red butttons.

### 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 $\inline&space;2^0$. 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)

Let k be the radius of gyration of the model about the longitudinal axis through G and let the metacentric height which equals GM be h

When oscillating, the unbalanced couple, when $\inline&space;\theta$ which acts on the model is:-

$W\times&space;GM\times&space;\theta$

$\therefore&space;\;\;\;\;\;\;W\times&space;GM\times&space;\theta=W\times&space;h\times&space;\theta=\frac{W\times&space;k^2}{g}\;\;\;\frac{d^2\theta}{dt^2}$

Re-writing
$\frac{d^2\theta}{dt^2}=\frac{g\times&space;h}{k^2}\times&space;\theta$

This is Simple Harmonic motion with a periodic time of:-

$t=2\pi\sqrt{\frac{k^2}{g\times&space;h}}$

Note. For more details on Simple Harmonic Motion please see" Maths: Differential Equations: Equations of Motion"

It can be assumed that the ball, on striking the model looses its momentum and produces an impulsive torque which starts the oscillations of the model.

To consider the impact of the ball on the model, let:-

• $\inline&space;W_1$ be the weight of the ball. (Note in the question, the weight is given in ounces and therefore must be divided by 16 to convert it to lb.)
• $\inline&space;V_1$ be the velocity just before impact.
• $\inline&space;\omega_1$ be the angular velocity of the model after impact.
• "a" be the distance of the point of impact, measure from the centre of gravity (G) of the model.
• A be the maximum angular velocity of the model during its Simple Harmonic Motion.
• t be the periodic time.

$\frac{a\times&space;W_1\times&space;V_1}{g}=\frac{W\times&space;k^2\times&space;\omega_1}{g}$

Substituting in given values:-

$2\times&space;\frac{1.6}{16}\times&space;15=80\times&space;k^2\times&space;\omega_1$

$\therefore&space;\;\;\;\;\;\;k^2\times&space;\omega_1=\frac{3}{80}$

As the model starts at the centre of its harmonic motion, it can be seen that $\inline&space;omega_1$ is the maximum angular velocity.

$\text{Then}\;\;\;\;\;\;\omega_1=2\pi\frac{A}{t}\;\;=\;\;2\pi\times&space;\frac{2}{57.3\times&space;1.5}=0.146\;rad./sec.$

From equations (26) and (27)

$k^2=\frac{3}{80\times&space;0.146}$

Substituting in equation (23)

$1.5=2\pi\sqrt{\frac{3}{80\times&space;0.146\times&space;32.2\times&space;h}}$

$\text{From&space;which\;\;\;\;\;\;h=0.14\;ft.}$