bifilar determination of earth

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Bifilar Determination of Earth's

Gravitational Field

Faysal Riaz, Feb 1999

Gravitation | Topics

By experimentation I aim to obtain an accurate value

for the gravitational field strength, g (N kg-1), of the

Earth. I will carry out an experiment using a Bifilar

Pendulum.

The Bifilar Pendulum

I will set a metal rod oscillating by giving it a small angular displacement,(rads),

about the vertical axis when it is fastened onto two strings that are held by corks

firmly clamped onto stands. The motion of the Bifilar Pendulum is Simple Harmonic

Motion, I will prove this later in the report.

The metal rod is uniform and of mass M (kg) and length L (m). The two strings are

of length y (m). I will make sure that both strings are parallel to each other, always

perpendicular to the rotating rod and the same distance from the rod’s centre of

mass (l/2 d metres). The inclination of the strings to the vertical will be given by

(rads).

I will vary the separation of the two strings, d (m), from d » 0.1m to d » 0.7m and

record the time for twenty oscillations of the rod, 20T (s), from this value I will

calculate the periodic time, T (s), ie the time taken for one oscillation.

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The diagram below shows the setup of the experiment:

I will not model the oscillating rod as a particle, instead I will regard it as a system

of connected particles moving in circles of different radii. The same torque applied

to different bodies produces diferent angular accelerations, indicating that each

body has an individual amount of Rotational Inertia that controls the degree ofchange in rotation. The measure of a body’s Rotational Inertia is called Moment of

Inertia. The more difficult it is to change the angular velocity of a body rotating

about a particular axis the greater is its Moment of Inertia about that axis. The

Moment of Inertia of a body is a function of the mass of the body, the distribution

of that mass, that is its size and shape, and the position of the axis of rotation.Therefore these factors must affect the rotational motion of the metal rod.



Values of Moment of Inertia for rigid bodies can be calculated using calculus. The

Moment of Inertia about an axis perpendicular through the centre of the metal

rod can be found by:

Mass per unit length =

Mass of elemental length = x

let I = Moment of Inertia

I = mr2 (for all the particles of the rod)

I = x2x

As x 0 Moment of Inertia becomes:

+1/2 L +1/2 L

-1/2 Lx2dx = [1

/ 3 x3]-1/2 L

I =1 / 3{[1

/ 8 L3 ] - [-

1 / 8 L

3 ]}

I =1 / 12 L

3

M = L



I =1 / 12 ML

2

I will now prove that the motion of the Bifilar Pendulum is in fact Simple Harmonic

Motion.

Vertically the metal rod is in equilibrium so the Tensions of the strings that holdthe rod are given by:

T =1 / 2 Mg

So the Tensions of the strings when they are inclined to the vertical by a (rads),

are given by:

T =1 / 2 MgCos

So the restoring forces, T R, which move the strings back to their original positionsare given by:

TR =1 / 2 MgSin

The angle a (rads) is very small so the restoring forces become:

TR =1 / 2 Mg

Since both the angle of inclination to the vertical of the strings, a (rads), and the

angular displacement of the metal rod, q (rads), are very small and then:

y =1 / 2d

= d

2y

TR = Mgd 4y

Therefore the restoring Couple, CR, which acts towards the equilibrium position so

is negative, is given by:

CR = -Mgd2

4y

Applying Newton’s Second Law for the rotational motion of the rod, which is of

constant mass:



Id2= -Mgd

2

dt2

4y

d2= -Mgd

2

dt2

4Iy

This equation is in the form of the standard equation for Simple Harmonic Motion,

which is shown below, therefore the motion of the Bifilar Pendulum is simple

Harmonic motion:

a = -2s

2= Mgd

2

4Iy

So because the motion of the metal rod is Simple Harmonic Motion its Time Period,T (s), is given by:

T = 2

T = 24Iy

Mgd2

I = MgT2d

2

162y

The Moment of Inertia of the metal rod, as I have previously proven, can also begiven by the following equation:

I =1 / 12 ML

2

ML2

= MgT2d

2

12 162y

T2

= 42L

2y

3g d2

T d

let k2

= 42L

2y

3g



T = k

d

This equation is in the form of y = mx + c, so if I plot T (s) against d-1 (m-1) I will

obtain a straight line graph through the origin, whose gradient will be found by:

Gradient = y

x

Gradient = T = Td d

-1

Gradient = k

So when I have a value for k (ms) I will use it to find the gravitational field

strength, g (N kg-1), of the Earth as shown below:

g = 42L

2y

3k2

Results

String Separation Time for 20 Oscillations Periodic Time d-1

d (m) 20T (s) T (s) (m-1)

0.164 65.58 3.29 6.10

0.212 50.79 2.53 4.72

0.316 33.59 1.68 3.16

0.399 27.22 1.36 2.50

0.501 21.78 1.06 2.00

0.620 7.26 0.84 1.61



The graph below shows the results for the Bifilar Pendulum experiment:

Now I am able to calculate a value for gravity using the above graph:

Gradient = 3.5 = 0.53 ms 6.6

g = 42L

2y

3k2

k2

= 0.532

m2s

2

L2

= 0.6622

m2

y = 0.478 m

g = 9.80 N kg-1

Conclusion



By experimentation I have obtained an accurate value for the gravitational field

strength, g (N kg-1), of the Earth. The value of gravity that I have obtained from

the Bifilar Pendulum experiment is 9.80 N kg-1. The value of gravity that I have

obtained from the light gates experiment is 9.77 N kg-1.

Taking into consideration England's latitude and longitude, our distance from the

Earth’s centre of mass and the effects of the rotation of the Earth the

approximate value of gravity in England is 9.8143 N kg-1.

So the percentage error in the Bifilar Pendulum experiment is:

Percentage Error = 9.81 - 9.80 x 100%

9.81

Percentage Error = 0.12 % However this limited the experiment because if the angular displacement, (rads),

was large, that is over 0.175 rads or 10° , then the motion of the Bifilar Pendulum

was no longer Simple Harmonic Motion. So if this occurred the Time Period, T (s),

formula sown below would no longer be valid for the motion of the metal rod:

T = 24Iy

Mgd2

If this occurred it would have definitely caused a significant error in the results

for this experiment because from the above formulae the gravitational fieldstrength, g (N kg-1), of the Earth was calculated.

The lengths of the two strings may not have been the same, or they may not have

been the same distance from the centre of mass of the metal rod (1/2 d metres),

so the rod may have been slightly slanted. Therefore the position of the axis of

rotation of the rod was not perpendicular through the centre of the rod, this

means that its Moment of Inertia is not exactly given by the formula shown below:

I =

1

/ 12 ML

2

This may have caused the slight error in the result of this experiment.

If the metal rod that was used was not a uniform rod (of uniform density) then its

centre of mass may not have been at the axis of rotation, which was perpendicular

through the centre of the rod. Therefore the Moment of Inertia may not exactly



be given by the formula shown above. This may have caused the slight error in the

result of this experiment.

The error in this experiment may have been caused by the incorrect measurement

of the length of the metal rod, L (m), the string separation, d (m), the string

length, y (m), or of the time for twenty oscillations, 20T (s). A faulty stopwatch or

human error may have caused this.

I hope you have found this page useful.

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bifilar determination of earth

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