BIFILAR SUSPENSION
GROUP 1
CLASS E
GROUP MEMBERS
1. IFEANYI NDUKWE
20751
2. JOE GODWIN IME
20630
3. DENNIS OGANGWU
20606
4. GARBA MOHAMMED
20649
5. MOSES SAWA
20696
6. ONYEBUCHI UCHENDU 20828
7. STANLEY OKONKWO
20540
BIFILAR SUSPENSION
AIM: To determine the moment of inertia of a horizontal rectangular drop bar about its
center of mass using the bifilar suspension technique.
INTRODUCTION
The bifilar suspension is a technique used to determine the moment of inertia of
any type of object about any point on the object. This is done by suspending two
parallel cords of equal length through the object examined. However, the
approach taken for this experiment is to determine the moment of inertia of a
drop by suspending the cords through the mass centre of bodies, obtaining an
angular displacement about the vertical axis through the centre of mass by a
sensibly small angle.
SYSTEM LAYOUT AND APPARATUS USED
The apparatus used for this experiment consists of a uniform rectangular drop bar
suspended by fine wires (assumed to have negligible weight contribution to the
system). This rectangular bar contains holes equidistant from each other and two
extra with equal masses of 1.85kg, are made to peg through these holes. Chucks
are also in place to alter length of suspended wires.
In addition an enclosed-type measuring tape and weight balance were also
used for the experiment.
A diagram on the arrangement of the apparatus used is illustrated below:
Suspended cord
Rectangular bar
1.85Kg Masses
Chucks
b
L
x
Ɵ
G
mg
PROCEDURE
1. With the bar suspended by the wires, the length L was adjusted to a
convenient extent and then distance, b, between the wires was measured.
2. The bar was then tilted through a very small angle about the vertical axis and
time taken for 20 oscillations of the bar, was recorded. From this, the periodic
time was also calculated.
3. The length L was further adjusted and the time taken for another 20
oscillations was recorded.
4. The inertia of the rectangular bar was then increased by including the “two
1.85kg masses” symmetrically on either side of the centerline distance x
apart.
5. Then step 3 was repeated but with different values of L
6. The length of the wire L was then fixed at a value and the time taken was
recorded for 20 oscillations at varying distances x, between the two 1.85kg
masses.
7. The rectangular bar was then detached from the apparatus arrangement
and taken to weight balance in order to determine the mass of the bar.
8. The internal diameters of the holes, the thickness of the rectangular were also
measured.
9. All measurements and data recorded were collated for experimental
analysis.
THEORY
THE GOVERNING EQUATION – (THE EXPERIMENTAL ANALYSIS)
The equation of the angular motion is:
𝐼𝑑 2 Ɵ
𝑑𝑡 2
𝑑2Ɵ
𝑑𝑡 2
= −
+
𝑚𝑔𝑏2
4𝐿
𝑚𝑔𝑏2
𝐼∗4𝐿
Ɵ
=0
------------ (1)
------------- (2) (dividing through by the moment of inertia, I)
Note that, 𝐼 = 𝑚𝑘 2 (where ‘k’ is the radius of gyration and ‘m’ is mass of the bar)
So the above equation (2) may be written, as:
𝑔𝑏2
 + 4𝑘 2𝐿 Ɵ = 0 ------------- (3)
From equation (3),
ὥ=√
Also T =
2п
ὥ
𝑔𝑏2
4𝑘 2 𝐿
Where ὥ is the angular velocity of the bar.
where T is the periodic time.
Therefore;
𝑇 = 4п√
𝑘2𝐿
𝑔𝑏2
Also;
𝑻 = 𝟒п√
𝑰𝑳
𝒎𝒈𝒃𝟐
---------- (4) (Since;𝐼 = 𝑚𝑘 2 )
However, Introduction of the two equal masses into the bifilar suspension system can
gives rise to a modification of its mechanism and the equation of the angular motion.
This also provides another approach to the determination of the moment of inertia of
the body. In this situation the length of the wires are kept constant and the distances
between the two equal masses are varied. This is called the auxiliary mass method but
for this experiment, our analysis would be on just the generic bifilar suspension
approach. Nonetheless, a description of its angular motion is enumerated below;
(𝐼 + 2𝐼. 𝑚) ∗
𝑑2Ɵ
𝑑𝑡 2
= −(𝑚 + 2𝑀) ∗
𝑔𝑏2
4𝐿
Ɵ ---------- (5)
𝐼. 𝑚, is the moment of inertia of the body is added and 𝑀 is the mass of each body
added.
1
𝐼. 𝑚 = 2 𝑀𝑟 2 + 𝑀𝑥 2
{r is the radius of the mass (it has a circular cross-section) and x is the distance between
the centre of the bar and this mass)
From equation (5)
 +
(𝑚+2𝑀)
(𝐼+2𝐼.𝑚)
∗
𝑔𝑏2
4𝐿
Ɵ=0
Following the same approach from the previous period derivation, the new periodic
time (inclusive of the new masses) becomes;
𝑻. 𝒎 = 𝟒п√
𝑳
𝒈𝒃𝟐
∗
(𝑰+𝟐𝑰.𝒎)
(𝒎+𝟐𝑴)
------------ (6)
THE ANALYTICAL APPROACH
The general approach for calculating the moment inertia of any type of body about
any axis on the body is given as:
𝐼 = ∫ 𝑅2 𝑑𝑚 = ∑ 𝛿𝑚𝑅2
Whereby individual moments of inertia from individual differential mass value 𝛿𝑚 and
distance X between the axis point and the object are summed up and its summations
gives rise to the final moment of inertia of that body.
For a solid object with a rectangular cross-section, the same general approach gave
rise to determination of the moment of inertia about its centre of mass G and this is
expressed as;
𝐼𝑠 =
𝑀𝑙 2
---------------- (7)
12
Since the rectangular bar has bored holes within it, therefore its moment of inertia is;
I = Is (solid rectangular bar) - Io (bored holes).
Where Io is the moment of inertia of the bored holes.
1
𝐼𝑜 = 15 ( 𝑀𝑜𝑅𝑜 2 ) + 2 ∗ 𝑀𝑜 ∑ 𝑋 2
2
------------- (8)
Where Mo is the mass of the hole bored, Ro is the radius of each hole bored and X is the
distance from each hole (on either side) to the centre of the rectangular bar. Note the
there are 15 holes bored, hence its introduction in equation (8)
Therefore, Moment of inertia of the bar, I is
𝐼=
𝑀𝑙 2
12
1
− [15𝑀𝑜{( 𝑅𝑜 2 ) + 2 ∗ ∑ 𝑋 2 }]
2
----------- (9)
However, 𝑀𝑜 = 𝜌п𝑅𝑜2 𝑡 where (‘ρ’ is the density of the rectangular bar and‘t’ is
the thickness of the rectangular bar)
Therefore the moment of inertia becomes,
𝑰=
𝑴𝑳𝟐
𝟏𝟐
𝟏
− [𝟏𝟓𝝆п𝑹𝒐𝟐 𝒕(𝟐 𝑹𝒐𝟐 + 𝟐 ∑ 𝑿𝟐 )] --------------- (10)
TABLE OF RESULTS
Basic Parameter Information;
Parameter
Length of bar l
Value
50.8cm
Mass of the bar m
1.4kg
Mass of each
added mass
Radius of each
added mass
Parameter
Thickness of the bar
Distance between
the wires, b
Density of the bar
(steel bar)
Radius of each
bored hole
1.85kg
3.8cm
Value
1.2cm
0.48m
7850kg/m3
0.5cm
Data at various L (no mass included) TABLE 1
τ=
Test
L (m)
X
(m)
t (s)
1
2
0.3
0.4
-
13.54
15.38
𝐭
𝟐𝟎
(s)
0.677
0.769
k (m)
(from
equation
above)
0.148
0.145
k2 (m)
m (kg)
I = mk2
(kgm2)
0.0219
0.0212
1.4
1.4
0.0306
0.0296
Data at various L (mass inclusive) TABLE 2
Test
L (m)
X
(m)
1
2
3
4
0.3
0.4
0.5
0.6
0.355
0.355
0.355
0.355
k2 (m)
m
(kg)
{bar}
0.689
0.765
k (m)
(from
equation
above)
0.150
0.145
0.0225
0.0209
1.4
1.4
0.929
1.103
0.157
0.170
0.0247
0.0289
1.4
1.4
𝐭
t (s)
13.77
15.29
18.58
22.05
τ= 𝟐𝟎
(s)
Data at various x and at a fixed L (i.e. L = 0.4m) TABLE 3
Test
L (m)
X
(m)
1
2
3
4
0.4
0.4
0.4
0.4
0.355
0.305
0.255
0.205
t (s)
16.59
14.72
13.26
11.77
τ=
𝐭
𝟐𝟎
(s)
0.689
0.765
0.929
1.103
M (kg)
I=
{the
2
added (m+2M)*k
(kgm2)
mass}
1.85
1.85
1.85
1.85
0.1153
0.1067
0.1260
0.1478
*Note: the extra parameters are not included in this table 3 because the radius of
gyration and moment of inertia are to be determined with the auxiliary mass method
but we are streamlined to work with the generic bifilar suspension equation of angular
motion.
In addition, the moment of inertia can be further determined by graphical
representation from data collated in the experiment.
Equation 4 aforementioned above can be modified to:
𝐼𝐿
𝑇 2 = 16п2 (𝑚𝑔𝑏2 )
Invariably, 𝑇 2 can be plotted against 𝑳, whereby the slope can be used to determined
𝐼
the moment of inertia, I. i.e. the slope = 16п2 (𝑚𝑔𝑏2 ). Since m g and b are known, and
then I can be determined accordingly.
Additional table 4 (extrapolated from table 2)
Square of periodic time T2 (s2)
T
0.689
0.765
0.929
1.103
T2
0.4747
0.5852
0.8630
1.2166
L
0.3
0.4
0.5
0.6
T2 versus L graph
1.4
y = 0.303x
1.2
1
0.8
0.6
0.4
0.2
0
0.3
0.4
0.5
0.6
Length of wire, L (m)
From the graph,
Slope (sec2/m)
0.303
𝐼
Since the slope = 16п2 (𝑚𝑔𝑏2 ) as explained earlier
Therefore,
0.303 =
0.303 =
16п2
𝑚𝑔𝑏2
∗𝐼
16п2
5.1∗9.81∗0.482
∗𝐼
𝑰 = 𝟎. 𝟎𝟐𝟐𝟏 𝒌𝒈𝒎𝟐
ANALYTICAL DETERMINATION OF THE MOMENT OF INERTIA
From equation 10 above,
𝑴𝑳𝟐
𝟏
𝑰=
− [𝟏𝟓𝝆п𝑹𝒐𝟐 𝒕( 𝑹𝒐𝟐 + 𝟐 ∑ 𝑿𝟐 )]
𝟏𝟐
𝟐
Therefore,
𝐼 = 1.4 ∗
0.5082
12
1
2
− [15 ∗ 7850 ∗ п ∗ 0.005 ∗ 0.012(2 ∗. 0052 + 2 ∑17.5
𝑋=2.5 𝑋 )]
𝑰 = 𝟎. 𝟎𝟏𝟎𝟔𝟖 𝒌𝒈𝒎𝟐
The moment of inertia calculated analytically is 0.01068 kg m2
PRECAUTION(S)
1. Measurement taken from the rule and the weight balance was done such that
the line of sight and the markings of the measuring equipments were in
alignment in to reduce errors due to parallax.
2. When taking down the time for the oscillations at various distances ‘x’ and a
fixed length ‘L’ for table 3 data, the length L was periodically checked after
each test in order to maintain the fixed length value of 0.4m
3. Precautionary methods were in place to keep the masses at a very comfortable
position so as to avoid slip or fall which could in-turn cause harm to our feet.
4. The experiment is done such that the oscillation was not dampened by carefully
tilting the bar before release for oscillations
CONCLUSIONS
OBSERVATION(S):
1. The periodic time significantly increased when the length of the wires also go
increased.
2. The periodic time also increased when the distances between the masses
added to system reduced
3. The moment of inertia determined using the analytical approach was
approximately equal to the value determined from test 2 in table 2 above.
4. The moment of inertia determined from the graph representation was greater
than the value gotten from the analytical approach indicating that the two
masses added during the experiment had a part to play in the increment of the
moment of inertia and also unavoidable human errors caused a variation in their
values.
5. The radius of gyration and moment of inertia reduced after the length of wire
was increased from test 1 to test 2 but increased right after till test 4.
FINAL DEDUCTIONS:
The bifilar suspension technique offers the opportunity to determine the radius of
gyration of a body by relating the readings gotten from the procedure in the
techniques and relating that into the equation of angular and this invariably provides
the determination of the moment of inertia for the same body. These readings
encompasses the distance between the wires used for the suspension, the length of the
wires, the time for the required number of oscillations, the distance between the masses
introduced into the experiment, and so on. All these and lots more provide the avenue
for determining the radius of gyration and the moment of inertia
Below is a tabular representation of the final value of the moment of inertia
determined from the analytical approach, the graphical approach and a selected
value of the moment of inertia from test 2 in table 2.
Methodology
Moment of inertia values
Analytical approach
Value from test 2 in table 2
Graphical approach
0.01068 kg m2
0.01067 kg m2
0.02210 kg m2