COMPOUND PENDULUM
Topic : Compound Pendulum
Objective : To determine the centre of gravity of a connecting rod, as well as the
Radius of gyration about the centre of gravity.
Theory : Moment at O:
Clockwise order
\ 
If 
is smaller
By using the
This equation is the same as the general equation which is “Simple Harmonic Motion”. Therefore, we can obtain the frequency by using this system as followed below:-
Meanwhile, the periodic time (T) is been taken to complete a swinging, therefore:- 
Seconds
For the rigid body, moment of inertia at point O is:
Where 
is the moment of inertia at centre of gravity.
Where 
is the radius of gyration at centre of gravity.
====================
We could obtain the radius of gyration of any rigid body with this equation.
Method :
- A rounded connecting rod was placed at the hinge with the smaller radius of circle at the top and the larger radius of circle at the bottom.
- Then, the rod was set at an angle of
. Next, the rod released and at the same moment, time was taken for ten swings. This action repeated at the opposite side and the average reading for ten swings was then calculated. - Later, the rounded connecting rod was turned upside down with the larger radius of circle at the hinge and the same steps as above were repeated.
- After the first experiment had been done, the steps were repeated by using rectangular connecting rod and also prismatic thick plane.
Apparatus : Stopwatch, Ruler, retort set, protector, rectangular shaped connecting rod, prismatic thick plane, circle shaped connecting rod.
Cylinder shaped connecting rod Rectangular shaped connecting rod Prismatic thick plane
Diagrams :
When the conrod is suspended at A, the periodic time is given by:-
(i)
When the conrod is suspended at B, the periodic time is given by:-
(ii)
At the same,total length between A and B is:-
(iii)
It can be seen that in equation (i),(ii) and (iii) there are three unknowns, i.e. k, 
and 
.Therefore, solving the three equations simultaneously, the three unknowns can be determined.
Result :
Pendulum swing
| Type | q° | Smaller hole at top,(s) | Average Time, (s) | Larger hole at top, (s) | Average Time, | ||
| Cylinder shaped conrod |
| 9.10 | 9.06 | 9.08 | 8.44 | 8.62 | 8.53 |
| Rectangular shaped conrod |
| 8.71 | 8.75 | 8.73 | 8.81 | 8.56 | 8.70 |
| Prismatic thick place |
| 9.11 | 9.18 | 9.15 | 8.59 | 8.67 | 8.63 |
Calculation (k,
,
)
Cylinder Shaped Conrod
Total Length 
= 0.29 m
Periodic Time smaller hole at top (
) = 0.908 s
Periodic Time larger hole at top (
) = 0.853 s
= 
9.81 
0.0209 = 
0.0594 – 0.2050 = 0.1808 + 0.0841 – 0.58
0.58 - 0.2050 - 0.1808 = 0.0841 – 0.0594
0.1942 = 0.0247
= 0.1272 m
From equation total length
= 
= 
= 0.29 – 0.1272
= 0.1628 m
For radius of gyration (k)
= 
= 
= 
= 
= 0.0826 m
Rectangular Shaped Conrod
Total Length = 0.29 m
Periodic Time smaller hole at top () = 0.873 s
Periodic Time larger hole at top () = 0.870 s
= 
9.81
0.0193 = 
0.055 – 0.1893 = 0.1881 + 0.0841 – 0.58
0.58 - 0.1853 - 0.1881 = 0.0841 – 0.055
0.2066 = 0.0291
= 0.1409 m
From equation total length
=
=
= 0.29 – 0.1409
= 0.149 m = 0.15 m
For radius of gyration (k)
=
= 
= 
= 
= 0.0815 m
Prismatic thick plane
Total Length = 0.2947 m
Periodic Time smaller hole at top () = 0.915 s
Periodic Time larger hole at top () = 0.863 s
= 
9.81 
0.0212 = 
0.0613 – 0.208 = 0.185 + 0.0868 – 0.5894
0.5894 - 0.208 - 0.185 = 0.0868 – 0.0613
0.1964 = 0.0255
= 0.1298 m
From equation total length
=
=
= 0.2947 – 0.1298
= 0.1649 m
For radius of gyration (k)
=
= 
= 
= 
= 0.0847 m
Discussion :
Any rigid body pivoted at a point other than its center of mass will oscillate about the pivot point under its own gravitational force. Such system is known as a compound pendulum.
The concepts of compound pendulum and center of percussion can be used in many practical qapplications. For instance a hammer can be shaped to have the center of percussion at the hammer head while the center of rotation is at the handle. In this case, the impact force at the hammer head will not cause any normal reaction at the handle.
Another one is in Izod (impact) testing of materials, the soecimen is suitably notched and held in a vice fixed to the base of the machine. A pendulum is released from a standard height, and the free end of the specimen is struck by the pendulum as it passes through its lowest position. The deformation and bending of the pendulum can be reduced if the center of percussion is located near the striking edge. In this case, the pivot will be free of any impulsive reaction.
Conclusion :
It can be concluded that from our experiment, we managed to obtain:
The centre of gravity of cylinder shaped conrod measured from LA is 0.1628m
Its radius of gyration, k is 0.0826m
The centre of gravity of rectanngular shaped conrod measured from LA is 0.15m
its radius of gyration, k is 0.0815m
The centre of gravity of prismatic thick plate measured from LA is 0.1649m
Its radius of gyration, k is 0.0847m
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