Unit 4 Compound Pendulum
Objectives
To study the physical properties of compound pendulum and calculate the free-fall
acceleration by this device.
Apparatus
Pendulum, rack, optical gate, optical timer, ruler, and section paper.
rack
Pendulum
Optical gate
Optical timer
Set-up of this experiment
Principles
P
θ
h
C
θ
Mg
Fig. 1
A compound pendulum is a pendulum that is not uniform in mass. As shown in Fig.
1, a rigid body of mass M hanged by P which is used as a pivot. The distance from P
to the center of mass C is h. When this object swings at a small angel, the torque from
the gravity is
τ = − Mgh ⋅ sin θ .
Furthermore, if the angel is extremely small, the above equation can be simplified as
τ = −(Mgh )θ ……………………..…………..(1)
In S.H.M, we know that
τ = −κθ …………….……………………….…(2)
and
T = 2π
I
κ
……………………………………… (3)
Due to the similarity of Eq. (1) and (2), the period of the compound pendulum can be
expressed as
T = 2π
⎛ d 2θ
Meanwhile, since τ = I ⎜⎜
2
⎝ dt
I
.
Mgh
⎞
⎟⎟ = I α , therefore
⎠
d 2θ τ
Mghθ
= =−
2
dt
I
I
where α is the angular acceleration
I is the moment of inertia.
In this experiment, we get the free-fall acceleration g by making the
compound pendulum oscillates through two different axes, even though we don’t
know the position of the center of mass.
D
A
L
h1
B
θ
C
h2
θ
C
L
h2
h1
B
Mg
Mg
D
A
Fig. 2.
The sketch of the device is shown in Fig. 2, a compound pendulum is composed of a
metal rod, and an adjustable pendulum D. Suppose that the mass of this system is M,
and the center of mass is at C, which is h1 from A. We get the period TA when the
system swings by using A as the pivot. On contrary, we get the period TB by using B,
which is h2 from C, as the pivot. Therefore, we have
I C + M h12
IA
T A = 2π
= 2π
Mg h1
Mgh1
TB = 2π
I + M h22
IB
= 2π C
Mgh 2
Mgh2
I A and I B are the moment of inertias corresponding to A and B respectively, and
we can get their relations with I C by parallel-axis theorem ( I C is the moment of
inertia corresponding to center-of-mass, and I C = Mr 2 . r is an unknown variable and
relates to the distribution of the mass.) Assume that length of AB is constant, i.e.
AB= h1 + h2 = L = constant . Position of C (center-of-mass) changes as we change the
position of the pendulum D; therefore, TA and TB change. If C is in the middle of
AB, then h1 = h2 and T A = TB .
If h1 ≠ h2 , it is still possible to find a value that T A = TB . In such a case,
r 2 + h12
r 2 + h22
.
T = 2π
= 2π
gh1
gh2
Where r 2 = h1 h2 , and TA = TB = T = 2π
h1 + h2
L
= 2π
…………………(4)
g
g
Where L is the distance between A and B, and T is the period. Hence, we can get the
free-fall acceleration g by this experiment. The point of this experiment is to find the
proper position of pendulum D in order to get the condition of T A = TB .
D
The position of pendulum D
Instructions
1. Measure the length of L. Fix the pendulum D at a position S. Use A and B as the
pivot, start the swing with 50 cycles for each, and measure the period T A and
TB by the optical timer.
Q1: What is the number for the timer to count that you should set on the optical
timer?
2. Shift the pendulum D for 1 cm and fix it. Repeat step 1.
3. Repeat step 2 until you have the data of 5 different positions of D.
4.
Do the plot of S versus T (S for x-axis and T for y-axis). Label 5 points of
（S ,TA） and 5 points of （S ,TB）. Connect each of the 5 points (as shown in the
following figure).
T
TB
S6
TA
S
5.
Check if your lines have an intersection. If not, extend them until the intersection
appears. Then place the pendulum D on the position ( position of the intersection
point), and repeat step 1.
Q2: Why do you need to find the intersection of two lines?
6. See if you have the same periods by using A and B as the pivot. If not, fine tune
the position of D and repeat step 1 until the difference of TA and TB is no more
7.
than 0.2sec under 50 cycles.
Calculate g by Eq. (4).
Remark
1. Place your compound pendulum firmly on the rack to reduce swaying. If it’s not,
2.
put a sheet of paper on the rack to reduce sliding between the pivot and the rack.
Bring section paper.