CLASSROOM
Sandeep Bala
1st Year B.Tech.
Indian Institute of Technology
Torsion Pendulum Experiment at the
International Physics Olympiad
Mumbai 400 076, India.
Email:[email protected]
Introduction
The International Physics Olympiad is held every year to test
the theoretical and experimental skills of the best physics students
from high schools all around the world. The theory part of the
examination has three long problems. to be solved in five hours.
The experimental part of the examination is given 40% weightage
and may consist of two short experiments or one long experiment to be carried out in five hours. This year the 30th Physics
Olympiad was held at Padua in Italy and 291 students from 62
countries participated.
Here, a long experiment on torsion pendulum was chosen to test
the skills of the students and the examination was conducted in
two batches. By choosing this problem the organisers had
ensured that the experimental set-up was simple, inexpensive
and could be quickly and easily manufactured with uniform
quality in large numbers ( ~ 150) so that every student faced the
same level of difficulty in conducting the experiment. There
were no delicate components and sufficient length of wire was
provided for replacement in case a careless student broke it
while assembling the torsion pendulum. The problem chosen
was based on secondary and higher secondary school level
physics. It tested the skills of the students (a) to assemble the
experimental set-up, (b) to make measurements using only the
given instruments, (c) to present the collected data appropriately
in the form of tables, (d) to derive the necessary equations for
estimating physical quantities from experimentally determined
data, (e) in choosing the best way of obtaining the results with
least errors and (f) in the estimation of errors to say something
about the reliability of the conclusions. (In fact more weightage
was given to the estimation of errors than to the actual obtained
value.)
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N
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Figure 1. Sketch of apparatus with horizonfDl torsion
axis, C, C2 • w-wir4 ov
vertical.
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PositivesIde, ~~
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Apparatus
Figure 1 shows the experimental set-up. The pendulum has an
outer cylinder A of mass MI. A is not longitudinally uniform
(part of it is hollow) and has hole 0 at one end. Through the hole
a steel wire passes. After clamping the steel wire at C l' the
handle H is used to pull and keep the wire under tension before
clamping at C 2" A is locked to the wire using the 'sunk' screw S 1.
Through the other end of A a uniform threaded inner rod B of
length 1 and mass M z can be screwed in and out along the
cylinder axis, to vary the length of the pendulum x (measured
with respect to 0) between Xl and xz. The inner rod cannot be
removed from the outer rod nor can it be completely screwed
inside. Asmall hex nut (hn) is used to lock the position of the
inner rod every time one moves the threaded rod. (The mass of
the hex nut is included in M 1 and is considered to be a part of the
outer cylinder.) The direction along the steel wire (C 1- C z) is the
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CLASSROOM
torsion axis and O-X is along the axis of the cylinders. When OX is along O-N the 'elastic restoring torque' is zero. The direction
of the gravitational force is of course the vertical.
Theory
The experimental problem required one to study the torsional
oscillations of the pendulum (i) with the torsion axis of the
pendulum along the vertical direction (not shown in the figure)
and (ii) with the torsion axis in the horizontal plane (as shown in
Figure 1).
The first case is quite simple and is a standard simple harmonic
motion problem. Here, a displacement of the pendulum in the
horizontal plane will twist the wire resulting in an increase in
the potential energy Ue associated with the 'shear elasticity' of
the wire. The elastic torque tries to restore the pendulum to its
original position. For small angular twists the elastic restoring
torque Te' is proportional to £)h' the angle between the cylinder
axis O-X and the direction O-N.
(1)
where K is the torsional constant of elasticity. If / is the
moment of inertia of the 'pendulum', the equation of motion is
similar to that for a simple harmonic motion and the expression
for the period (= 'l') of torsional oscillations is
r (x)
=
2n (/ / K)0.s.
(2)
The second case is more complex. With the apparatus set-up as
shown in Figure 1, if the pendulum is displaced in a plane
perpendicular to the wire in the plane of the paper, not only is
there an increase in the elastic potential energy but also there is
a change in the gravitational potential energy. The expressions
for the restoring torque T and the potential energy U in this case
can be written as follows.
T
= -K (£)- £)0) + (M l + M) g R(x) Sin£),
(3)
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v (0) =
- fTdO =
0.5 K (0- ( 0 )2 + (M l + M) g R(x) CosO,
(4)
after choosing the constant of integration as the reference point
of zero energy with respect to which the potential energies are
given. 0 is the angle made by the pendulum axis O-X with the
vertical, 00 is the angle between O-N and the vertical and R(x) is
the distance of the centre of mass of the pendulum with respect
to O. Depending on the extent of insertion of the inner rod into
the outer cylinder, x and R(x) vary.
The expression for V(O) shows some interesting features: for
-n/2 < 0 < n/2, the second term of V( 0) decreases with increase
in I0 I (i.e. on either side of the vertical), whereas the first term
decreases until 0 = 0 0 and increases on further increasing 0 on
the positive side of the vertical or always increases with its
increase on the negative side of the vertical. Thus, on the
positive side of the vertical we will always get a 0 = 0 e for which
V( 0) is a minimum. But on the negative side of the vertical we
may get a minimum only if the second term is sufficiently large.
Figure 2 shows plots of U*(O) (= V(O) / {(M l + M 2)gR(x)}) for
two choices of f3 (f3 = K/{2(M l + M 2 )gR(x)}). The doubling of
the potential energy minimum is. seen for a sufficiently small
1.4
80
= 0.02
11'
2
"c:l
1.2
>-
~
..0
0
.£
ctI
....... 1.0
* ::>
0.8
_ _----L_ _ _- - ' -_ _ _- ' - -_ _- - - - '
-1
-2
o
2
~
8 in radians
Figure 2.
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value of f3 and is termed 'bifurcation'. The 'bifurcation' can be
observe~ only if the 'symmetry of the system is broken' by
having a non zero 8 o' This sort of bifurcation is seen in particle
physics and statistical mechanics with various kinds of symmetry
breaking. We note that for the set-up of Figure 1, the 'minimum'
on the positive side of the vertical is more stable, located farther
from the vertical than the 'minimum' on the negative side. With
the set-up shown in Figure 1, we see the existence of two equilibrium positions ('bifurcation') by loading the free end of the
threaded rod with a heavy nut so that both Ml + M2 and R(x)
increase. At the equilibrium position corresponding to 8 = 8 e'
the net restoring torque is zero and hence using (3) we can write,
(5)
The equation of motion of the pendulum can be written as
follows:
Since the equilibrium position does not change with time
d 2 8 e /dt2 = 0 and (6) can be rewritten as follows using 8 =
8e
+ 8d •
J(x)(d 2 8d /dt 2)=-K (8e + 8d - 80) + (M l + M)g R(x) Sin(8e + 8d)
= - K (8e - ( 0 ) + (M l + M 1 ) g R(x) Sin8e Cos8d
-K 8d + (M l + M 2 )g R(x) Cos8e Sin8d .
For small deviations 8d from the equilibrium position, Cos8d :::::
1 and Sin8d ::::: 8d and we can use (5) to cancel the first two terms.
Equation 7 is also similar to that for simple harmonic motion.
Interestingly the period of oscillation r depends on the
equilibrium angle 8.e
r=2n[l(x)/{K-(M 1 +M)gR(x)Cos8)]0.s.
(8)
Since x can be varied, we must also know how R(x) and lex) vary
with x. Both the cylinders constituting the pendulum are
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laterally uniform. Therefore, we can take their centres of mass
to lie on the pendulum axis O-X. Also, since the inner cylinder
part is uniform its centre of mass will be at its centre, i.e. at a
distance R2 (= x -1/2) from O. If Rl is the distance of the centre
of mass of the outer cylinder part from 0, R (= the distance of
the centre of mass of the pendulum as a whole from 0) can be
written as follows making use of the expression for R2 :
R(x)
= {R 1 Ml / (M 1 + M 2) -I M2 / 2(M 1 + M 2)} +
x M2 / (M 1 + M)
(9)
The moment of inertia of the pendulum I(x) is the sum of the
moments of inertia of the outer cylinder part (= II) and the inner
cylinder part (= I). Since the outer cylinder is fixed II does not
vary with x. Assuming the lateral dimensions of the inner
cylinder to be negligible we can write the following expressions
for 12 (using the parallel axis theorem) and I(x) ( = II + 12),
12 = Mi 2 / 12 + M2 (x -l/2)2
I(x) = II + (Mi 2 / 3) - x Mi + x 2 M2
(10)
Experiments
The main aim of the experiment was to determine the periods of
oscillation for different choices of x for the torsion pendulum
assembled with its torsion axis in the horizontal plane and then
to compare it with the functional form expected using the theoretical equation relevant for that specific pendulum. Only
the total mass Ml + M2 (~ 40 gms) was given as data. The
measurements had to be carried out using only some adhesive
tape (if necessary), aT-shaped rod, millimeter graph papers, a
timer, a graduated ruler and a right triangular plate. No weighing
scale was provided and the inner cylinder could not be separated
from the outer cylinder. Since only a non-programmable
scientific calculator was allowed, the slopes and intercepts of
linear plots could not be determined by feeding the data to the
calculator and running a least square analysis programme.
The problem in the above form seems a bit difficult for high
school students. Therefore, at the Olympiad the problem was
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divided into several sub-parts and the student was asked to
follow the 'steps' given in the instruction sheet. Thus, the
student was asked first to write an equation for R(x) as a function
of x and the parameters Ml' M 2, Rl and I, as in (9). Then the
student had to measure R(x) for several values of x (at least 3)
(obviously before attaching the pendulum to the steel wire).
After preparing suitable tabular columns, by the screw like
motion of the inner threaded rod, convenient and nearly equally
spaced values x could be chosen. The pendulum is balanced on
a T -shaped rod. The distance of the point of balance from 0
gives R(x). Equation (9) tells us that a plot of R(x) against x is
going to be linear and the slope of the straight line plot can be
used to get M/(M 1 + M 2 ). Since the total mass is known, the
masses M 1 and M 2 can be determined making use of the slope of
the plot.
The third step required the student to write the equation for the
total moment of inertia I(x) as a function of x and the parameters
M 2, II and 1as in (10). In the fourth step with the torsion axis in
the horizontal plane the pendulum equation of motion had to be
obtained as a function of 8, x, K, 80, Ml' M 2, I(x) and R(x) as in
(6).
In the fifth step the pendulum had to be assembled with its
torsion axis in the horizontal plane and K had to be estimated
from the equilibrium angle 8.e This requires one to first write
equation (5). From this equation it is obvious if some how we set
up 80 = 0, the equation will be satisfied for 8e = whatever may
be the value of x and we will not be able to get K by plotting the
data. Therefore, it is prudent to fix the pendulum to the wire
(using S 1) with a substantial value for 80 , The instructions given
also say that the equilibrium angle 8e should substantially deviate
from the vertical and should be measured for five or more values
of x. As no protractor was available, 8e had to be determined
from Cos8e which could be obtained by measuring h, the height
of the tip of the pendulum from the level of 0 (cf. Figure 1) for
the chosen x and using Cos8e = h/x. h could be measured using
the right triangular plate and the graduated ruler. Alternatively,
°
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one could determine Sin Be (= p/x) by measuring p, the horizontal distance between 0 and the topmost point of the pendulum
(cf. Figure 1). P could be measured by fixing a graph paper on the
table using adhesive tape and taking projections of 0 and the tip
of the pendulum on this paper with the help of the right
triangular plate and the ruler.
The next step is to use the data collected (as described above) to
determine K with the help of (5). It is important to note that
the analysis is easier if the data can be represented by a linear
equation. Equation (5) can be considered to be a linear equation
of the type a y = b z + c , where y (= R(x)SinB) and z (= Be) are
variables, a (= (M l + M)g), b (= K) and c (= - KBo) are
constants. Using the experimentally determined Be for n choices
of x we can set up n such equations. R(x) needed for this purpose,
if not already determined could be calculated with the help of
the slope and intercept of the earlier plot corresponding to (9).
By solving these simultaneous equations we get b/a and cia using
which K and Bo can be determined. The solutions of the
simultaneous equations can be obtained algebraically by taking
two equations at a time. Choosing different pairs of such equations we can get several estimates of K and Bo and also the
average of these estimates. However, such a method is not only
time consuming but may also lead to greater errors in the quantities determined. A graphical method is therefore preferred.
The simultaneous equations can be solved graphically by plotting y against z. Since a, band c are constants, ideally this plot
should be a straight line. Therefore, the slope and intercept of
the straight line which deviates least from the experimental
points of this plot give estimates of b/a and cia, respectively from
which K and 80 can be determined. Nearly 42.5% of the marks
were allotted for the fifth and sixth steps indicating the importance of careful choice of x, careful measurements and the plot in
these steps.
In the seventh step the torsional pendulum is set up with its
torsion axis along the vertical. With such an arrangement]l and
1 had to be estimated from the periods of torsional oscillations r
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determined for several values of x (at least 5). For this situation
the expression for r is given by (2). Equation 2 can also be made
to look like a linear equation by using (10) and rewriting it as
Note that it is not physically possible to have x = 0 or I. After
rearrangement, (11) also can be recognised to be a linear equation,
y = b z + c, with a new set of variables: y = (K ,.,;z / 4n2 - X Z M 1 ),
z = x and constants: b = -Mzl andc = (11 + M1lz /3). Again the
linear simultaneous equations set up using the experimental
data can be solved algebraically or graphically. In the graphical
method, the slope of the straight line which deviates least from
the experimental plots of y against z give -M z I. This can be
used to get I since M z is already determined as described above.
Equating the intercept of the straight line plot to (11 + M z [2/3)
we can then determine 11 also.
At the end of the seventh step all the system parameters needed
to predict the period of torsional oscillations as ~ function of x,
with the torsion axis in the horizontal plane would have been
determined, see (8). The eighth and final step is to determine r
as a function of x in this configuration of the pendulum and
discuss whether it is a decreasing or increasing function of x.
This can be compared with the r predicted using (8).
It was also necessary to discuss the reliability of the results
obtained. The obvious thing to do is to obtain error estimates
from the plots and use them in the discussion. It would be
interesting to discuss the extent of errors which might have been
introduced by the assumptions used in the analysis. One such
assumption pertains to the tension in the steel wire since we do
not have precise control over this factor. Is it possible that K
depends on the tension to which the steel wire is subjected
either during the experiment or at the time of setting up the
apparatus? Ifwe had stretched the wire close to the elastic limit
or beyond the elastic limit, can K and e be affected? To what
extent do these parameters reflect the 'history' of the wire and
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reflect the 'fatigue'? Tom Morfett of England pondered about
such· questions and got the special prize for the 'most original
experimental solution' even though he did not get 100% marks.
It was not impossible to score 100% in this examination.
Konstantin Kravstov of Russia showed this by scoring 100%
marks. Finally, one may also ponder over other practical
applications of such a set-up. Since the time period of the
pendulum depends on the inclination of torsion axis, by
measuring 'f accurately using modern electronic timers one can
determine the direction of the 'vertical' and the magnitude of g.
Such measurements may be of use in aircraft and artificial
gravity experiments.
Ingenhouszian Motion?
What we call Brownian motion was first observed by a
Dutch pl~ysician, Jan Ingenhousz (in 1785), by looking at
trajectories of finely divided charcoal on the surface of
alcohol. He was also apparently the court physician of
Maria Theresa, for which he received the princely sum of
5000 gold gulden, thus giving him time off to study the
motion of charcoal on alcohol. Incidentally, he is probably
better known in the biological community as the person
who first noted that sunlight was essential for the process
by which plants converted carbon dioxide to oxygen, or
photosynthesis. He also vaccinated the family of George
III of England against smallpox. A reference is 'Dictionary
of Scientific Biology, ed. C C Gillespie, Scribners, NY,
(1973), p.11. Perhaps Ingenhousz did not publish in a
refereed journal.
Gautam Menon
Institute for Mathematical Sciences, Chennai
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