Physical pendulum experiments to enhance the understanding of moments of inertia and

simple harmonic motion

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Physical pendulum experimentsto enhance the understanding ofmoments of inertia and simpleharmonic motionTim H Richardson and Stuart A Brittle

Department of Physics, University of Sheffield, Sheffield, UK

E-mail: t.richardson@sheffield.ac.uk

AbstractThis paper describes a set of experiments aimed at overcoming some of thedifficulties experienced by students learning about the topics of momentsof inertia and simple harmonic motion, both of which are often perceivedto be complex topics amongst students during their first-year universitycourses. By combining both subjects in a discussion about physical pendula,in which the oscillation time periods for the periodic motion of severalobjects (a tennis ball, a thin beam, a hoop and a solid disc) are measuredand compared, students are able to understand both topics at a higher leveland also experience the synergistic effect of combining two or more physicsthemes in order to accelerate their learning whilst simultaneously raisingtheir motivation. Special attention is given to the ‘ball and stick’ pendulumin which a block of material (treated as a point mass) can be moved along ashaft to create a composite pendulum whose time period exhibits a minimumvalue at a certain separation between the block and the rotation axis.

Introduction

Two topics that students in the age group 16–19find challenging, be they A-level students or Level1 university students, are oscillations and thephysics of rotation. The former is taught withinmost A-level specifications in the UK [1] in theform of simple harmonic motion (SHM) but nowin a far more qualitative manner than in the 1980s,for example; at Level 1 in most university physicsprogrammes worldwide, SHM is covered in theform of the simple (point mass) pendulum andthe physical pendulum [2]. The latter requiresan appreciation of the concept of moment ofinertia, which again is perceived as a difficult

subject to understand. The physics of rotation,more generally, is covered only at a very basiclevel within A-level syllabuses [3]; in the firstyear of university courses, the idea of momentof inertia is most commonly introduced initiallyin the context of point masses separated from anaxis of rotation around which they are revolving.At a deeper level, this subject is developed inthe context of the rotation of a mass distributionwhich itself defines the geometry (shape) of theobject, and thus provides an appreciation of thefact that the moment of inertia, and thereforerotational kinetic energy, of an object is dependenton this mass distribution rather than simply itsmass.

0031-9120/12/050537+08$33.00 c© 2012 IOP Publishing Ltd P H Y S I C S E D U C A T I O N 47 (5) 537

T H Richardson and S A Brittle

Here, a demonstration lecture is described inwhich the authors bring these two topics togetherin order to illustrate practically how the timeperiod of oscillation of various objects (differentmass distributions) can be accurately predictedfrom an understanding of the fundamental physicsof the simple and the physical pendulum. Fromthe basic measurement of the oscillation timeperiod of a simple pendulum of length L whichconsists of a tennis ball tethered to a lengthof string, the oscillation periods, T , for aswinging rod, hoop and solid disc are predicted.Subsequent measurement of the latter three valuesof T are found to agree very well with thosepredicted values, so demonstrating the value ofthe physics learnt, and how a deeper, synergisticunderstanding of both topics can be gained.

Background to the simple pendulum andthe physical pendulumA laboratory simple pendulum can be made mosteasily using a length, L, of ordinary shop-boughtstring and a tennis ball. A triple knot tied at theend of the string is forced into a very small slitcut in the rubber of the tennis ball to form thependulum ‘bob’. The free end of the string istied to a clamp which forms the pivot or axis ofrotation. Figure 1 shows a schematic diagram ofthe pendulum.

In the figure, the bob is shown with anangular displacement, θ , from the vertical; therestoring force which causes the tennis ball tomove towards its equilibrium position is given byFres, where:

Fres = −mg sin θ (1)

in which m is the mass of the bob and g isthe acceleration due to gravity. From Newton’ssecond law, this restoring force causes the linearacceleration such that:

− mg sin θ = md2x

dt2. (2)

This is valid in the case when θ is small so thatsin θ ∼ θ and x ∼ Lθ and hence:

− gθ = Ld2θ

dt2. (3)

Finally, this equation can be compared to the

general equation for SHM, d2φ

dt2=−ω2φ, in which

ω is the angular frequency given by ω = 2πTsimple

,

Figure 1.where Tsimple is the oscillation time period ofthis simple pendulum, which can be expressed inrelation to its length by:

Tsimple = 2π

√L

g. (4)

Theoretically, this equation is valid when thependulum bob is a point mass separated fromthe rotation axis by a distance, L. In practice,for lengths of L greater than around 50 cm thetennis ball acts effectively as a point mass, inthe context of this experiment. Educationally, it isimportant to emphasize that the mass itself of thependulum bob does not appear in the expressionfor its time period. Students initially find this adifficult concept.

A physical pendulum in which the bob is nowan object whose mass is distributed over a certaingeometry to give the object its characteristic shapecan be analysed in a similar manner, as supportedby figure 2. Here, the restoring torque, τ , can bewritten as:

τ = −mg sin θd (5)

in which the distance d is shown in figure 2. For

538 P H Y S I C S E D U C A T I O N September 2012

Enhancing the understanding of moments of inertia and simple harmonic motion

Figure 2.small angle oscillations, equation (5) reduces to:

τ = −(mgd)θ. (6)

From elementary rotational kinematics, thetorque causes the angular acceleration via τ = Iα,where I is the moment of inertia of the oscillatingobject and α is the angular acceleration givenby α = d2θ

dt2. Thus, the equation for SHM of the

physical pendulum can be written as:

Id2θ

dt2= −(mgd)θ (7)

from which the oscillation period for the objectwhose mass is distributed over its own particulargeometry or shape is:

Tphysical = 2π

√I

mgd. (8)

In equation (8), the mass m is the total massof the object and I is the total moment of inertiaof the object. Generally, the moment of inertia isdefined as I =

∑imir2

i for all point masses mi thatcomprise the object. Thus, for a point mass, M,revolving around an axis of rotation a distanced away, the moment of inertia is simply Md2.However, for any object for which the mass isdistributed over a well-defined region of space(e.g. a beam), the more general integral definition

of moment of inertia is required, which is I =∫r2 dM. Reference to texts such as [2] provides

the moments of inertia expressions required forthe four objects discussed in this paper.

Furthermore, if the oscillating body iscomposed of two of more contributing masses (forexample, a metal rod and a metal block throughwhich the rod was placed), then both massescontribute to the total mass, m, and both momentsof inertia would contribute to I. This will becomeimportant later in the paper.

The case of the simple pendulum can be seento fit the general, physical pendulum analysis. Ifthe tennis ball is treated as a point mass, thenits moment of inertia, Iball, is simply mL2, but asthe pendulum length is also the distance betweenthe tennis ball centre and the rotation axis (pivot),L = d, and so Iball = md2. Thus, this leads back tothe formula for the oscillation time period of theball, Tball, which is that of the simple pendulum:

Tball = 2π

√md2

mgd= 2π

√d

g= 2π

√L

g. (9)

Furthermore, equation (8) allows the oscilla-tion time period to be predicted for an oscillatingobject of any shape, provided its moment ofinertia is known. From the physics educationpoint of view, this facilitates an excellent way oflinking together two topics (moment of inertiaand SHM) with a view to providing a lesscompartmentalized learning experience, and alsothe way in which mathematics—albeit at a simplelevel initially—is used as a tool for modellingreal physical behaviour that can be observed withthe intention of putting theory to the test, as willbecome evident later.

Oscillation time period measurements forobjects of different shapesMost commonly, 16–19 year old students ex-perience as a classroom demonstration only theoscillation of a simple pendulum. Most willmeasure the time taken for several completeoscillations to occur in order to obtain an accuratevalue for Tsimple. From this value, hopefully theyare encouraged to use equation (4) to calculate avalue for the acceleration due to gravity for theEarth (9.81 m s−2) [4].

However, with very little additional expenseand effort, this learning experience can beextended and enhanced dramatically.

September 2012 P H Y S I C S E D U C A T I O N 539

T H Richardson and S A Brittle

Hoop or ring

A circular aluminum hoop made with analuminum rod of diameter ∼8 mm is the firstgeometry discussed in which the mass of theobject is distributed over a well-defined region ofspace. The diameter of the hoop is chosen to bethe same as the length of the simple pendulumcomprising the tennis ball for reasons which willbecome clear shortly, and is considerably largerthan the aluminum rod diameter. The momentof inertia of a hoop when rotating about an axisthrough its centre is simply mR2, where R isits radius. From a teaching point of view, thiscould either be accepted or derived [5]. However,when the hoop is hung from the clamp and set tooscillate, the axis of rotation is at one edge of thehoop rather than at its centre, and so the parallelaxis theorem (PAT) [6] is required to account forthis displacement, leading to a moment of inertia,Ihoop, given by:

Ihoop = mR2+ md2

= 2mR2 (10)

since the distance between the centre of mass andthe rotation axis, d, is also equal to the radius ofthe hoop, R. Thus, substitution of equation (10)into (8) yields the time period of oscillation of thehoop, given by:

Thoop = 2π

√2mR2

mgR= 2π

√2R

g. (11)

Since the diameter, 2R, of the hoop has beenchosen to be equal to the length of the simplependulum consisting of the string and tennis ball,the oscillation time period of the hoop is expectedto be equal to Tball. Thus, the theory makes thisfirst prediction, Thoop = Tball which is verifiedexperimentally later.

Solid disc

A solid circular disc of radius R made fromplywood forms the second object whose massis distributed over a region of space. The exactmaterial from which the disc is made is not criticalsince again its mass itself will not appear in thefinal equation for the time period of oscillation.The moment of inertia of a solid disc about anaxis through its centre is 1

2 mR2 [7], but of coursewhen suspended from the clamp its rotation axisis displaced a distance d = R from the centre of

mass, so that its moment of inertia then is obtainedusing the PAT as:

Idisc =12 mR2

+ md2=

32 mR2. (12)

Substitution of equation (13) into (8) pro-vides the time period of the oscillating disc:

Tdisc = 2π

√(3/2)mR2

mgR= 2π

√(3/2)R

g. (13)

Since the radius of the disc, R = L/2 (whereL is the length of the simple pendulum comprisingthe tennis ball), the relation between the value ofTdisc and Tball is simply:

Tdisc =√

3/4Tball. (14)

Thus, the time period of the disc is predictedtheoretically to be 86.6% of that of the tennis ballpendulum.

Solid beam

A solid beam, whose length is the same as thelength of the tennis ball pendulum, is suspendedfrom one end via a small hole drilled as closeto the end as possible. This forms the thirddistributed mass in this investigation. A similaranalysis to the hoop and the disc leads to amoment of inertia of 1

3 mL2 [8]. Substitution intoequation (8) leads to the time period for theoscillating beam:

Tbeam = 2π

√(1/3)mL2

mg(L/2)= 2π

√(2/3)L

g. (15)

In terms of the time period of the tennis ballpendulum, the time period of the beam is givenby:

Tbeam =√

2/3Tball. (16)

Thus, the time period of the beam is predictedtheoretically to be 81.6% of that of the tennis ballpendulum.

Measurements

The analysis above allows the time periods of allfour pendula to be predicted according to:

Tball = Thoop = 1.155Tdisc = 1.225Tbeam. (17)

It is now necessary to measure experimentallythe time periods and compare them to the

540 P H Y S I C S E D U C A T I O N September 2012

Enhancing the understanding of moments of inertia and simple harmonic motion

Table 1. Oscillation time period measurements for thetennis ball, hoop, disc and beam pendula.

Pendulum T30 (s) T (s) Tball/Tshape

Tennis ball 52.88 1.76 1.00Hoop 52.63 1.75 0.99Disc 45.78 1.53 1.16Beam 43.20 1.44 1.22

theoretically predicted values. This is done usinga regular laboratory stopclock to time ∼20–30full swings of each pendulum—students shouldbe encouraged to do this∼3 times and to calculateaverage values. The length of the pendulum used(tennis ball and string) was 0.77 m; thus, atheoretical time period of 1.760 s is predicted.Table 1 shows the experimental data collectedfrom such an investigation and confirms thatthe relationships between the time periods foreach pendulum are consistent with the theoreticalpredictions to an accuracy of <2%. The finalcolumn in the table gives the ratio Tball/Tshape,where Tshape is the time period for each of thefour objects used to make the pendulum. Thesemeasured values are in close agreement with thoseseen in equation (17) above, as expected.

Having investigated these four mass distri-butions, students should grasp the fundamentalconcepts that (a) the time period of oscillationfor a pendulum undergoing SHM depends onits moment of inertia and (b) the moment ofinertia itself depends on the shape or geometryof the object over which its mass is distributed.Beyond this, the use of simple mathematicalsubstitution is emphasized as an important toolin understanding physical principles and physicalobservations. This latter aspect is taking to a muchhigher level in the section ‘Time period modellingfor the ball and stick pendulum’.

Time period modelling for the ball andstick pendulumDevelopment of time period expression

The basic understanding of the oscillationbehaviour of simple objects, as described above,can be developed significantly if a pendulumis made up of more than one object. Figure 3shows a photograph and a schematic diagramof a ball and stick pendulum in which theshaft of the pendulum is made of a 9 mm

Figure 3.diameter wooden dowel (the ‘stick’) onto whichis threaded a wooden block (the ‘ball’), which canbe positioned anywhere along the length of theshaft.

The oscillations of the shaft are governed bythe fact that its moment of inertia is equivalent tothe beam described above (Istick =

13µL2), where

µ is the mass of the dowel shaft and L is its length.However, the oscillations of the wooden block canbe modelled in the same way that the tennis ball(point mass) was modelled earlier (Iblock = Ml2),where M is the mass of the block and l is thedistance between the centre of mass of the blockand the axis of rotation (pivot).

Thus, the total moment of inertia that isrequired for substitution into equation (8) in orderto obtain the time period for the ball and stickpendulum is Itotal, where

Itotal = Ml2 +µL2

3. (18)

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T H Richardson and S A Brittle

Figure 4.However, the distance d in equation (8)is also required—this is the separation of therotation axis and the centre of mass of the entiresystem. This can be found by considering figure 4in which the positions of the centres of mass ofboth the stick and the ball are indicated, and alsothe centre of mass of the entire system is shownat point C. Taking moments about C reveals thefollowing equation:

M(d − l) = µ

(L

2− d

), (19)

which can be rearranged to yield an expression ford itself:

d =

(Ml+ µL

2

M + µ

). (20)

At this point, it is instructive to testequation (20) to ensure that it makes physicalsense for specific values of l. Firstly, if the ballis temporarily ignored (i.e. the mass M is giventhe value zero), it can be seen that the valueof d is consistent with that of the stick alone,d = L

2 . Similarly, if the ball is reintroduced andpositioned at l = L

2 , then the value of d is alsoconsistent with the centre of mass of the systemoccurring at the half-way point, d = L

2 . Finally,if we set the mass of the stick to equal the massof the block (M = m) and position the ball atl = L, then d takes the value d = 3L

4 , which is,as expected, half-way between the centre of massof the stick and the centre of mass of the ball.It is very useful that students develop a cultureof testing their developed models during theirevolution so as to gain confidence in their analysisor to identify errors early.

Combining equations (8), (18) and (20) leadsto an expression for the time period of the balland stick pendulum in which the ball is placed ata distance l from the pivot, given by:

Tbs = 2π

√√√√ Ml2 + µL2

3

g(Ml+ µL2 ). (21)

Again, it is now important to test thisequation for specific conditions. Firstly, if thestick is ignored (i.e. the value of µ is set to zero)then for a value l = L, the time period is the sameas that for a simple (point mass) pendulum asexpected. Secondly, if the ball is ignored (i.e. M =0) then for l = L, the time period is that found

for the beam, Tbeam = 2π√(2/3)L

g . Finally, if themasses of the ball and stick are equal and the ballis positioned at l = L

2 , the time period is given by:

Tbs = 2π

√√√√ M(L2 )

2 +µL2

3

g(M(L2 )+

µL2 )= 2π

õL2( 1

4 +13 )

gµL

= 2π

√7

12L

g. (22)

Thus, with suitable apparatus (M = m, l = L)it can be verified that the time period of the balland stick pendulum under these conditions is afactor of 0.764 times that of a simple pendulumof the same length.

Measurements and results

Experimentally, the time period Tbs is measuredas a function of the distance between the pivotand the ball, l. Figure 5 shows a plot of Tbsversus l for measured data and also the theoreticalcurve based on equation (21) using the actualvalues of M and µ for the real pendulum. TheTbs value is obtained by timing the durationof 30 complete oscillations of the pendulumfor each value of l. It is clear from the plotthat a minimum in Tbs occurs at a specificvalue of l. The realization of this feature wouldtypically not be immediately obvious qualitativelyif students were asked to predict the behaviour ofthe pendulum prior to any mathematical analysis.Furthermore, few students would recognize thisminimum upon simple inspection (without furtheranalysis) of equation (21), yet it is very clear inthe experimental measurements.

This provides the exciting opportunity here todemonstrate to students the role of calculus [9]

542 P H Y S I C S E D U C A T I O N September 2012

Enhancing the understanding of moments of inertia and simple harmonic motion

Figure 5.in (a) identifying the existence of a turningpoint in the Tbs versus l curve, (b) finding theexact position of the ball needed to observe theminimum value of Tbs and (c) thus demonstratingthe power of using mathematics as a tool foraiding the understanding of the physics.

The role of calculus

In order to make equation (21) slightly moremanageable, firstly the constants α and β are

introduced, where α = µL2

3 and β =µL2 , thus

allowing equation (21) to be written as:

Tbs = 2π

√Ml2 + α

g(Ml+ β). (23)

Thus, in order to find the minimum value ofTbs, its first derivative with respect to l, dT/dl,must be found. Solving dT/dl = 0 will generatethe value of l for which Tbs takes its minimumvalue.

Some simplification to the analysis can bemade here since equation (23) can be recast as:

Tbs =2π

g1/2

(Ml2 + α)1/2

(Ml+ β)1/2. (24)

A constant γ = 2πg1/2 can be introduced as

can the functions p(l) and q(l), where p(l) =(Ml2 + α)1/2 and q(l) = (Ml + β)1/2, allowingthe following expression to be written:

Tbs = γp(l)

q(l). (25)

The quotient p(l)/q(l) can be differentiatedusing the quotient rule for differentiation such thatthe first derivative of Tbs is:

dTbs

dl= γ

{q(l)p′(l)− p(l)q′(l)

[q(l)]2

}. (26)

It can now be seen that the condition dTbsdl =

0 is met provided q(l)p′(l) − p(l)q′(l) = 0 andthus there is no need to evaluate every term inequation (26). The derivatives p′(l) and q′(l) aregiven by:

p′(l) =12(Ml2 + α)−1/22Ml =

Ml

(Ml2 + α)1/2

q′(l) =12(Ml+ β)−1/2M =

M

2(Ml+ β)1/2.

(27)

Thus:

q(l)p′(l)− p(l)q′(l) =(Ml+ β)1/2Ml

(Ml2 + α)1/2

−(Ml2 + α)1/2M

2(Ml+ β)1/2(28)

and for the condition q(l)p′(l) − p(l)q′(l) = 0, itcan be seen that:

(Ml+ β)2l = (Ml2 + α). (29)

Rearranging (29) yields a simple quadraticequation Ml2 + 2βl− α = 0 which can be solvedfor l as below:

l =−2β ±

√4β2 + 4Mα

2M. (30)

Only the positive root is meaningful in the contextof this experiment given by l+ where:

l+ =−β +

√β2 +Mα

M. (31)

Again, this equation can be tested; firstly, thevalue of l+ can be confirmed using the particularvalues of M and µ for the actual pendulum used.However, the mass of the ball can be changed

September 2012 P H Y S I C S E D U C A T I O N 543

T H Richardson and S A Brittle

Figure 6.simply by using different materials to make theball. In this way, the value of l+ can be predictedtheoretically for different values of M and thenconfirmed experimentally. For the ball and stickpendulum used in the development of this paper(L = 1.0 m), M = 0.061 kg and µ = 0.025 kg,thus α = 0.0088 kg m2 and β = 0.0128 kg m.Equation (31) predicts a value of l+ of 0.223 m;inspection of figure 5 shows that experimentallythe value obtained is 0.22 ± 0.01 m in verygood agreement with theory. Figure 6 showshow l+ is expected to vary as a function ofthe mass of the ball for a value of stick massof µ = 0.025 kg. Several different balls weremade using materials of different density so thata few experimental data points could be added toillustrate the excellent fit between experiment andtheory that can be seen in the plot.

SummaryThese experiments are useful in demonstrating(a) how basic theory can be developed to explainreal observations made by the physicist, (b)the importance of basic mathematical tools insupporting the development of the physical theoryand (c) the fact that physics education at this

level does not need to be reliant on expensiveapparatus. This latter point emphasizes that theprogramme of experiments in this paper couldbe followed in any country, however rich orpoor, throughout the world, and, as a result,students can expect to acquire several skillsimportant to theory development, experimentdesign, measurement and analysis.

Received 2 February 2012, in final form 27 February 2012doi:10.1088/0031-9120/47/5/537

References[1] Pople S 2009 AS and A Level Physics Through

Diagrams: Oxford Revision Guides (Oxford:Oxford University Press)

[2] Young and Freedman 2011 University Physicswith Modern Physics 13th edn (Harlow:Pearson)

[3] Lowe T L and Rounce J F 2002 Calculations forA-Level Physics Fourth Edition 4th edn(Cheltenham: Nelson Thornes)

[4] Baker G L and Blackburn J A 2008 ThePendulum: A Case Study in Physics (Oxford:Oxford University Press)

[5] Muthuraman S 2011 Engineering Statics: A Studyof the Basics of Statics, Centre of Gravity andMoment of Inertia (Saarbrucken: VDM Verlag)

[6] Hulse R and Cain J 2000 Structural Mechanics2nd edn (Basingstoke: Palgrave Macmillan)

[7] Benson H 1996 University Physics (New York:Wiley)

[8] Knight R D 2007 Physics for Scientists andEngineers: A Strategic Approach with ModernPhysics (Harlow: Pearson)

[9] Spivak M 2006 Calculus 3rd edn (Cambridge:Cambridge University Press)

Tim Richardson is a Reader inNano-Science at the University ofSheffield (UK). His research interestshave centred on nano-materialsengineering, but have recently broadenedto encompass educational physics. Timobtained his BSc in 1985 from DurhamUniversity (UK) and his PhD fromOxford University (UK) in 1989.

Stuart Brittle is a Research Associate atthe University of Sheffield (UK) workingin polymer physics. He also obtained hisBSc and PhD from the University ofSheffield in 2005 and 2009, respectively.

544 P H Y S I C S E D U C A T I O N September 2012