The kinematics of a superball bouncing between two vertical surfaces

The kinematics of a superball bouncing between two vertical surfacesBrian T. Hefner Citation: Am. J. Phys. 72, 875 (2004); doi: 10.1119/1.1701843 View online: http://dx.doi.org/10.1119/1.1701843 View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v72/i7 Published by the American Association of Physics Teachers Related ArticlesA demonstration condenser microphone Phys. Teach. 50, 508 (2012) Dashboard Videos Phys. Teach. 50, 477 (2012) Peer Assessment with Online Tools to Improve Student Modeling Phys. Teach. 50, 489 (2012) Analyzing spring pendulum phenomena with a smart-phone acceleration sensor Phys. Teach. 50, 504 (2012) Locating the Center of Gravity: The Dance of Normal and Frictional Forces Phys. Teach. 50, 456 (2012) Additional information on Am. J. Phys.Journal Homepage: http://ajp.aapt.org/ Journal Information: http://ajp.aapt.org/about/about_the_journal Top downloads: http://ajp.aapt.org/most_downloaded Information for Authors: http://ajp.dickinson.edu/Contributors/contGenInfo.html

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The kinematics of a superball bouncing between two vertical surfacesBrian T. Hefnera)

Applied Physics Laboratory, University of Washington, 1013 NE 40th Street, Seattle, Washington 98105

~Received 14 October 2002; accepted 6 February 2004!

When a superball is thrown so that it bounces between two horizontal parallel surfaces, the ball willbounce from both surfaces and then return to the person who threw it. If the surfaces are vertical anda ball is thrown downward so that it bounces between both surfaces, it also will return to the personwho threw it. In the general case of a superball bouncing between two extended vertical surfaces,the ball becomes trapped instead of falling and moves up and down the length of the channel withan oscillation frequency that depends on the ball’s moment of inertia. For this vertical channel theintroduction of a tangential coefficient of restitution produces motion that is similar to both adamped harmonic oscillator and a falling body. ©2004 American Association of Physics Teachers.

@DOI: 10.1119/1.1701843#

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I. INTRODUCTION

When a superball is thrown so that it bounces fromfloor to the underside of a surface, such as a table, itbounce back to the person who threw it instead of continuto bounce between the table and the floor@see Fig. 1~a!#.Strobel1 and Garwin2 independently analyzed this sequenof collisions and their analysis has been revisited many timin undergraduate laboratories, physics demonstrations,textbooks. This paper explores a variation on this demonstion: instead of bouncing a superball from the underside otable, what happens when the superball is thrown downa vertical channel composed of two parallel walls? If thdemonstration were performed with a steel bearing, the bing would ricochet back and forth as it falls down the chanel under the force of gravity. However, the superball seeto defy gravity and bounces back out and returns to theson who threw it, as in Fig. 1~b!.

A comparison of the ball’s path in both the horizontal avertical channels suggests that these two sets of collismay be related. I explore this similarity for the more genecase of a superball bouncing between the walls of an initely long vertical channel. The matrix description of thsuperball motion is the most useful and straightforwardproach to this problem and the equation of motion whichdeveloped in Sec. III captures the dynamics of the balboth the vertical and horizontal channels. When gravity aalong the length of the channel~the vertical channel!, the balldoes not continuously fall, but instead is trapped in a lenof the channel, repeatedly rising and falling with a fixfrequency. This frequency does not depend on gravity buthe ball’s moment of inertia. Hence the motion in both tvertical and horizontal channels is essentially the same, wgravity affecting the amplitude and the displacement ofoscillation.

You will probably find that even if the ball is thrown sthat it enters the channel, it sometimes does not quite maout. Even in this failed demonstration, however, the trajtory of the superball is unexpected and instead of immeately falling down the channel, it undergoes a decayingcillation as it descends. This motion is a consequence ofinelasticity of the collision, and hence during the collisiothere is not a perfect exchange between the linear motionthe spin. As a result, the ball may not be able to climb outhe channel, but it will certainly resist the descent.

To accurately describe the ball’s motion, it is necessary

875 Am. J. Phys.72 ~7!, July 2004 http://aapt.org/ajp

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take into account the energy losses in the tangential bouof a realistic superball. This analysis was done by Cross, wpresented a modification of Garwin’s superball model to tainto account the tangential coefficient of restitution.3 SectionII examines this modified model and its implications for tdynamics of the superball. Because of the simplicity of tcollision matrix with tangential and normal coefficientsrestitution, the equations of motion for the fully elastic sperball, which are developed in Sec. III, can be easilyplied, with only minor modifications, to the behavior ofrealistic superball. The discussion in Sec. IV examines thgeneral equations for the special case of a ball with tangtial loss, but perfect restitution normal to the surface. For tcase, the motion of the ball has the properties of bothdamped harmonic oscillator and a body falling throughresistive medium. The ball falls down the channel at a ternal velocity that depends on the tangential coefficient of rtitution and the moment of inertia of the ball. Before reacing this terminal velocity, however, the ball undergoes eitha decaying oscillation or an exponential decay, once agdepending on the restitution and the moment of inertia. Tdependence is similar to the underdamped and overdammotion of a harmonic oscillator.

II. KINEMATICS OF A PARTIALLY INELASTICSUPERBALL

The simple model of an ultraelastic rough ball or superbbegins with two assumptions about the ball during its cosion with a flat surface:1,2 ~a! the kinetic energy is conserveduring the collision, and~b! there is no slip at the point ocontact. Because the collision is assumed to occur at a pon the ball’s surface, all of the forces act through this poand the torque is zero. As a result, the angular momenabout the point of contact is conserved during the collisio

From these two conservation laws, the spin and velocparallel to the surface after the collision are expressedterms of the initial spin and velocity of the ball:

vx52~a21!

~a11!vx02

2a

~a11!c0 , ~1!

and

c522

~a11!vx01

~a21!

~a11!c0 , ~2!

875© 2004 American Association of Physics Teachers

license or copyright; see http://ajp.aapt.org/authors/copyright_permission

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wherevx is the velocity parallel to the surface,I 5amr2 is

the moment of inertia of the ball,c5r u is the peripheral

velocity of the ball’s surface,u is the rotational velocity orspin of the ball, andr is the radius of the ball. In this notation, a ball withc.0 would have backspin while a ball witc,0 would have topspin. As a result of these conservquantities, there also is a reversal of the surface velocitythe ball at the point of contact,

~vx1c!52~vx01c0!. ~3!

The change in the surface velocity of the ball is the tangtial analog of the change in the velocity perpendicular tosurface of the ball during the bounce,

vy52vy0 . ~4!

For a partially inelastic, partially smooth ball, both thassumption of energy conservation and no slip at the poincontact do not necessarily apply. In realistic collisions, thexists the possibility of sliding at the point of collision awell as a loss of tangential strain energy that is stored duthe collision.4–7 In many studies of the dynamics of systemof inelastic spheres, such as in granular flows8 and granulargases,9 the models of the particle collisions attempt to tathese rotational energy losses into account. The simpmodel involves the addition of a coefficient of tangentrestitution into the change in the surface velocity, Eq.~3!,such that there is no longer a perfect reversal of the veloof the ball’s surface at the point of contact. This simplismodel does not capture the more complicated dynamicslip and elastic restitution over the area of contact betwthe ball and the surface,5,10 but it does provide a simple extension of the original superball description to capture ad

Fig. 1. The motion of a superball thrown~a! between two parallel, horizontal surfaces and~b! between two parallel, vertical surfaces.

876 Am. J. Phys., Vol. 72, No. 7, July 2004


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tional aspects of the ball’s motion. Although the addition otangential coefficient of restitution to Garwin and Strobeoriginal equations was recently presented by Cross,3,11 it isrevisited here with particular attention to its implications fthe dynamics of the superball.

To account for energy losses associated with the changvelocity along they direction, the simplest description of thcollision is

vy52evy0 , ~5!

where 0<e<1 is the coefficient of normal restitution. Thintroduction of a coefficient of restitution into Eq.~4! sug-gests a similar modification for the change in the surfavelocity at the point of contact. Equation~3! becomes

~vx1c!52b~vx01c0!, ~6!

where21<b<1 is the coefficient of tangential restitution8

As b decreases from the fully elastic case,b51, the reboundvelocity of the point of contact decreases untilb50 forwhich no rebound occurs and there is instantaneous rolof the ball. Forb521, the ball has a perfectly smooth suface and there is no change in the spin or the velocity. Forintermediate case of21,b,0, there still is slipping be-tween the ball and the surface, but energy is assumed tlost primarily to friction. The use of a coefficient of tangetial restitution is a simplified approach to modeling the sling contact of the ball, which, in a more realistic modewould include Coulomb friction. In a more realistic modelthe collision, the energy losses would depend on bothnormal and tangential velocities of the ball.7,12,13

Even with this change in Eq.~3!, the forces on the balcontinue to act through the point of contact and the torqabout this point remains zero. Thus, angular momentumconserved while the energy is not. With conservation ofgular momentum and Eq.~6!, thex component of the veloc-ity and the spin of the ball after the collision become

vx52~ba21!

~a11!vx02

a~b11!

~a11!c0 , ~7!

and

c52~b11!

~a11!vx01

~a2b!

~a11!c0 . ~8!

Note that thex direction will always be taken as parallel tthe surface regardless of whether or not the surface is treas horizontal or vertical. If we introduce the generalized vlocity vector,

V85S vx

cvy

D , ~9!

the velocity and spin before and after the collision are relaby

V85M 8•V08 , ~10!

where the collision matrix is

M 8521

~a11! S ~ba21! a~b11! 0

~b11! 2~a2b! 0

0 0 e~a11!D .

~11!

876Brian T. Hefner


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In this matrix notation, it is easier to examine complicattrajectories of the ball such as the channel geometrysented in Sec. III.

Even without examining a complicated trajectory, the clision matrix can provide some insight into the behavior omore realistic superball. Consider what happens to a suball that is bouncing along a flat surface under the actiongravity. Because gravity is the only external force, there ischange in the horizontal velocity or spin between collisioBecause the velocity in they direction has no effect on thexcomponent of the velocity or the rotation of the ball, tmatrix equation describing the horizontal componentscomes

V5M "V0521

~a11! S ~ba21! a~b11!

~b11! 2~a2b!D S vx0

c0D .

~12!

After one bounce, the velocity vector follows from Eq.~12!.For the second bounce, the velocity vector becomesV2

5M2•V0 and for thenth bounce of the superball,Vn5Mn

•V0 , whereMn is given by

Mn51

~a11! S 11abn 2a~12bn!

2~12bn! a1bn D , ~13!

whereb[2b. For the completely elastic and rough sup

ball, b521, the multiple collision matrix reduces to

Mn5H 1 ~n50,2,4!, . . .

M , ~n51,3,5!,...,~14!

which means that after two bounces, the ball returns toinitial state. Figure 2~a! illustrates this sequence for a bawith zero initial horizontal velocity andc,0, dropped froman arbitrary height. This result also is a consequence ofsymmetry under time-reversal14,15 and an explicit expressionfor M is not necessary in order to derive this result.

For 21,b,1, becauseb8n→0 asn→`, the collisionmatrix becomes

limn→`

Mn5M ub5051

~a11! S 1 2a

21 a D . ~15!

Whenb50, it is straightforward to show thatv52c, whichmeans that the ball undergoes simple rolling at the poincontact after the first bounce. Thus for21,b,1, Eq. ~15!indicates that the ball undergoes a transition to simple rolas n→`. Figure 2~b! shows this transition for the initiaconditions used in Fig. 2~a!. This transition captures an aspect of superball motion that many find very useful; itmuch easier to pick up a dropped ball if you wait until terratic bouncing decays to simple rolling.

Finally, consider the eigenvalues and eigenvectors ofM .The first eigenvalue for this matrix isl1511 and the cor-responding eigenvector is

V(1)5S 2cc D . ~16!

This result can be understood by noting that the velocitythe point of contact for this eigenvector is (vx1c)5(2c1c)50. The point of contact is stationary with respect to tsurface, and therefore there is no exchange of energytween the translation motion parallel to the surface and



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rotational motion. This result holds for all values of the tagential restitution coefficient. The second eigenvalue isl2

52b, and the corresponding eigenvector is

V(2)5S acc D . ~17!

When b51, the ball simply rebounds back along the sampath with both the spin and velocity reversed. Likewiswhenb521, the motion of the ball is unchanged, whichexpected because the interaction with the surface is frictless slipping. The most interesting situation is whenb50. Inthis case, both the spin and horizontal velocity become zand remain so for all subsequent bounces.

III. IDEAL SUPERBALL MOTION IN A CHANNEL

To understand the dynamics of a superball thrown betwtwo vertical or horizontal surfaces, first consider the genecase of a fully elastic superball confined to a vertical chanof infinite extent. In this case, the ball has an initial velocacross the channel and bounces back and forth while graacts to pull it down the length of the channel. If we use tcollision matrix defined in Sec. II, the subsequent motionthis confined ball is described by a series of repeated maoperations. By solving this matrix expression, the ball’s mtion assumes a familiar and transparent analytic solutioncaptures the ball’s behavior. This solution also is applicato the horizontal channel motion~with g50).

The geometry of the confined superball is shown in FigThe channel is infinite in thex direction and has widthw in

Fig. 2. Motion of a superball dropped withvx05vy050 andc,0 from aheight above a flat surface. The ball has a coefficient of normal restitute50.95, and coefficients of tangential restitution,~a! b51.00 and~b! b50.90. Note that the vertical spikes in~a! are bounces for whichvx050.These spikes broaden in~b! indicating that as the inelastic ball bounces,approaches the pure rolling condition,vx52c.

877Brian T. Hefner


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the y direction. Gravity acts in the negativex direction. Theball starts in the channel at the left wall aty50 and leavesthe wall with some initial velocity,v5vx0x1vy0y, and an

initial spin, c05r u0 . Because the only external forcegravity and the ball is assumed to be completely elastic,ball’s velocity across the channel does not change. The tsit time of the ball across the channel is therefore a constDt5w/vy0 . Becausevy0 is constant, for collisions with thesides of the channel, the simplified collision matrix, givenEq. ~12!, can be used.

After thenth bounce of the ball from the side of the chanel, the ball travels across while accelerating in the negax direction. The change in velocity isDvx52gDt, wheregis the acceleration due to gravity, and the total velocity vtor just prior to the (n11) collision is

Vn2G5S vxn

cnD2S gDt

0 D . ~18!

If the (n11) collision occurs with the left side, the velocitvector after the collision becomes

Vn115M•~Vn2G!. ~19!

If the collision occurs with the right side, the coordinasystem of the ball must be rotated such that they componentof the ball’s coordinate system corresponds to the norvector of the wall as shown on the right side of Fig. 3. Throtation is necessary to properly apply the collision matrThe velocity vector at the right wall is therefore modifiedthe rotation matrix,

D25S 21 0

0 1D , ~20!

and the velocity vector after the (n12) collision becomes, inthe original coordinate system,

Vn125D2MD2•~Vn112G!. ~21!

With the starting position of the ball atn50, repeated appli-cation of Eqs.~19! and ~21! yields the velocity vector aftethe nth bounce:

Fig. 3. Coordinate system used to describe the motion of the ball withinvertical channel.



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Vn5D2nF ~MD2!nV02(

j 51

n

~21!n2 j~MD2! jGG . ~22!

Although Eq. ~22! determines the velocity and spin foany collision of the ball with the channel wall, it does ngive any immediate insight into the behavior of the baHowever, it does show that the general motion of the balthe superposition of two special cases. The first correspoto the motion of a superball in a horizontal channelg50), in which case Eq.~22! reduces to the first term,

Vn5D2n~MD2!nV0 , ~23!

while the second corresponds to the initial conditionV0

50, for which Eq.~22! reduces to

Vn52D2n(

j 51

n

~21!n2 j~MD2! jG. ~24!

To determine how the position of the ball relates to tvelocity vector, note that the height of the ball at a collisiis related to the previous height at the opposite wall by

Xn5Xn211Vn21Dt2 12 GDt, ~25!

where the position vector is

X5S xru D . ~26!

The quantityx is the position of the ball in the channel, anru is the distance along the surface of the ball which isconsidered here. If the height of the ball at the left wall isx0 ,then the position vector aftern collisions becomes

Xn5X01F (j 50

n21

V j2n

2GGDt, ~27!

whereV j is given by Eq.~22!.The motion of the confined ball using Eqs.~23!, ~24!, and

~27! is shown in Fig. 4. These results are for a solid sph(a52/5) which has a velocity across the channel such tDt5w/vy050.05 s. Figure 4~a! shows the velocity of theball along the channel both in the presence of gravity~solidline! and without gravity~dashed line!, while Fig. 4~c! showsthe position of the ball in the channel. These quantitiesplotted versus the normalized time which is found frot/Dt5t(vy0 /w) and corresponds to the number of crossinof the ball across the channel.

For the horizontal channel, the ball initially beginstravel down the channel but reverses direction, which is csistent with Refs. 1 and 2 as well as the motion shown in F1~a!. Because the channel is now infinite, however, the bmoves back and forth within the channel. For the vertichannel, the superball undergoes the same type of oscillaconsistent with the demonstration discussed earlier but ctrary to the usual expectation. Although the amplitude ofoscillation is different than in the horizontal channel, tfrequencies are the same.

To understand how the superball seems to defy gravconsider how the velocity and the spin are related to oanother as the ball travels through the channel. Figure 4~b!shows the spin of the ball for both the vertical and horizonchannels. From Fig. 4~b! it isn’t clear how the spin is relatedif at all, to the ball’s velocity. However, if the peripheravelocity in the coordinate system of each wall,cn8

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878Brian T. Hefner


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5(21)ncn , is plotted as in Fig. 5, it becomes clear that tspin also oscillates at the same frequency as the velocityout of phase with the velocity byp/2. To illustrate how therelationship between the spin and the velocity leads tooscillation in the channel, Fig. 6 shows the path of the bfor the first five bounces as well as the spin of the ball aeach collision.

In the horizontal channel, the ball starts with no spin anpositive horizontal velocity. After the first collision with thfloor, it gains topspin that is actually backspin relative to tupper surface. When it strikes the upper surface, the spsufficient to send the ball in the opposite direction andproceeds back down the channel to repeat the cycle. FFig. 6 the spin is obviously greatest when the ball is chaing direction along the channel, that is,vx50, which ex-plains the phase shift in Fig. 5~a! and is consistent with energy conservation.

In the vertical channel, the same transfer between spinvelocity occurs as the ball moves down the channel. W

Fig. 4. ~a! Vertical velocity,~b! spin,c5r u, and~c! height of the ball aftereach collision with the walls of the channel withg59.8 m/s2, vx050 m/s~solid line! andg50, vx0520.5 m/s~dashed line!.



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Fig. 5. Vertical velocity~solid line! and the spin in the coordinate systerelative to each wall,cn85(21)ncn ~dashed line! for ~a! g59.8 m/s2, vx0

50 m/s, andc050 m/s and~b! g50, vx0520.5 m/s, andc050 m/s.

Fig. 6. Path of the ball for the first five bounces in~a! the horizontal channeland ~b! the vertical channel. The direction and magnitude of the spin aeach collision is depicted by the ball and arrow next to each point of imp

879Brian T. Hefner


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the ball first strikes the wall, it gains topspin relative to tright wall, which both slows the downward velocity after thinitial collision and acts to reverse the velocity of the ballthe next collision. This sequence continues until the bgains enough spin to push itself back up the channel.spin once again decreases as the ball reaches the top ochannel and the ball begins to fall back down. The spin avelocity are both zero simultaneously at the top of the chnel, whereas in the horizontal channel, the spin is greateboth ends of the channel where the velocity is nearly ze

This different behavior is due to the difference in the oset between these two oscillations. For the horizontal chnel, the oscillation occurs aboutvx'0. In the vertical chan-nel, an examination of Fig. 5~b! indicates that the periodicmotion occurs about some non-zero steady state velocitthe ball leaves the wall with an initial velocity vector thcorresponds to this nonzero velocity, the ball will bounback and forth, colliding with the wall at the same pointeither side of the channel. This velocity vector should thefore satisfyV25V0 , or

Vss5~MD2!2Vss2~MD2!2G1MD2G. ~28!

The solution of Eq.~28! for Vss yields

Vss(n)5

1

2gDtS 1

~21!n/a D . ~29!

To understand this result, note that in order for the ballbounce from one wall and strike the other at the same heithe initial velocity must bevx051/2gDt. For the ball tocollide with the wall and travel back along the same path,velocity vector must also correspond to the second eigentor, Eq.~17!, of the collision matrix. The relation between thspin and velocity is thereforec052vx0 /a, where thechange in sign reflects the action of the rotation matrixthe collision with the right wall.

The velocity vector aboutVss is

Vn* 5S vxn*

cn*D 5Vn2Vss

(n) . ~30!

With this substitution, Eq.~22! reduces to

Vn* 5Dn2~MD2!nV0* , ~31!

which is identical to the equation of motion for the horizotal channel. This behavior is similar to the motion of a mahanging on a spring. The force of gravity introduceschange in the equilibrium position which can be removeda shift in the coordinate system. As it is for the mass ospring system, the frequency of oscillation of the ball is tsame regardless of the presence of gravity.

With these considerations in mind, the trial solution to E~31! becomes

Vn* 5S A cos~nvDt1f!

~21!nB sin~nvDt1f! D , ~32!

where the choice of cosine and sine reflects the phasebetween the two components. The amplitude of the velocomponent isA, the amplitude of the spin component isB,andv is the radial frequency of the oscillation. If we substute Eq.~32! into Eq. ~31!, we obtain the amplitude ratio,

A

B5Aa, ~33!



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which also follows from energy conservation. The frequenof the oscillation is found from

cos~vDt !512a

11a. ~34!

The productf Dt5vDt/2p is the normalized frequency ooscillation within the channel and is the number of cyclescrossing of the channel. This normalized frequency depeonly on the ball’s moment of inertia, while the actual frquency,f , is inversely proportional to the transit time of thball across the channel and hence the channel width.

For the solid superball,a52/5 andvDt51.1279. For areasonable choice ofDt50.05 s, the period of the oscillatiois T50.2785 s. For the extreme case of a super ring,a51andvDt5p/2. This period is simplyT50.2 s. The momentof inertia facilitates the exchange of energy between thelocity and spin. Hence the larger the value ofa, the greaterthe ‘‘restoring force’’ in the oscillation.

For a given set of initial conditionsvx0 andc0 , the finalexpression becomes

Vn* 5AS cos~nvDt1f!

~21!n21a2 1/2sin~nvDt1f! D , ~35!

where A5Avx0*21ac0*

2 and tanf5Aac0* /vx0* . If we useEq. ~35! with the equation for the evolution of the positiovector, Eq.~27!, the vertical position of the ball is describeby

xn5ADt

2A11a

a@cos~nvDt1u1f!2cos~u1f!#,

~36!

where the phase difference between the position oscillaand the oscillation in the velocity is given by tanu5a21/2.

When the ball is thrown into the channel as in the demstration shown in Fig. 1~b!, it enters with some initial negative vertical velocity, an initial positive horizontal velocityand an arbitrary spin. By using Eq.~36!, we can determineapproximately how many bounces it takes for the ballleave the channel for a given set of initial conditions. Supose that the ball enters the channel very close to the tothe left wall. The ball will then leave the channel whenxn

'x0 . From Eq.~36!, the condition for the ball to leave thchannel is

nvDt52~p2u2f!. ~37!

The ball in Fig. 1 was thrown withvx0520.05 m/s andc0

50. The value ofn that satisfies Eq.~37! is n53.79, whilethe actual number of bounces isn854. @The number ofbounces,n8, treats the last bounce as the point at whichball passes through the plane of the left wall upon leavingchannel, whereas Eq.~37! treats the last bounce as the poiwhere the ball reaches the initial height.# This analysis sug-gests a range of possible experiments. For example,possible to throw the ball such that it enters the channelnever collides with the left wall? Also, consider a superbwith a heavy core. If the ball has a mass at the center,value ofa decreases and it takes longer for the ball to coout of the channel. If the ball is thrown with the initial conditions given in Fig. 1 for the extreme case ofa51/10, thenn58.76. Hence this ball will remain in the channel neatwice as long as a normal superball.

880Brian T. Hefner


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IV. DECAYING SUPERBALL MOTION IN ACHANNEL

The discussion in Sec. III does not consider the effecenergy losses on the ball’s motion within the channel.have seen that in the absence of any losses, whenever this thrown into the channel, it should rebound back up aout. It is easy to demonstrate that this motion does notways occur; often the ball comes partially up before begning to fall in a seemingly complicated way.

With the introduction of the coefficients of normal antangential restitution as in Sec. III, the motion of an inelasball can be approximated, thus capturing the dynamics odescent. The coefficient of tangential restitution enters~22! in the collision matrixM , which, for b,1, takes theform given by Eq.~12!. The coefficient of normal restitutionhowever, has a much more significant effect on the equaof motion. Because the velocity across the channel decreafter each collision, the transit time between collisionscomes

Dtn5w

env0y. ~38!

As expected, asn→`, the velocity of the ball decreaseuntil it no longer crosses the channel. For a slightly inelasball, wheree'1, the velocity will decrease very slowly anfor a number of bounces the coefficient can be approximaby e51. This exceptional case,e51 andb,1, captures theeffect of the tangential restitution on the ball’s motion andthe focus of this section.

The velocity, spin, and position of the ball are shownFig. 7 for several values of the restitution coefficient in trange 0<b<1. In these graphs,g59.8 m/s2 and the initialconditions of the ball arevx05c05x050 andDt50.05 s. InFig. 7~a!, the velocity of the ball once again shows similaties to the mass–spring system. Just as the introductiodamping produces a decaying oscillation of the mass, sodoes the tangential restitution produce a decaying oscillain the channel for certain values ofb. For small values ofb,the velocity of the ball undergoes an exponential decay wout any oscillation. This motion is similar to an overdamposcillator, which suggests that there exists a critical point tmarks this transition, analogous to the critically dampedcillator.

To explore this analogy, it is reasonable to introducetrial solution, Eq.~32!. However, because the motion is dcaying, the frequency is now assumed to be complexv5vR2 iv I . With this substitution, the real and imaginacomponents of the frequency are found from

e2v IDt5b1/2, ~39!

and

cos~vRDt !52~b11!

2Ab

~a21!

~a11!. ~40!

The oscillation decays exponentially, as seen in Figs. 7~a!and 7~b!, and is a function of the tangential restitution, nthe moment of inertia. The frequency of oscillation is afected only weakly by the tangential restitution and is aproximately equal to the oscillation of the fully elastic ba

As b decreases, there is a point where cos(vRDt)51. Themotion at this point can, following the analogy with th



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damped oscillator, be considered critically damped andcritical restitution coefficient is related to the moment of iertia by

bc5~Aa21!2

~Aa11!2. ~41!

For the solid sphere, the critical restitution isbc50.051.Notice that regardless of the value ofa, the critical restitu-tion is always positive. It is only zero whena51, for whichvRDt5p/2, and hence the frequency is unaffected bytangential losses.

Fig. 7. ~a! Velocity and~b! spin relative to the wall coordinate systems, a~c! height of the ball for the initial conditions,vx050 m/s, c050 m/s, andDt5w/vy50.05 s. The coefficients of tangential restitution shown areb51 ~solid line!, 0.75 ~dotted line!, 0.5 ~dash-dot line!, 0.25 ~dashed line!,and 0~solid line with circles!.

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For b,bc , the frequency becomes purely imaginary ahence the motion is an exponential decay. The frequencythis range ofb is found from

e2vDt52b1/2~b11!

2Ab

~a21!

~a11!

3S 12A122Ab

~b11!

~a11!

~a21!D . ~42!

As the value ofb decreases below the critical restitution, tvelocity vector approaches the steady state value mslowly until b521, where this motion is never reached,would be expected for a frictionless ball falling down a chanel.

The similarity between the motion of the ball and the bhavior of a damped oscillator comes as a consequencincorporating tangential losses into the ball. This analohowever does not fully capture the ball’s behavior, as is sin Fig. 7. The oscillation in the vertical velocity does ndecay to the steady state velocity discussed in Sec. IV,instead approaches a velocity that is less than the steadyvalue. From the explanation of the steady state velocitycussed in Sec. IV, if the new steady state velocity is less tin the elastic case, it no longer has an upward velocity thasufficient to overcome the effect of gravity. The ball wtherefore fall down the channel.

The superball thus exhibits behavior that is similar to aother simple physical system: a falling body. For a bofalling through a resistive medium, the acceleration ofbody decreases exponentially until the body reaches a tenal velocity which depends on the resistance of the mediLikewise, in Fig. 7~a!, the superball also approaches a termnal velocity that depends on the tangential losses. The bterminal velocity is a generalization ofVss and hence also isa solution to Eq.~28! with b<1:

Vterm51

2gDtS b~2a11!21

a~b11!

~21!n/aD . ~43!

The vertical component of the terminal velocity decreaseb decreases. Asa decreases, the coupling between thelocity and spin of the ball becomes weaker and consequethe resistance to the fall decreases as well.

Although the vertical velocity component ofVterm dependson the restitution coefficient, the spin component doesand is identical to the spin component ofVss, as is apparenin Fig. 7~b!. For this velocity vector there is obviouslybalance between the kinetic energy gained between csions and the energy lost at the collision. However, it isclear whether this vector can be explained as simply asdone forVss whenb51. In both cases, although gravity anthe tangential restitution coefficient introduce an offset to‘‘steady-state velocity,’’ this offset can be removed byappropriate choice of the velocity origin as in Eq.~30!.

The terminal velocity also is reflected in the positionthe falling ball, as in Fig. 7~c!. After the initial oscillationdecays, the height of the ball when it collides with each wdecreases at a constant rate which is not equal to the termvelocity, but to the average velocity between collisions,

vxav5

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a~b11!. ~44!



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Although the terminal velocity of the ball can be positive, taverage velocity of the ball is always negative forb,1,which shows that with any amount of tangential loss, thedisplacement of the ball will be down the channel.

V. CONCLUSION

The description of the superball given originally in Refs.and 2 provides insight into a simple system in which therean exchange between two degrees of freedom. For mobetween two parallel surfaces, this exchange producesoscillation in the motion, similar in many respects tosimple harmonic oscillator. When tangential losses are incporated into the superball model, the motion is similar todamped harmonic oscillator, exhibiting underdamped, crcally damped, and overdamped behavior. All of these aspof the model can be seen by simple demonstrations witsuperball, with the security of knowing that if you fail at ondemonstration, you will succeed at presenting the other.

ACKNOWLEDGMENTS

I wish to thank James S. Walker for his helpful discusions. I also wish to thank Robert J. Kruhlak and JenniferMorey for their critical readings of the manuscript.

APPENDIX: DEMONSTRATION NOTES

To demonstrate the superball kinematics discussed intext, you need a superball and two vertical surfaces as in1~b!. It is not necessary for one surface to be shorter, bumakes the demonstration easier. Unlike the original demstration, where two horizontal surfaces are easy to find~forexamples, tables and park benches!, two closely spaced vertical surfaces are not as common. For a classroom demstration, it should be straightforward to build a simple vercal channel that is sturdy enough to withstand the impacthe ball as it bounces between the two surfaces. For a qand dirty demonstration, I have found that a cinder blostanding on end next to a wall works well. The cinder blois useful because it can be easily moved toward or awfrom the wall to adjust the channel width, while the blockweight helps it to resist the force of the superball bouncbetween the block and the wall. Similarly, two desktops faing one another could be used or one table top and a tpiece of wood.

This demonstration seems to work best when the surfaare only a few ball diameters apart. If the cinder blockresting on the floor, stand several feet from the wall andblock. It should be possible to throw the ball down into tchannel and have it return. For this simple demonstrathowever, there are no exact requirements for the width ofchannel. The technique for throwing the ball into the chanis also a matter of practice.

It is inevitable, however, that on some throws the ball wnot be able to escape from the channel as discussed in SI and IV. The subsequent motion is interesting in itself awith a little practice it should be possible to determine whkind of throw ~topspin, backspin, etc.! will give rise to thismotion. It should be possible to observe the ball moveand down the channel several times before it falls downchannel. The number of oscillations also can be increaseinclining one of the surfaces. It should be fairly easy to gthe ball to bounce 15–20 times before the ball falls downchannel.

882Brian T. Hefner


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The analysis of the horizontal channel also suggestsother demonstration that is worth discussing briefly. If aperball is thrown in the middle of a long horizontal channit should be possible to see the ball move up and downchannel once or twice before the ball loses so much enethat it is unable to bounce against the upper surface. Fordemonstration to work well, it is necessary for the ballstart with a high velocity so that it can undergo a numberbounces and for the surfaces to be close to one another. Isurfaces are close, however, it would be difficult to throwball with very much force. This requirement suggestspossibility of using a spring-loaded launcher to fire the bat an angle through a small hole in one of the surfaces. Tmay provide enough velocity for the ball to undergo a nuber of collisions and hence go through several cycles ofcillation along the channel.

a!Electronic mail: [email protected]. L. Strobel, ‘‘Matrices and superballs,’’ Am. J. Phys.36, 834–837~1968!.

2R. L. Garwin, ‘‘Kinematics of the ultraelastic rough ball,’’Am. J. Phys.37,88–92~1969!.

3R. Cross, ‘‘Measurements of the horizontal coefficient of restitution fosuperball and a tennis ball,’’ Am. J. Phys.70, 482–489~2002!.

4H. Brody, ‘‘That’s how the ball bounces,’’ Phys. Teach.22, 494–497~1984!.



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5N. Maw, J. R. Barber, and J. R. Fawcett, ‘‘The oblique impact of elasspheres,’’ Wear1, 101–114~1976!.

6N. Maw, J. R. Barber, and J. R. Fawcett, ‘‘The role of tangential compance in oblique impacts,’’ Trans. ASME, J. Lub. Tech.20, 327–344~1981!.

7R. Cross, ‘‘Grip-slip behavior of a bouncing ball,’’ Am. J. Phys.70, 1093–1102 ~2002!.

8J. T. Jenkins and M. W. Richman, ‘‘Kinetic theory for plane flows ofdense gas of identical, rough, inelastic, circular disks,’’ Phys. Fluids28,3485–3494~1985!.

9S. McNamara and S. Luding, ‘‘Energy nonequipartition in systems ofelastic, rough spheres,’’ Phys. Rev. E58, 2247–2250~1998!.

10K. L. Johnson, ‘‘The bounce of a ‘superball,’ ’’ Int. J. Mech. Eng. Ed.11,57–63~1983!.

11It is interesting to note that the equations of motion for a ball withtangential coefficient of restitution were originally presented nearly thyears before the invention of the superball and Garwin’s analysis. SeGoldsmith,Impact ~Dover, New York, 2001!, pp. 18–19.

12J. Duran,Sands, Powders, and Grains: An Introduction to the PhysicsGranular Materials~Springer, New York, 2000!.

13S. Luding, ‘‘Granular materials under vibration: Simulations of rotatispheres,’’ Phys. Rev. E52, 4442–4457~1995!.

14F. S. Crawford, ‘‘Superball and time-reversal invariance,’’Am. J. Phys.50,856 ~1981!.

15F. S. Crawford, ‘‘A theorem on elastic collisions between ideal rigid boies,’’ Am. J. Phys.57, 121–125~1987!.

THE TWO CULTURES

‘‘It’s rather odd,’’ said G. H. Hardy, one afternoon in the early Thirties, ‘‘but when we hearabout ‘intellectuals’ nowadays, it doesn’t include people like me and J. J. Thomson and Ruther-ford.’’ Hardy was the first mathematician of his generation, J. J. Thomson the first physicist of his;as for Rutherford, he was one of the greatest scientists who has ever lived. Some bright youngliterary person~I forget the exact context! putting them outside the enclosure reserved for intel-lectuals seemed to Hardy the best joke for some time. It does not seem quite such a good jokenow. The separation between the two cultures has been getting deeper under our eyes; there is nowprecious little communication between them, little but different kinds of incomprehension anddislike.

C. P. Snow,The New Statesman and Nation~Cambridge University Press, 1956!. Reprinted inThe World Treasury ofPhysics, Astronomy, and Mathematics~Little, Brown and Company, Boston, MA, 1991!, p. 741.

883Brian T. Hefner


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The kinematics of a superball bouncing between two vertical surfaces