The falling pencil: a "Divertimento" in four movements
aa r X i v : . [ phy s i c s . c l a ss - ph ] J a n The falling pencil:a
Divertimento in four movements
Nicola
Cufaro Petroni ∗ Dipartimento di
Matematica and
TIRES , University of Bari (
Ret )and
INFN
Sezione di Bari; via E. Orabona 4, 70125 Bari, Italy
Abstract
The dynamics of a simple pencil with a tip laid on a rough table and set free tofall under the action of gravity is scrutinized as a pedagogic case study. Thefull inquiry is anticipated by a review of three other simplified movementsforeshadowing its main features. A few exact and general results about thesliding angles and the critical static coefficient of friction are established
Keywords : Newton laws; Rigid bodies; Friction
When he was a beginner in his physics studies the author of these lines was not veryadroit in solving exercises. That notwithstanding he managed to pass his exams andhe subsequently acquired the usual skills – and even some zest – in designing andanswering problems: this was of course also a result of his first acquaintance withthe teaching. In those years he posed to himself some seemingly simple questionsthat he could not immediately answer and that he did not happen to find discussedon his handbooks; but then he dropped them and went along his way without caringtoo much, even if every now and again they popped up in his head. He remembersin particular asking himself what exactly happens to a simple pencil with a tip laidon a table and set free to fall under the action of gravity: would the tip on thetable stay put at its initial position, or will it begin to slide, and when? And whatis its subsequent movement? The author didn’t spend in fact too much effort onthat, and he eventually gave up, but for some unrelated reason this query resurfacedrecently in his thoughts and now – being today retired – he decided to devote sometime in finding an elementary, but satisfactory answer: a pursuit prompted by sheercuriosity and to him comparable to a
Divertimento that hopefully could also be ofsome interest for students and scholars ∗ [email protected] The falling pencil: a
Divertimento in four movements movements of growingdifficulty: in the first one (Section 2) the pencil tip is hinged in a point and thesystem is free to rotate without friction around it sweeping an arbitrary fall angle0 ≤ θ ≤ π (in this section there is no table to speak about). This simplified settingwill lend the possibility of studying the hinge reaction forces without making anyreference to the friction. This smoothness requirement is carried on also in the twosubsequent sections where the second and third movement are investigated: in theSection 3 the pen tip is restrained to slide along a horizontal frictionless rail (hereagain θ is allowed to go from 0 to 2 π ) so that a first idea of what happens in thislimiting case is acquired. Then in the Section 4 the horizontal table appears (sothat now 0 ≤ θ ≤ π / ): it is still frictionless, but featuring a step that forbids anearly sliding of the pencil on one side. This third movement allows to recognizethat beyond an angle θ r = arccos / ≃ . π the pencil tip begins to slide on thestep-free side. In the Section 5 we finally turn our attention to the fourth movementof the free pencil on a rough table where µ s and µ κ respectively are the static andkinetic coefficients of friction. In this case it is found that there is a precise criticalvalue µ s = 2 − / · · / ≃ .
371 of the static coefficient beyond which no earlysliding is allowed (much as if the step of the third movement was in place). A furtherfallout of this finding is that there are exact angles θ = arccos 911 ≃ . π θ = arccos 48 √ − ≃ . π such that an early sliding (for µ s ≤ µ s ) can happen only at θ ∗ ≤ θ , while a latersliding on the opposite side (for µ s ≥ µ s ) only starts at θ ∗∗ ≥ θ . It is worthwhileto remark that the values of θ r , µ s , θ and θ are universal for every idealized barused as a pencil and for every kind of rough table used to perform the experiment.The values either of θ ∗ or of θ ∗∗ on the other hand apparently depend on µ s . Thetrajectories of the center of mass of the pencil for the third and fourth movementare also investigated, those for the first and second movement being utterly trivial.A few final remarks are ultimately added in the last Section 6 Consider a homogeneous, rigid rod (the pencil ) of mass m and length L with oneof its extremities in contact with a horizontal surface ((the table ) and suppose that µ s and µ κ respectively are the static and kinetic friction coefficients (see Figure 1).Let θ be the angle between the pencil and the vertical to the surface, and x, y thecoordinates of the middle point (the center of mass, CM ) in a plan containing thepencil and the vertical so that (when the pencil tip stays still in the axes origin) x = L θ y = L θ ≤ θ ≤ π / (1) Cufaro Petroni: The falling pencil: a
Divertimento in four movements N and F respectively the vertical and horizontalcomponents of the ground reaction force: apparently F is non-zero only if a frictionis there. The aim of the present paper is a discussion of the dynamics of the fallingpencil, and in particular of its behavior when it also possibly slips on the surfacebefore touching the groundWe will suppose for simplicity at first that the pencil is not allowed to movealong the surface: for instance we can imagine it hinged at the axes origin andfree to rotate without friction around it. We will also admit that it can go fullcircle – as if the table were not there – so that now 0 ≤ θ ≤ π . This wouldenable us to study the reaction forces N and F in detail in an initially simplifiedsetting that will be useful in the subsequent discussion. We have indeed in this casejust a physical pendulum (an extended rigid body) performing swings of arbitraryamplitude. The topic is very well known and has been widely studied, for instance as inverted pendulum w.r.t. the stabilization of its equilibrium (see for instance [1], [2]and [3]): we will however skip these topics altogether by confining ourselves just toa simplified discussion of the circular pendulum.The Newton equations of motion, with a fixed point in the origin, can be simplywritten in this case as m ¨ x = F m ¨ y = N − mg I ¨ θ = mgx = mg L θ (2)where I = mL / is the moment of inertia of the pencil w.r.t. its fixed end. Neglect- Cufaro Petroni: The falling pencil: a
Divertimento in four movements θ = ω ⊥ θ ω ⊥ = r gL (3)There is not an explicit elementary solution of this non linear equation, but thatnotwithstanding we can study it in some detail. It is easy to see indeed that ddt (cid:0) ˙ θ (cid:1) = 2 ˙ θ ¨ θ = ω ⊥ ˙ θ sin θ = − ω ⊥ ddt (cos θ )and therefore ˙ θ = − ω ⊥ cos θ + c (4)where c is an arbitrary integration constant depending on the initial conditions. Letus make at first (a bit naively) what seems to be the simplest choice, namely θ (0) = 0 ˙ θ (0) = 0 (5)In this case apparently we have c = ω ⊥ and hence˙ θ = ω ⊥ (1 − cos θ ) ≥ ≤ θ ≤ π or in another form ω ( θ ) = dθdt = ω ⊥ √ − cos θ This non-linear, first order equation – which also shows that ω ⊥ is the angularvelocity at θ = π / – can be easily solved by separating the variables, namely Z θ dφ √ − cos φ = ω ⊥ Z t ds = ω ⊥ t but it can be seen that the left hand integral diverges because the integrand functionhas a non integrable singularity in the origin:1 √ − cos φ = O (cid:0) φ − (cid:1) φ → φ → φ √ − cos φ = 2 lim φ → √ − cos φ sin φ = 2 lim φ → s − cos φ − cos φ = 2 lim φ → √ φ = √ θ = 0 Cufaro Petroni: The falling pencil: a
Divertimento in four movements t (in seconds) needed to reach an angle θ according to (9) forthree different values of ǫ and ω ⊥ = 10 sec − (corresponding to an L of roughly 30cm). The pencil is allowed to go full circle from 0 to 2 π .would diverge. We need therefore to take a slightly different (and more realistic)initial condition, for instance with a gentle push onward θ (0) = 0 ˙ θ (0) = ω > ω can be chosen small and even infinitesimal to approach the ideal (butsingular) condition (5). With this new assumption the integration constant in (4)becomes c = ω o + ω ⊥ and the equation takes the form˙ θ = ω + ω ⊥ (1 − cos θ ) (7)to wit ω ( θ ) = dθdt = ω ⊥ √ ǫ + 1 − cos θ ǫ = ω ω ⊥ This equation can be solved again by separating the variables Z θ dφ √ ǫ − cos φ = ω ⊥ Z t ds = ω ⊥ t (8)but now (see [4] 2.571.5) the left hand side integral converges for ǫ > r
21 + ǫ F arcsin r (1 + ǫ ) 1 − cos θ ǫ − cos θ , r
11 + ǫ ! = ω ⊥ t (9)where (see [4] 8.111.2) F ( ϕ, b ) = Z ϕ dα p − b sin α b < The falling pencil: a
Divertimento in four movements F / mg and N / mg of (14) and (15) (for ǫ = 0) as functions of the position θ .is the elliptic integral of the first kind. As a matter of fact the equation (9) gives thefunction θ ( t ) in an implicit form that is not easy to invert, and this form moreoveris not much more manageable than the original integral formulation (8) because thefunction F ( ϕ, b ) is nothing else than a name for another integral. Since however theseintegrals are nowadays numerically performed by the usual mathematical software,the results (8) and (9) can easily be used to plot the function t ( θ ), time needed toreach an angle θ , as in the Figure 2 where, with an exchange of the coordinate axes,we would also get a graphical representation of θ ( t )We can next take advantage of the first two equations (2) to find the reactionforces N and F : since from (1) it is¨ x = L (cid:16) ¨ θ cos θ − ˙ θ sin θ (cid:17) ¨ y = − L (cid:16) ¨ θ sin θ + ˙ θ cos θ (cid:17) (10)from the first two equations in (2) we have F = mL (cid:16) ¨ θ cos θ − ˙ θ sin θ (cid:17) N = mg − mL (cid:16) ¨ θ sin θ + ˙ θ cos θ (cid:17) and, since we know that, with the initial conditions (6), the equations (3) and (7)hold, after a little algebra we find how the reaction forces vary as functions of θF ( θ ) = 3 mg (cid:18)
32 cos θ − − ǫ (cid:19) sin θ (11) N ( θ ) = mg mg (cid:18)
32 cos θ − − ǫ (cid:19) cos θ (12)From (7) moreover it is also possible to show that the angular velocity ω = ˙ θ varieswith the position θ according to the formula ω ( θ ) = ω ⊥ √ ǫ + 1 − cos θ (13) Cufaro Petroni: The falling pencil: a
Divertimento in four movements ω / ω ⊥ of (13) as a function of θ for twodifferent values of ǫ .It is interesting to remark at this point that, while the time formula (8) is singularfor ǫ → + , the equations (11), (12) and (13) continuously go into their ǫ = 0 forms F ( θ ) = 3 mg (cid:18)
32 cos θ − (cid:19) sin θ (14) N ( θ ) = mg mg (cid:18)
32 cos θ − (cid:19) cos θ (15) ω ( θ ) = ω ⊥ √ − cos θ (16)corresponding to the null initial conditions (5): these limiting formulas can nowbe properly used to represent the simplest behavior of the reaction forces and ofthe angular velocity at every possible position θ . In the Figure 3 we have plottedthe dimensionless functions F / mg and N / mg of (14) and (15) (with ǫ = 0), whilethe velocity (13) of (16) in its dimensionless form ω / ω ⊥ is plotted in the Figure 4in the interval [0 , π ]: for ǫ > ω ( θ ) turns out to be a smooth function even atthe angles θ = 0 , π, π . . . Apparently when we also plug into these formulas thefunction θ ( t ) implicitly defined in (9) we also get the time dependence of F, N and ω , but, needless to say, this would be a cumbersome task that we will neglect hereIt is worthwhile to remark finally that the two reaction components N, and F also take negative values: for F this is apparent from the Figure 3 and we seefrom (14) that – even with ǫ = 0 and remaining just in the interval [0 , π / ] – wehave F < > cos θ ≥ (cid:0) / (cid:1) < θ ≤ π (cid:0) / (cid:1) ≃ . π As for the normal component we find instead from (12) that N can be negative only Cufaro Petroni: The falling pencil: a
Divertimento in four movements ǫ >
0: more precisely we have N ≤ (cid:16) ǫ + p ǫ (1 + ǫ ) (cid:17) ≥ cos θ ≥
13 + 23 (cid:16) ǫ − p ǫ (1 + ǫ ) (cid:17) namely for θ falling in an interval that shrinks to the single point arccos( / ) ≃ . π for ǫ = 0 + . These negative values will be of some consequence in the sequelbecause they will suggest where an un -hinged pencil will begin a sliding movementwhen the available constraints will be unable to provide a negative reaction Before going ahead to our pencil with one end laid on a horizontal rough table andfree to move along it, we will stop for a while to consider two more frictionlesscases. In order to allow again for a full swing of the system from 0 to 2 π , moreover,in the first of these examples we will suppose that the pencil tip on the x axis inthe Figure 5 is in fact constrained to slide along a rail without leaving it while thecenter of mass goes from L / to − L / and back again. At variance with the caseof the previous section, however, now there is no horizontal force F because neitherfriction nor hinges in the axes origin are present. As a consequence the pencil CM Cufaro Petroni:
The falling pencil: a
Divertimento in four movements t (in seconds) needed to reach an angle θ according to (23) forthree different values of ǫ and ω ⊥ = 10 sec − . The pencil is allowed to go full circlefrom 0 to 2 π .will simply move along the y axis with x = 0, if its movement starts with this initialcondition. On the other hand there is no longer a fixed point of the system so thatnow its rotational dynamics is better accounted for by looking at its motion aroundthe CM . Therefore the Newton equations now are m ¨ x = 0 m ¨ y = N − mg I CM ¨ θ = N L θ (17)where I CM = mL / is the moment of inertia of the pencil w.r.t. its CM , while thegeometrical relations among the coordinates become x = 0 y = L θ (18)To tackle our problem we can now retrace a path similar to that followed in theSection 2: from (17) the rotational acceleration around the CM is¨ θ = 6 NmL sin θ (19)while on the other hand again from (17) and from (10) we find N = m (¨ y + g ) = mg − mL (cid:16) ¨ θ sin θ + ˙ θ cos θ (cid:17) (20)so that altogether it is ¨ θ = sin θ h ω ⊥ − (cid:16) ¨ θ sin θ + ˙ θ cos θ (cid:17)i Cufaro Petroni:
The falling pencil: a
Divertimento in four movements N / mg of (24) and (25) as a function ofthe position θ for two values of ǫ .It is easy to see now that ddt (cid:0) ˙ θ (cid:1) = 2 ˙ θ ¨ θ = 2 ˙ θ sin θ h ω ⊥ − (cid:16) ¨ θ sin θ + ˙ θ cos θ (cid:17)i = − ddt (cid:20) ω ⊥ cos θ + 3 (cid:16) ˙ θ sin θ (cid:17) (cid:21) to wit ˙ θ (cid:0) θ (cid:1) + 4 ω ⊥ cos θ = c (21)and since with the slightly off-equilibrium initial conditions (6), and keeping thesame notations, it is c = 4 ω ⊥ + ω o = 4 ω ⊥ (cid:16) ǫ (cid:17) ǫ = ω ω ⊥ we finally have ˙ θ = ω ⊥ ǫ + 4(1 − cos θ )1 + 3 sin θ = ω ⊥ ǫ + 4(1 − cos θ )4 − θ (22)This equation can be integrated again by separating the variables giving Z θ s − φ ǫ + 4(1 − cos φ ) dφ = ω ⊥ t = t r gL (23)and while this implicit solution has no elementary inverse function it is possible tonumerically evaluate the integral to calculate the time t needed to reach an angle θ : the results plotted in the Figure 6 show a qualitative behavior similar to that ofthe Figure 2. Here too, however, the time t diverges when ǫ → + Cufaro Petroni:
The falling pencil: a
Divertimento in four movements ω / ω ⊥ of (26)and (27) as a functionof θ for two values of ǫ .To study next the reaction force N we plug (19) and (22) into (20) obtaining theequation N = mg − mL (cid:18) NmL sin θ + 3 gL ǫ + 4(1 − cos θ )1 + 3 sin θ cos θ (cid:19) that is easily solved providing N = mg θ − ǫ + 2) cos θ (4 − θ ) (24)For ǫ = 0 this simply becomes N = mg θ − θ (4 − θ ) (25)The dimensionless function N / mg is plotted in the Figure 7 for two different initialconditions ǫ , and it is interesting to remark that now – at variance with the casediscussed in the Section 2 – its values are always positive, and that to have alsonegative values the initial angular velocity ω should in fact exceed a fairly largethreshold. More precisely it would be possible to see that the Mexican-hat shapedred curve of the Figure 7 bends its tails under the x -axis only for ǫ > / , namelyfor ω > p g / L sec − : for example, for L = 0 . ω >
10 sec − . Finally from (22) we have the angular velocity ω ( θ ) = ˙ θ = ω ⊥ s ǫ + 4(1 − cos θ )1 + 3 sin θ (26) Cufaro Petroni: The falling pencil: a
Divertimento in four movements F / mg and N / mg (14) and (15) of the hinged pencil of the Section 2 as functions of the position 0 ≤ θ ≤ π / for ǫ = 0.From (14) we see that the force F reverses its sign when θ ≥ θ r = arccos / .that is for ǫ = 0 ω ( θ ) = 2 ω ⊥ r − cos θ θ (27)In its dimensionless form ω / ω ⊥ this angular velocity is reproduced in the Figure 8for two values of ǫ , and the functions turn out to be smooth again even at θ =0 , π, π, . . . for every non-zero ǫ > In our third frictionless case we begin first by looking back to the reaction forcesdiscussed in the Section 2. When it begins to fall, indeed, the hinged pencil rotates asin the Figure 1 with a fixed point and hence the reaction forces vary with θ as in theFigure 3. To be more precise we have reproduced in the Figure 9 the forces F ( θ ) / mg and N ( θ ) / mg in the interval 0 ≤ θ ≤ π / in the limiting case of ǫ = 0 presented in (14)and (15). From this picture and the corresponding equations we see in particularthat, while N never goes negative, the horizontal component F of the reactions inthe Figure 1 reverses its sign beyond an angle θ r = arccos / ≃ . π suggestingthat some force is needed to keep the pencil tip from moving to the right when θ > θ r . Suppose then now that – without being hinged at the origin – our pencil isjust laid on a frictionless table and allowed to fall as in the Figure 1, but also that itscontact tip is forbidden to slide leftwards (as instead it was allowed in the Section 3)by the presence of some obstacle, for instance a step as in the Figure 10. From theprevious remarks it follows then that when θ exceeds θ r the pencil tip starts sliding rightwards because – being now unhinged – no negative horizontal reaction forcecan arise to prevent that. The pencil reaches the angle θ r at a time t r that can be Cufaro Petroni: The falling pencil: a
Divertimento in four movements θ > θ r = arccos / on a frictionless surface with a step the pencilalso starts drifting rightwards.explicitly calculated from the integral (8) for a small initial destabilizing condition ǫ >
0, while at that point its angular velocity from (16) is ω r = ω ( θ r ) = ω ⊥ √ r gL We will now investigate the movement of our system for θ r < θ < π / from the time t r until the instant T of the impact on the table.If, according to the Figure 10, z is the position of the contact point on the tablethe relationships among the variables are now x = z + L θ y = L θ ≤ θ ≤ π / (28)while the Newton equations of motion are m ¨ x = 0 m ¨ y = N − mg I CM ¨ θ = N L θ (29)where again I CM = mL / . As a consequence the second equation (10) togetherwith the equations (19) and (20) still hold, and hence also (21) can be deduced.Imposing then the conditions at t = t r we find that the integration constant now is c = ω r (1 + 3 sin θ r ) + 3 ω ⊥ cos θ r = 329 ω ⊥ Cufaro Petroni:
The falling pencil: a
Divertimento in four movements N / mg and angular velocity ˙ θ / ω ⊥ (continuouslines when θ ≥ θ r ) on a frictionless surface with a step, compared with the samequantities in the case of the hinged pencil (dashed lines) that also accounts for themovement when θ < θ r .and therefore we get ˙ θ = 4 ω ⊥ − θ − θ (30)with its new corresponding time equation Z θθ r s − φ − φ dφ = 2 ω ⊥ t − t r ) (31)that can be numerically evaluated to calculate the time t needed to reach an angle θ ∈ [ θ r , π / ]: for instance the time T when the pencil hits the floor will be T = t r + 32 ω ⊥ Z π / θ r s − φ − φ dφ ≃ t r + 0 . ω ⊥ ω ⊥ = r gL (32)where t r comes from (8) choosing a small initial condition ǫ >
0. As for the reactionforce N on the other hand, from (19), (20) and (30) we have that N = mg − mL (cid:18) NmL sin θ + 4 g L − θ − θ cos θ (cid:19) that eventually gives N = mg θ − cos θ + 4(4 − θ ) (33)The plot of N / mg in the Figure 11 shows in particular that N always stays positiveeven in the interval [ θ r , π / ] signaling that the pencil tip never leaves the table Cufaro Petroni: The falling pencil: a
Divertimento in four movements z ( θ ) / L of the pencil tip laid on a frictionless tablewith a step, as a function of θ : as long as θ ≤ θ r it is z ( θ ) / L = 0, but when θ ≥ θ r the function z ( θ ) = ζ (cos θ ) should be calculated from (36).surface. In the same Figure 11 also the dimensionless angular velocity ˙ θ / ω ⊥ isdisplayed in the same interval.To investigate next the behavior of x, y and z of (28) we begin by remarking thatthe first equation in (29), m ¨ x = 0, clearly entails that ˙ x = c for t ≥ t r , while to findthe integration constant c it is enough to remark that˙ x ( t ) = c = ˙ x ( t r ) = L ω r cos θ r = ω ⊥ L √ t r ≤ t ≤ T (34)As a consequence we will have x ( t ) = L θ r + ω ⊥ L √ t − t r ) = L ω ⊥ L √ t − t r ) t r ≤ t ≤ T The chronological equations of y and z , instead, can not be deduced so simply: thesecond equations (29) for y , for instance, would be nothing new w.r.t. the angularequation, in the sense that if we know θ ( t ) we also can find y ( t ) by taking advantageof the second equation (28). But we have seen that the angular equation (30) cannot be integrated in an elementary way, and hence even y ( t ) has not a manageableform. Shunning however this chronological issue, we can at least gain some insightinto the shape of the trajectory of the CM of coordinates x and y .It is apparent indeed that until the sliding begins (namely when 0 ≤ t ≤ t r and0 ≤ θ ≤ θ r ) the CM follows a circular path of radius L / around the origin; assoon as t > t r and θ > θ r , however, the CM parts way from the aforementionedcircumference following a different flight that can be scrutinized by looking againinto the equations (28): by eliminating indeed cos θ between the equations( x − z ) = L θ = L − cos θ ) y = L θ Cufaro Petroni:
The falling pencil: a
Divertimento in four movements CM xy -trajectory: it coincides with the circular path ofthe hinged pencil for θ ≤ θ r , but as soon as θ ≥ θ r it follows a different flight withthe parametric equations (35).we have y = r L − ( x − z ) pointing to the fact that now the CM treads along a circle, but with a moving centerin z . The t -parametric equations of this trajectory then are x ( t ) = z ( t ) + L p − cos θ ( t ) y ( t ) = L θ ( t )and if we define a function ζ ( s ) such that z ( t ) = ζ (cid:0) s ( t ) (cid:1) s ( t ) = cos θ ( t )by adopting s as a new parameter the parametric equations become x ( s ) = ζ ( s ) + L √ − s y ( s ) = L s s = cos θ ∈ [0 ,
1] (35)We are therefore prompted to study ζ ( s ): from (28), (30) and (34) we know that˙ z = ˙ x − L θ cos θ = ω ⊥ L √ − r θ − θ − θ ! and since ˙ z = ζ ′ ( s ) ˙ s = − ζ ′ ( s ) ˙ θ sin θ , using (30) again we find ζ ′ ( s ) = − L √ s − s (1 − s )(8 − s ) − r s − s ! ≤ s ≤ / Cufaro Petroni:
The falling pencil: a
Divertimento in four movements θ ∗ <θ = arccos / ≃ . π if the static friction coefficient is not large enough ( µ s <µ s ≃ .
37) to forbid it.namely ζ ( s ) = L √ Z / s s − u (1 − u )(8 − u ) − r u − u ! du ≤ s ≤ / (36)This integral, that can be performed at least numerically, lends now the possibilityof plotting both z ( θ ) = ζ (cos θ ) (Figure 12), and the trajectory parametric equa-tions (35) (Figure 13) where it is understood that ζ ( s ) = 0 when / ≤ s ≤ x T of x when the pencil finallyhits the floor: if T , as provided by (32), is the impact time, we of course have θ ( T ) = π / , namely s ( T ) = 0 and therefore the CM x -coordinate when the pencillands on the table is x T = z T + L L ξ T ) z T = z ( T ) = ζ ( s ( T )) = ζ (0) = L ξ T where ξ T = 1 √ Z / s − u (1 − u )(8 − u ) − r u − u ! du ≃ . L = 20 cm, this roughly means that z T ≃ . The falling pencil: a
Divertimento in four movements | F | / mg may exceed its maximum µ s N / mg at several possible angles according to the value of µ s . If µ s < µ s a slippingtakes place toward the left beyond an angle θ ∗ ; when instead µ s > µ s the pencil tipstarts sliding toward the right at a later time past the angle θ ∗∗ . We finally go back to our initial problem of a pencil with one end laid on a horizontalrough table and free to slide along it while falling, Because of the presence of friction,when the pencil starts its movement the extremity in contact with the surface doesnot move, but it can possibly slide (on both sides as we shall see) at a later time t ∗ when it passes beyond a position θ ∗ : to understand how this happens we musttherefore first of all look again at the reaction forces discussed in the Section 2and 4. When it begins to fall, indeed, the pencil rotates as in the Figure 1 with afixed point and hence the reaction forces – in the limiting case of ǫ = 0 presentedin (14) and (15) – vary with θ as in the Figure 9 with 0 ≤ θ ≤ π / . We alreadyremarked in the Section 4 that F (the horizontal component of the reactions in theFigure 1) is supposed to reverse its sign beyond an angle θ r = arccos / ≃ , π (suggesting that some force is needed to keep the pencil tip from moving to the right when θ > θ r ), but only if it does not start to slide to the left at an earlier time t ∗ atan angle θ ∗ as in the Figure 14. This second occurrence must indeed be taken intoaccount because now – in absence of the step of the Section 4 – the static frictionforce must always satisfy the condition | F | ≤ µ s N , and it may happen that thisrequirement is not met beyond some angle θ ∗ < θ r .In order to understand if and when this happens, a (dimensionless) comparisonbetween | F | and µ s N has been displayed in the Figure 15 wherefrom we see thatwhenever µ s is smaller than a critical value µ s the pencil starts slipping to the leftat an angle θ ∗ . From the equation (8) it is also possible to find the time t ∗ of thisoccurrence for every non zero initial condition ǫ >
0. When instead µ s > µ s , theslipping happens toward the right at a later time t ∗∗ when the absolute value of the Cufaro Petroni: The falling pencil: a
Divertimento in four movements s = cos θ such that R ( s ) = µ s . The critical friction coefficient µ s – beyond which no left-slipping takes place – coincides with the maximum of R ( s ). The particular valuesand the notation are carried over from those of the Figure 15.(now reversed) friction force exceeds the critical value at a larger angle θ ∗∗ .To find the numerical values of these quantities we must first of all look (witha given µ s ) for the values of the angle θ such that | F | = µ s N namely, from (14)and (15), such that (cid:12)(cid:12)(cid:12)(cid:12) (cid:18)
32 cos θ − (cid:19) sin θ (cid:12)(cid:12)(cid:12)(cid:12) = µ s (cid:18)
14 + 32 (cid:18)
32 cos θ − (cid:19) cos θ (cid:19) (37)when 0 ≤ θ ≤ π / . Squaring both sides and defining for simplicity s = cos θ ∈ [0 , − s )(9 s − = µ s (3 s − (38)and to search for its solutions in [0 ,
1] we recast it in the form R ( s ) = (1 − s )(9 s − (3 s − = µ s (39)that is represented in the Figure 16 with the same values of µ s adopted in theFigure 15. It is apparent therefrom that s ∗ = cos θ ∗ and s ∗∗ = cos θ ∗∗ are the valuesfor slipping toward the left and toward the right respectively, while s r = cos θ r = / corresponds to the sign inversion of F in the case of the hinged pencil discussed inthe Section 2. The critical value µ s of the friction coefficient, beyond which noleft-slipping is possible, can moreover be deduced as the maximum value of R ( s ) byrequiring that R ′ ( s ) = 0: a little algebra would show indeed that the maximum of R ( s ) is attained at s = / , namely at θ = arccos / ≃ . π corresponding tothe following critical value of the static coefficient of friction µ s = R (cid:0) / (cid:1) = 2 − ≃ . µ s = 1564 r ≃ . The falling pencil: a
Divertimento in four movements θ ∗ , θ ∗∗ can finally be deduced by numerically solvingthe equation (38): it is easy to show for instance that with the values of µ s used inthe Figures 15 and 16 we would have µ s = 0 . < µ s ≃ .
37 cos θ ∗ ≃ . θ ∗ ≃ . πµ s = 0 . > µ s ≃ .
37 cos θ ∗∗ ≃ . θ ∗∗ ≃ . π It is also possible to see by direct calculation that at the critical friction coefficient µ s of (40) the equation (39) in [0 ,
1] also has its smallest solution in s = 48 √ − θ = arccos 48 √ − ≃ . π By summarizing: the limiting angles are0 < θ < θ r < θ < π / θ = arccos / ≃ . πθ r = arccos / ≃ . πθ = arccos √ − ≃ . π and when µ s < µ s the pencil tip slips leftwards past an angle θ ∗ ≤ θ , while if µ s > µ s a rightward sliding starts only later beyond an angle θ ∗∗ ≥ θ : the particular values of θ ∗ and θ ∗∗ depend on µ s and can be calculated numerically from the equation (38).It also goes without saying that θ ∗ grows from 0 to θ when µ s grows from 0 to µ s ,while subsequently θ ∗∗ starts growing from θ > θ when µ s exceeds µ s : no slippingangle (either θ ∗ or θ ∗∗ ) can be found instead in the interval [ θ, θ ], namely betweenarccos / ≃ . π and arccos √ − ≃ . π Beyond these slipping angles, either θ ∗ or θ ∗∗ , the pencil dynamics is ratherdifferent and we will study in some detail only the case µ s < µ s with a leftwardslipping beyond θ ∗ represented in the Figure 14: the case µ s > µ s with a rightwardslipping beyond θ ∗∗ is not really different ad its discussion – combining elements ofthe following treatment and of the case of Section 4 – is left to the interested reader.When µ s < µ s we already know that the leftward sliding of the pen tip begins pastan angle θ ∗ = arccos s ∗ where s ∗ is the largest solution of the equation (38) in [0 , t ∗ that can be calculated from (8) with θ = θ ∗ and in fact depends on the initial conditions: we recall from the discussionof the Section 2 that in fact t ∗ diverges when we choose the zero initial condition ǫ → + , but also that this is not an insurmountable hindrance if we leave aside thecomplete chronological equations and focus instead on the trajectory shape.In order to analyze the movement in the intervals t ∗ ≤ t ≤ T and θ ∗ ≤ θ ≤ π / we recall first that the coordinates of the system of Figure 14 still satisfy the Cufaro Petroni: The falling pencil: a
Divertimento in four movements z = 0 for 0 ≤ θ ≤ θ ∗ , and z ≤ θ ∗ ≤ θ ≤ π / . If moreover µ κ is the kinetic coefficient of friction between the penciland the rough surface, the Newton equations of motion now are m ¨ x = µ κ N m ¨ y = N − mg I CM ¨ θ = N L θ (41)with I CM = mL / : these equations coincide with the (29) of Section 4 but forthe first one that now accounts for the kinetic friction force. Therefore the secondequation (10) and the equations (19) and (20) still hold and hence we can deducethe equation (21) again as we did in the Sections 3 and 4: here however, to find theintegration constant c , we must impose new conditions at t = t ∗ . We have indeedfirst from (16) that θ ( t ∗ ) = θ ∗ ˙ θ ( t ∗ ) = ω ∗ = ω ( t ∗ ) = ω ⊥ √ − cos θ ∗ ω ⊥ = r gL and then that x ( t ∗ ) = L θ ∗ ˙ x ( t ∗ ) = L ω ⊥ cos θ ∗ √ − cos θ ∗ y ( t ∗ ) = L θ ∗ ˙ y ( t ∗ ) = − L ω ⊥ sin θ ∗ √ − cos θ ∗ z ( t ∗ ) = 0 ˙ z ( t ∗ ) = 0We are therefore able to calculate c and after a little algebra we find˙ θ = 4 ω ⊥ − cos θ ∗ (1 − cos θ ∗ ) − θ − θ (42)that replaces (30) with its corresponding time equation which is now Z θθ ∗ s − φ − cos θ ∗ (1 − cos θ ∗ ) − φ dφ = 2 ω ⊥ t − t ∗ ) (43)This integral can be numerically evaluated to calculate the time t needed to reachan angle θ ∈ [ θ ∗ , π / ]: for instance, if we take cos θ ∗ = 0 . > / = cos θ (thatcorresponds to µ s ≃ . < .
371 = µ s ), the time T when the pencil hits the floornow becomes T = t ∗ + 32 ω ⊥ Z π / θ ∗ s − φ − cos θ ∗ (1 − cos θ ∗ ) − φ dφ ≃ t ∗ + 2 . ω ⊥ (44)where t ∗ comes from (8) choosing a small initial condition ǫ >
0. As for the reactionforce N on the other hand, from (41) (namely (19) and (20) as in the Sections 3and 4) and from (42) we have now Cufaro Petroni: The falling pencil: a
Divertimento in four movements N / mg and angular velocity ˙ θ / ω ⊥ (continuouslines) on a rough surface with µ s < µ s , compared with the same quantities in thecase of the hinged pencil (dashed lines): the values coincide when θ < θ ∗ . Here wetook cos θ ∗ = 0 .
95 corresponding to µ s ≃ . N = mg θ − (cid:2) − (1 − cos θ ∗ ) cos θ ∗ (cid:3) cos θ + 4(4 − θ ) θ ∗ ≤ θ ≤ π / (45)while for 0 ≤ θ ≤ θ ∗ it takes the same values of the hinged case of Section 2. Theplot of N / mg in the Figure 17 shows in particular that N is discontinuous at θ ∗ signaling the transition from the static to the kinetic friction. In the same Figure 17also the dimensionless angular velocity ˙ θ / ω ⊥ is displayed in the same intervals.We come finally to give some detail about the CM trajectory and the position z of the tip, but at variance with the discussion of the Section 4, ˙ x ( t ) no longeris a constant as in (34) since we must now take into account the kinetic frictionforce in the first equation (41). A quest for a simple chronological equation x ( t ),however, would still be doomed because of the rather involuted form (45) of N .We can nevertheless gain some insight into the trajectories by looking again to ourquantities rather as functions of the angle θ , as we already did in the previoussections. While apparently for 0 ≤ θ ≤ θ ∗ it is z = 0 and the CM follows a circularpath of radius L / around the origin, as soon as θ > θ ∗ it will follow a path ofparametric equations (35) with ζ ( s ) = 0 for s = cos θ ∈ [ s ∗ ,
1] ( s ∗ = cos θ ∗ ): in orderto complete the trajectory we are therefore left just with the task of calculating ζ ( s )for s ∈ [0 , s ∗ ]. In order to do that we first remark that from (28) we have˙ z = ˙ x − L θ cos θ On the other hand, within the notations of the Section 4 with s = cos θ , it is˙ z = ζ ′ ˙ s = − ζ ′ ˙ θ sin θ = − ζ ′ ˙ θ ( s ) √ − s Cufaro Petroni:
The falling pencil: a
Divertimento in four movements z ( θ ) / L of the pencil tip laid on a rough table fortwo different kinetic friction coefficients µ k , as a function of θ : as long as θ ≤ θ ∗ itis z ( θ ) / L = 0, but when θ ≥ θ ∗ the function z ( θ ) = ζ (cos θ ) should be calculatedfrom (48): its value is now in the negative. Here again we have chosen cos θ ∗ = 0 . v ( s ) such that ˙ x ( t ) = v (cid:0) s ( t ) (cid:1) , we get ζ ′ ( s ) = − v ( s )˙ θ ( s ) √ − s + L s √ − s ζ ( s ∗ ) = 0 (46)We see moreover from the definitions that¨ x = v ′ ˙ s = − v ′ ( s ) ˙ θ ( s ) √ − s and hence the first dynamical equation (41) becomes v ′ ( s ) = − µ κ m N ( s )˙ θ ( s ) √ − s v ( s ∗ ) = v ∗ = L ω ⊥ s ∗ √ − s ∗ to wit v ( s ) = v ∗ + µ κ m Z s ∗ s N ( r )˙ θ ( r ) √ − r dr (47)By assembling (46) and (47) we finally have ζ ( s ) = Z s ∗ s " θ ( q ) p − q (cid:18) v ∗ + µ κ m Z s ∗ q N ( r )˙ θ ( r ) √ − r dr (cid:19) − L q p − q dq (48)where, with β ∗ = s ∗ √ − s ∗ , it is understood from (42) and (45) that˙ θ ( s ) = ω ⊥ r − β ∗ − s − s N ( s ) = mg s − − β ∗ ) s + 82(4 − s ) Cufaro Petroni: The falling pencil: a
Divertimento in four movements CM xy -trajectory: it coincides with the circular pathof the hinged pencil for θ ≤ θ ∗ , but as soon as θ ≥ θ ∗ it follows different flightaccording to the kinetic coefficient of friction: the parametric equations are (35)together with (48) to calculate ζ ( s ).The integral (48) can be calculated numerically and lends again the possibility ofplotting both z ( θ ) = ζ (cos θ ) (Figure 18), and the trajectory parametric equa-tions (35) together with (48) (Figure 19) where it is understood that ζ ( s ) = 0when s ∗ ≤ s ≤
1. In both the plots we have chosen θ ∗ = arccos 0 .
95 (correspondingto the static coefficient of friction µ s ≃ . µ κ = 0 . µ κ = 0 .
10 . Remark that now,at variance with what we have found in the similar discussion of the Section 4, z ( s )takes negative values for 0 ≤ s ≤ s ∗ accounting for the fact that the pencil tip slidesleftward. From (48) we can also calculate the point x T where the pencil CM hitsthe floor at the time T : since it is θ ( T ) = π / , namely s ( T ) = 0 we will have x T = z T + L ζ (0) + L L − ξ T ) ξ T = − ζ (0) L ≥ θ ∗ = arccos 0 .
95, it is (cid:26) ξ T ≃ . , for µ κ = 0 . ξ T ≃ . , for µ κ = 0 .
10 Cufaro Petroni:
The falling pencil: a
Divertimento in four movements In this paper we have given an elementary treatment of a mechanical case study:the dynamics of a pencil with a tip laid on a rough table and set free to fall un-der the action of gravity. Despite its seeming modesty and lack of pretention wehave shown that a discussion of this simple problem still conceals many details of(maybe) unexpected – but never unsurmountable – intricacy that may turn out tobe pedagogically edifying. Along our exploration we also had the occasion to pointout a few small results of a broader scope, as for instance some critical values ofthe sliding angles and of the static coefficients of friction. We hope that this
Diver-timento could eventually prove to be both profitable and entertaining for all thosewilling to stop for a while to listen at it
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G.F. Franklin, J.D. Powell and A. Emami-Naeini , Feedback Control ofDynamic Systems (Pearson, Boston )[2]
D. Liberzon , Switching in Systems and Control (Birkh¨auser, Boston, )[3] ∼ npetrov/joe-report.pdfrobotics.ee.uwa.edu.au/theses/2003-Balance-Ooi.pdf [4] I.S. Gradshteyn and I.M. Ryzhik , Table of Integrals, Series and Prod-ucts (Academic Press, Burlington2007