International Journal of Non-Linear Mechanics 82 (2016) 24–36
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Dynamics of the rolling and sliding of an elongated cylinder C. Criado a, N. Alamo b,n a b
Departamento de Fisica Aplicada I, Universidad de Malaga, 29071 Malaga, Spain Departamento de Algebra, Geometria y Topologia, Universidad de Malaga, 29071 Malaga, Spain
art ic l e i nf o
a b s t r a c t
Article history: Received 17 March 2015 Received in revised form 19 February 2016 Accepted 20 February 2016 Available online 3 March 2016
A snap of a finger on an elongated cylinder produces on it a surprising fast spinning motion during which its mass center rises with very large oscillations. After that, the mass center goes to a long quasistationary oscillations state and eventually goes down very slowly. To explain this behavior we present a theoretical and numerical analysis of the dynamics of a spinning elongated cylinder moving with a single point of contact on a horizontal plane under the action of gravity. The study has been made taking into account the rolling and sliding dissipation as well as the Kutta–Joukowski airflow effect. The results of the simulations are in agreement qualitatively with the observed real motion. & 2016 Elsevier Ltd. All rights reserved.
Keywords: Non-linear dynamics Rolling and sliding cylinder Kutta–Joukowski Numerical simulation
1. Introduction The origin of this paper was the observation of the surprising fast spin motion that acquires an AA battery after a snap of a finger (click on the figure below to see a slow motion video of the initial instants of the motion, or see the video on youtube: www.youtube.com/ watch?v¼ 4LYv4NsiliQ). Video S1 As in the amazing tippe top or in the spinning egg (see [1–4]), we observe that the mass center of the battery rises with a few very large oscillations, after what the battery goes to a quasi-stationary state while the mass center goes down slowly. To explain this behavior, we have studied the motion of an elongated cylinder with a single point of contact on a horizontal surface under the action of gravity. This problem is a special case of the motion of a rigid body of revolution over a plane, which is a classical mechanics problem that was studied in the past by Routh [5] and Appell [6]. In particular, the motion of a disk has been widely studied (see for instance [7–16]). In the nondissipative case the non-holonomic constrained motions of both disk and cylinder are integrable and, using this fact, several authors have studied their stability as well as the associated bifurcations [7–9]. The dissipative case for a disk of finite thickness has been studied in [10,11] with both sliding and rolling friction. We remark that in all the articles we have found in the literature for the disk motion with friction, the initial conditions are such that the disk always falls over its base. In contrast, we deal with a problem whose initial conditions ensure that the cylinder always falls over one of its edges. Related with this problem is the study of the motion for a conical frustum with sliding friction but without rolling friction in [17]. Also, an interesting study about the rocking by rolling of a cylindrical container has been made in [18]. In our study of the cylinder motion we have considered the sliding and rolling cases, both with dissipation, although without air viscous dissipation. To simulate the pivoting dissipation we also introduce a friction torque which realizes that we have not a contact point but a small contact region. To obtain more accurate equations for the case of an elongated cylinder, we have also introduced the airflow effect using the Kutta–Joukowski lift theorem. On the other hand, our work can be easily generalized to the case of a rotationally symmetric body whose contact point with the horizontal plane is in a circle. These mechanical problems are present in many real situations. Section 2 is divided into three parts. In the first part, we set the notation and establish the general equations of the motion. Then in the following parts, we specialize to the equations for the sliding and rolling cases. In the sliding case, we find six differential equations giving the evolution of the two components of the velocity of the contact point, the three components of the angular velocity, and the inclination angle. In the rolling case, the velocity of the contact point is null, thus we have to determine only four differential equations giving the evolution of the three components of the angular velocity, and of the inclination angle. The details of the calculations are given in the Appendices. n
Corresponding author. E-mail address:
[email protected] (N. Alamo).
http://dx.doi.org/10.1016/j.ijnonlinmec.2016.02.006 0020-7462/& 2016 Elsevier Ltd. All rights reserved.
C. Criado, N. Alamo / International Journal of Non-Linear Mechanics 82 (2016) 24–36
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Video S1. Motion that acquires an AA battery after a snap of a finger (slow motion version).
Fig. 1. Euler angles ðϕ; θ; ψÞ and the coordinate axes used to define the orientation of the cylinder. Below, the AA battery for which we have made the numerical simulation, before and after we give it a fast spin with a snap of a finger.
Section 3 is also divided into three parts. In the first part, we show the results of the numerical integration of the previously obtained differential equations. In order to simulate the rolling and sliding motion of the cylinder, it is assumed that the change in the motion from rolling to sliding occurs when static friction is less than kinetic friction, whereas the change from sliding to rolling occurs when the velocity of the contact point is less than a small fixed value. We consider initial conditions that match with our experiment (initial inclination angle C 0 and initial spinning velocity not too high). We compare the behavior of the cylinder with and without airflow effect and analyze all the relevant dynamical variables of the motion. In particular, our study shows that after a transitory state in which the cylinder inclination angle rises with large oscillations, it goes to a quasi-stationary state during which the rotation and precessional angular velocity remains very high whereas the inclination angle decreases very slowly. We have concluded that the results of the simulations are in good agreement qualitatively with the observed real motion. In the second part of the section we show that, in general, there exits three regions for the inclination angle, for which the cylinder has different qualitative dynamical behavior. In the last part of the section, we give some simple qualitative arguments about the dynamics of the rolling and sliding cylinder.
2. Equations of motion 2.1. General equations of motion Consider a cylinder that rolls or slides, under the action of gravity, with a single point of contact with the horizontal plane. Let m be the mass, R the radius, and L the length of the cylinder. The position of the cylinder is given by the coordinates of its center of mass G with respect to an inertial reference frame and the associated Euler angles. Following Goldstein [19], let us denote by ðϕ; θ; ψ Þ the Euler angles. We will consider three vectorial bases A, B, and C (see Fig. 1) with the following properties: (a) A ¼ fa1 ; a2 ; a3 g denotes a right oriented orthonormal basis, with a3 in the vertical direction, so that the gravity field is g ¼ ga3 . (b) B ¼ fb1 ; b2 ; b3 g is the basis obtained by rotating A around a3 an angle ϕ.
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(c) C ¼ fc1 ; c2 ; c3 g is the basis obtained by rotating B around b1 an angle θ. In Fig. 1, we have represented these bases in a particular position of the contact point M between the cylinder and the horizontal plane. Let O be a fixed point of the horizontal plane, so that fO; fa1 ; a2 ; a3 gg is an inertial reference frame. In the reference frame fG; fc1 ; c2 ; c3 gg the angular velocity of the cylinder is entirely along the symmetry axis through G with direction c2 , and is given by ψ_ , where the dot denotes the derivative with respect to time t. In this context, ϕ is the precession angle, θ is the angle between the horizontal plane and the axis of the cylinder, and ψ is the spin rotation angle. Let r denote the position vector of the center of mass G, and v ¼ r_ its velocity. At any time t, the position of the cylinder is determined by r ¼ ðxðtÞ; yðtÞ; zðtÞÞ, and its orientation is given by the Euler angles ðϕðtÞ; θðtÞ; ψ ðtÞÞ. To obtain these six functions we must integrate the system of six differential equations associated to Newton's second law and the torque equation. Newton's second law is given by: _ F ¼ P;
ð1Þ
where P ¼ mv and F ¼ Pe þ N þ Fd þ FK . Here, (i) Pe ¼ mga3 is the cylinder weight; (ii) N is the normal force acting at the contact point M; (iii) Fd is the friction force acting at M, which for the sliding case we assume to be Fd ¼ μk Nu=j uj , with μk being the coefficient of kinetic friction, and u the velocity of the contact point M. (iv) FK is the total force due to the relative motion of the air (Kutta–Joukowski effect) that will be calculated below (see Eq. (10)). On the other hand, the torque equation is given by: MG ¼ J_ G
ð2Þ
where JG is the angular momentum of the cylinder about the center of mass G and MG is the total moment of all external forces about the center of mass G. Therefore, ! MG ¼ GM 4 ðN þ Fd Þ þ Md þ MK :
ð3Þ
Here Md is the friction torque opposite to the rotation motion, which is introduced because the contact between the cylinder and the horizontal plane is a small region D rather than a single point. This torque is the plane reaction on the cylinder, opposite to the torque that the cylinder rotation exerts on this small region of the plane. We assume that Md is perpendicular to the horizontal plane, proportional to N, and such that, if ω is the angular velocity of the cylinder, ω Md r 0. Therefore we can write Md ¼ sμp Nb3 , where μp is a positive constant and s ¼ sgnðω b3 Þ. We must note that in general, the sliding friction force Fd and the friction pivoting torque Md are not independent of each other (see [20,21]). Namely, Fd is the integral of the force of friction per unit area f on D, whereas the total torque about G due to the friction, MT , is ! the integral on D of the moment of f about G, and so, MT ¼ GM 4 Fd þMd , where Md is the moment of f about the center point M of D, that we assume to be a disk. Because of the complexity of calculating Fd and Md in this way (see [22] for the case of a sphere motion on a horizontal plane), we have made the above assumption on Fd and Md . We denote MK the total moment due to the airflow (Kutta–Joukowski effect), which we calculate below (see Eq. (11)). We shall make now some kinematic considerations. Let ω ¼ ω1 c1 þ ω2 c2 þ ω3 c3 denote the angular velocity of the cylinder in the basis C. Since ω can also be expressed as
ω ¼ ϕ_ b3 þ θ_ c1 þ ψ_ c2 ¼ ϕ_ ð sin θc2 þ cos θc3 Þ þ θ_ c1 þ ψ_ c2 ;
ð4Þ
we obtain
ω1 ¼ θ_ ;
ω2 ¼ ϕ_ sin θ þ ψ_ ;
ω3 ¼ ϕ_ cos θ:
ð5Þ
On the other hand, the cylinder angular momentum JG about the center of mass G is given by JG ¼ I 1 ω1 c1 þ I 2 ω2 c2 þ I 3 ω3 c3 ;
ð6Þ
where Ii is the inertial momentum with respect to the axis through G and parallel to ci , i ¼ 1; 2; 3. Thus, I2 is the inertial momentum with respect to the symmetry axis of the cylinder, and, by symmetry, I 1 ¼ I 3 . For the case of an homogeneous cylinder I 2 ¼ mR2 =2 and I 1 ¼ I 3 ¼ ð3mR2 þ mL2 Þ=12. ! Moreover, the velocity u of the contact point M is given by u ¼ v þ ω 4 GM , and since ! GM ¼ ðL=2Þc2 Rc3 ;
ð7Þ
using Eq. (5) for ω, we deduce that u ¼ v Ab1 þ ω1 D2 b2 ω1 D1 b3 ;
ð8Þ
where A ¼ Rω2 ðL=2Þω3 , D1 ¼ ðL=2Þ cos θ R sin θ, and D2 ¼ ðL=2Þ sin θ þ R cos θ. Let us calculate now the Kutta–Joukowski effect on a spinning cylinder. According to the Kutta–Joukowski theorem (see [23]), the lift force per unit length exerted on a spinning cylinder by an airflow of velocity v orthogonal to the cylinder axis is orthogonal to the velocity v of the flow and to the cylinder axis, and has modulus f K ¼ ρΓ v, where ρ is the air density, and Γ, which is called vortex strength, is given in terms of the angular velocity of spin of the cylinder ψ_ and its radius R by Γ ¼ 2π ψ_ R2 . In our case, we do not consider the effect of the airflow on the cylinder tops, and therefore we can approximate the lift force on the cylinder integrating the lift force on a cylinder element
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of width dl, that is: Z FK ¼
L=2 L=2
f K dl:
ð9Þ
To calculate f K we assume that the airflow on this cylinder element in the reference frame fG; fc1 ; c2 ; c3 gg is uniform and is given by the orthonormal component to c2 , v0Q of the velocity vQ for Q the center of the cylinder element with minus sign. Thus we find f K ¼ ! ! 2πρR2 ð vQ 4 ψ_ Þ ¼ H ψ_ vQ 4 c2 , where we denote H ¼ 2πρR2 . Note that v0Q 4 c2 ¼ vQ 4 c2 . Since vQ ¼ vM þ ω 4 MQ and MQ ¼ ððL=2Þ þ lÞc2 , integrating we obtain L ð10Þ FK ¼ H ψ_ L u2 sin θ ω1 c1 þ ðu1 þ AÞc3 : 2 In the same way, we can calculate the total moment MK due to the forces f K with respect to G: Z MK ¼
! L3 GQ 4 f K dl ¼ H ψ_ ð ω3 c1 þ ω1 c3 Þ: 12 L=2 L=2
ð11Þ
We must note that the above calculation is an approximation in which we have not taken into account the effect of the airflow on the cylinder due to its acceleration, so FK and MK depend only on u and ω but not on their derivatives. A more precise study of the airflow effect would be extremely complicated and is out of the aim of this paper. 2.2. Sliding cylinder For the sliding motion the velocity u of the contact point lies in the horizontal plane, i.e., u ¼ u1 b1 þu2 b2 , or equivalently, u b3 ¼ 0. On the other hand, N ¼ N b3 ¼ ðmv_ þ mgb3 Fd FK Þ b3 ¼ mv_ b3 þ mg FK b3 :
ð12Þ
Fig. 2. Evolution with time for different initial conditions of the inclination angle θ with (gray (red in the online version)) and without (black (blue in the online version)) Kutta–Joukowski airflow effect. In all the cases we see that after a transitory state in which θ rises with large oscillations, it goes to a quasi-stationary oscillations state. Thick and thin lines for sliding and rolling, respectively. (For interpretation of the references to color in this figure caption, the reader is referred to the web version of this paper.)
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_ 1 D1 ω21 D2 , and then we get Moreover, from Eq. (8), we deduce that v_ b3 ¼ ω _ 1 D1 ω21 D2 Þ þHLψ_ ðu1 þ AÞ cos θ: N ¼ mðg þ ω
ð13Þ
Using this value for N in Newton's second law and in the torque equation we obtain the differential equations that give the evolution of the velocity u of the contact point M, the angle θ, and the angular velocity ω. They are six non-linear first order differential equations involving u1 ; u2 ; θ; ω1 ; ω2 , and ω3 that we can write in matrix form: 0 10 1 0 1 u_ 1 B1 1 0 0 a14 a15 a16 B C B u_ C B C 0 1 0 a 0 0 B C B 2 C B B2 C 24 B CB C B C B0 0 1 0 C B θ_ C B B3 C 0 0 B CB C B C ð14Þ B C: B C¼B C _ 1 C B B4 C 0 0 C Bω B 0 0 0 a44 B CB C B C B 0 0 0 a54 a55 B _ C B B5 C 0 C 2A @ A @ω @ A 0 a66 B6 0 0 0 a64 ω_ 3 where the values of the coefficients and the details of the calculations are given in Appendix A. 2.3. Rolling cylinder For a rolling cylinder the velocity u of the contact point is zero, and hence Eq. (8) can be rewritten as v ¼ Ab1 ω1 D2 b2 þ ω1 D1 b3 . Substituting this expression for v in Eq. (1) we obtain Fd ¼ F d1 b1 þ F d2 b2 , where 2 L _ H L ψ_ ω ; _ 2 ω _ 3 þ ω1 D2 ϕ ð15Þ F d1 ¼ m Rω 1 2 2 _ ω _ 1 D2 ω21 D1 Þ HLψ_ A sin θ: F d2 ¼ mðAϕ
ð16Þ
Fig. 3. Evolution with time of the inclination angle θ. Thick and thin lines for sliding and rolling, respectively. After a transitory state in which θ rises with large oscillations, it goes to a quasi-stationary oscillations state with decreasing frequencies (bottom left) and amplitudes (bottom right).
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Fig. 4. Time evolution of (a) precessional velocity, (b) spin velocity, (c) energies E; ER and ET þ Ep , (d) energies Ep and ET, (e) number of precessional turns, and (f) the ratio between first and second members of Eq. (21). In the evolution of the number of precessional turns ϕ, thick and thin lines correspond to sliding and rolling, respectively.
Fig. 5. Top view of the motion of the center of mass G corresponding to the three cases represented in Fig. 2.
Now, substituting the above expressions in Eq. (25) and using the torque equation, we obtain four non-linear first order differential equations involving θ; ω1 ; ω2 , and ω3 that can be written in matrix form: 0
1 B0 B B B0 @ 0
0
0
a044 a054 a064
a055
0 a065
1 0 0 1 B3 θ_ C B C B 0 C 0 C Bω _ 1 C B B4 C C: B C B C B _ C ¼ B B0 C a056 C A @ω @ 5A 2A a066 B06 ω_ 3 0
10
where the values of the coefficients and the details of the calculations are given in Appendix B.
ð17Þ
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3. Results and discussion of the numerical integration 3.1. General results We have solved numerically the differential equations obtained in the previous sections to simulate the motion of the cylinder. We have found that they are numerically very unstable, so we have used a variable step size method, and many hours of computing time. Beside at each step of the simulation we have to decide if the cylinder is sliding or rolling and then use the corresponding differential equations (14) (or 17) to calculate the motion. It is assumed that the change in the motion from rolling to sliding occurs when static friction is less than kinetic friction, that is, when μs N o F d , where μs is the coefficient of static friction. Otherwise, we change from sliding to rolling when the velocity of the contact point is less than a small fixed value u0. When we change from rolling to sliding we take u0 as the initial value of u, with the opposite direction of Fd . After obtaining uðtÞ; θðtÞ, and ωðtÞ, we use Eq. (8) to calculate the velocity of the center of mass, vðtÞ, and then by integration, we obtain the evolution of the remaining Euler angles ϕ and ψ, as well as of the center of mass rðtÞ. In the inertial reference frame fO; fa1 ; a2 ; a3 gg the coordinates of the velocity r_ ðtÞ of the center of mass are given by 8 _ > < x ¼ ðu1 þ AÞ cos ϕ ðu2 D2 ω1 Þ sin θ y_ ¼ ðu1 þ AÞ sin ϕ þ ðu2 D2 ω1 Þ cos θ ð18Þ > : z_ ¼ D ω : 1 1
Video S2. Simulation of the fast spin motion that acquires an AA battery after a snap of a finger. We have used the numerical integration data of the differential equations associated to the sliding and rolling motions (slow motion version).
Video S3. Simulation of the fast spin motion that acquires an AA battery after a snap of a finger. We have used the numerical integration data of the differential equations associated to the sliding and rolling motions. Side isometric view, slow motion version.
Fig. 6. Regions I, II, and III for the inclination angle θ as a function of the ratio k of the cylinder dimensions. In region II, the stationary motion is not possible. k1 corresponds to the value of k for which θ1 ¼ θ2 , and k2 corresponds to our cylinder dimensions.
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For the simulations we have used the dimensions and weight of an AA battery (see Fig. 1), which is also the cylinder with which we have carried out our observations. These data are R¼ 0.7 cm, L ¼4.8 cm, m ¼11 g. We used the following coefficients of friction: μs ¼ 0:35; μk ¼ 0:3; μp ¼ 1:4 10 4 ; ρ ¼ 1:18 kg=m3 for the air density, and u0 ¼ 10 5 m=s. We have made simulations for many different initial conditions but, to compare with the experiment in which we give a fast spin with a snap of a finger on an AA battery, we present the results for the following initial conditions: rð0Þ ¼ 0, vð0Þ ¼ 0, θ_ ð0Þ ¼ 0 rad=s, ϕð0Þ ¼ ψ ð0Þ ¼ 0 rad, θð0Þ ¼ 0:01 rad. With respect to the initial rotation and precessional velocities ψ_ ð0Þ and ϕ_ ð0Þ, we have considered three cases which are within the intervals of the initial conditions that we have determined through trial and errors for the finger snap: (a) ψ ð0Þ ¼ 280 rad=s, ϕ_ ð0Þ ¼ 80 rad=s, (b) ψ_ ð0Þ ¼ 375 rad=s, ϕ_ ð0Þ ¼ 84 rad=s, (c) ψ_ ð0Þ ¼ 400 rad=s, ϕ_ ð0Þ ¼ 70 rad=s. We show in Fig. 2 the variation of the inclination angle θ with respect to time t for the above three initial conditions. In each of these graphs the gray and black lines (red and blue in the online version) correspond to whether or not the Kutta–Joukowski effect is considered. We see that after a transitory state, in which the angle rises with large oscillations, the cylinder goes to a state of quasi-stationary oscillations. We note that both the transitory time and the amplitudes of the oscillations are larger for larger initial rotation velocities ψ_ ð0Þ. We also see that the Kutta–Joukowski effect mainly affects the transitory time, reducing the amplitudes of the oscillations. In these graphs we have represented the sliding and the rolling of the cylinder by thick and thin lines respectively. Along the transitory time interval, the cylinder slips most of the time while in the quasi-stationary state the cylinder is mainly rolling. In Fig. 3 we have represented the whole evolution of the inclination angle for the case (a), that is the case nearest to a standard finger snap. We see that after a transitory state (interval ½0; 0:3 s), in which the angle rises with large oscillations, the cylinder goes to a state of quasi-stationary oscillations (interval ½0:3; 7:1 s). Along the transitory time interval, the cylinder slips most of the time, while in the quasistationary state the cylinder is mainly rolling in the interval ½0:3; 4:4 s (thin line), and sliding in the interval ½4:4; 7:1 (thick line). The finding that, at intermediate times, the cylinder is mainly rolling coincides with what is found experimentally in the case of a short cylinder (disk) in [11]. Note also that during the rolling time interval ½0:3; 4:4 s, there are several rolling subintervals separated by very short sliding time intervals. Along the quasi-stationary state, the frequency of the oscillations decrease (see Fig. 3 bottom left). Observe also (Fig. 3 bottom right) the light increasing of the amplitude of the oscillations during the rolling subintervals; the stair decrease of θ at the short sliding steps; and the monotonous decreasing of both, frequency and amplitude, at the final sliding time interval. We have integrated Eqs. (14) and (17) for many different initial conditions and, in all the cases, the pattern is the same as described. We have also tried with different coefficients of kinetic friction μk, and we have seen that for higher values of μk the initial rising of θ is larger, from which it follows that the sliding friction is the cause of the initial rising of the center of mass. _ and the spin velocity ψ_ is shown in Fig. 4(a) and (b). The decrease of both The evolution with time of the precessional velocity ϕ follows a similar oscillations pattern, with decreasing frequencies and amplitudes. In Fig. 4(c) we have represented the monotonous decrease of the energy E, the rotation kinetic energy ER, and the sum of the potential energy Ep and the kinetic translational energy ET. Note that this sum is one order of magnitude lower than ER. In Fig. 4(d) we have represented Ep and ET; observe that ET is one order of magnitude lower that Ep. All these energies have been calculated using the expression: E ¼ ET þER þ Ep
E¼
1 m 2
"
ð19Þ
L u1 þ Rω2 ω3 2
2
# þ ðu2 ω1 D2 Þ2 þðω1 D1 Þ2 þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 I 1 ðω21 þ ω23 Þ þ I 2 ω22 þ m g R2 þ ðL=2Þ2 ð sin ðθ þ βÞ sin βÞ: 2
ð20Þ
We have also represented the number of precessional turns ϕ (Fig. 4(e)) to be able to compare with the number observed in the _ depends on θ: experiment. In the rolling quasi-stationary state for the ideal non-friction case, the ratio ψ_ =ϕ
ψ_ L cos θ sin θ ¼ _ 2R ϕ
ð21Þ 90 80 70
θ (deg)
60 50 40 30 20 10 0
0
1
2
3
4
5
6
7
8
9
t (s) Fig. 7. Time evolution of the inclination angle θ in which can be seen the three different qualitative dynamical behavior. Thick and thin lines represent sliding and rolling motions respectively.
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C. Criado, N. Alamo / International Journal of Non-Linear Mechanics 82 (2016) 24–36
as follows from Eq. (22) of next section. This equation has been derived previously in [11]. We have verified this equality approximately for the results of the numerical integration, as can be seen in Fig. 4(f), in which we have represented the time evolution of the ratio (rat) between the first and the second member of Eq. (21). Finally in Fig. 5 we have represented the top view of the center of mass for the three cases represented in Fig. 2 of the cylinder motion. Its coordinates have been calculated integrating Eq. (18). The gray and black lines correspond to be considering or not the Kutta–Joukowski airflow effect. As can be seen, there are very small differences between the two cases. Because of the imprecision of the initial conditions in our experiments we are not able to decide which fits better to the observed motion. Note that, after the initial transitory state, the movement of the center of mass G is very small, with oscillations of about 2 mm. The different initial displacement of the center of mass is _ ð0Þ and ψ_ ð0Þ makes larger the transitory time interval and then larger disassociated to different transitory time intervals. Larger ϕ placement of the mass center, that in the considered cases varies from 1 to 26 cm. Using numerical integration data we have made two animations of the motion of the cylinder. Both videos are slow motion versions, so we can appreciate the cylinder lifting from the floor. The second is a side isometric view of the first. Click on the images below or watch the videos www.youtube.com/watch?v ¼yVlnYBB4Rpc, and www.youtube.com/watch?v¼ sAPMe_AeUuo on youtube. Video S2 Video S3 The above simulated results fit well qualitatively with the observed motion in a video made with a camera Casio EX- ZR100 using 480 frames per second. We have not made a precise quantitative measurement of the variables of the cylinder motion in real experiments. We have only approximated data about the total time until the cylinder impact the horizontal plane, the total number of precessional turns, the transitory time interval, and the displacement of the mass center, and all of them agree well with the simulation. It would be very interesting to design an experiment to have precise real data of all the variables. 3.2. Regions with different qualitative dynamical behavior The above study has been made for energies for which the inclination angle does not exceed 451, that is, those for which we have been able to make observations with the battery. However we have found that, in general, there are three regions for the inclination angle, for which the qualitative dynamical behavior of the cylinder seems different. The values of θ that separates these regions depend on the ratio L=R of the cylinder dimensions. Fig. 6 shows these three regions for the ideal non-frictional case. To find these regions for the ideal non-frictional case, consider the stationary case, in which the center of mass is at rest ðv ¼ 0Þ. Given ! that in the rolling case the speed of the contact point is zero ðu ¼ 0Þ, we obtain from Eq. (25), ω 4 GM ¼ 0. _ Therefore, using Eq. (4) with θ ¼ 0, and Eq. (7), we get L Rðϕ_ sin θ þ ψ_ Þ þ ϕ_ cos θ ¼ 0: 2
ð22Þ
_ of precession The above equation relates, for the rolling case, the angular velocity ψ_ of the rotation of the cylinder with the velocity ϕ around the vertical axis. On the other hand, for the non-friction case and non airflow effect, we have the following expressions: ! L MG ¼ GM 4 mgb3 ¼ mg R sin θ cos θ c1 ; ð23Þ 2 J_ G ¼ ððI 1 I 2 Þϕ_
2
sin θ cos θ I 2 ψ_ ϕ_ cos θÞc1 :
ð24Þ
_ for a homogeneous cylinder: Thus, from the torque equation and Eq. (22), we can determine ϕ
ϕ_ ¼ 7
6 gð2R tan θ LÞ
ð3R2 þ L2 Þ sin θ 3RL cos θ
!1=2 :
ð25Þ
The above equation has also been derived in [11] and used together with Eq. (21) to test experimentally whether slipping occurs or not during the motion of a disk.
_ The torque M1 which Fig. 8. Illustration of the friction force Fd1 that is the responsible of the decreasing of the spin velocity ψ_ and the increasing of the precession velocity ϕ. is the responsible of the rising of the cylinder is also depicted. For opposites Fd1 and M1, the conclusion is reversed.
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Let
33
θ1 and θ2 be the values of θ for which the numerator and denominator of the above equation, respectively, vanish, that is:
θ1 ¼ arctan k;
θ2 ¼ arctan
6k 3 þ 4k
2
;
with k ¼ L=2R. To ensure that the radicand of Eq. (25) is positive, it has to occur one of the following two cases: ðaÞ
π 2
4 θ 4 θi ; i ¼ 1; 2
or
ðbÞ 0 o θ o θi ; i ¼ 1; 2:
pffiffiffi pffiffiffi pffiffiffi pffiffiffi Observe that θ1 ¼ θ2 for k ¼ k1 ¼ 3=2 or equivalently for L ¼ 3 R. For L 4 3 R (L o 3 R respectively) we have θ1 o θ2 (θ1 4 θ2 , respectively). pffiffiffi In Fig. 6 we have represented θ1 and θ2 for the case L 4 3 R that is the case of the battery of our observations. We have considered three regions: region I, where θ o θ1 ; region II, where θ1 o θ o θ2 ; and region III, where θ2 o θ. In the region II, because the radicand of Eq. (25) is negative, the stationary motion is not possible. We have tested experimentally the existence of these three regions in the frictional case doing some observations and simulations. We have found that the quasi-stationary oscillations never appear in region II. We have observed that the cylinder can be initially in any of the three regions depending on the initial conditions, but if it is in the region II, then after a short transition time interval, the cylinder reaches the quasi-stationary oscillations in the region I; and if the cylinder is initially in region III and has enough spinning velocity ψ_ , it makes quasi-static sliding oscillations and then, after some time, θ decreases very fast through region II and goes to the quasi-stationary rolling oscillations of region I. In Fig. 7 we show the graphic of a simulation of the cylinder motion with the evolution of the inclination angle along the three regions. _ ð0Þ ¼ 40 rad=s. We can observe that the cylinder rises doing quasi-stationary The initial conditions are θð0Þ ¼ 601, ψ_ ð0Þ ¼ 880 rad=s, and ϕ sliding oscillations of increasing amplitude (region III), and after being almost vertical, the cylinder goes down very fast (region II), and then begin to do rolling quasi-stationary oscillations (region I). We have not made observations in the region III because we cannot produce in our battery the necessary high spin velocity ψ_ , but in all the simulations the results are similar to that obtained in Fig. 7.
Fig. 9. Time evolution of (a) spin velocity, (b) precessional velocity, (c) inclination angle (thick and thin lines represent sliding and rolling motions respectively), (d) torque M1, (e) friction force for sliding, (f) friction force for rolling, (g) potential energy, and (h) rotation kinetic energy. The vertical line has been drown to make easier the comparison.
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3.3. Some simple qualitative explanations In this section we discuss qualitatively the dynamics of the rolling and sliding of an elongated cylinder in region I by means of arguments similar to that used in [1,3]. ! We begin analyzing the influence of the velocity of the contact point, u ¼ v þ ω 4 GM on the evolution of the angular precession velocity ϕ_ , the spin velocity, ψ_ , and the inclination angle θ. From Eq. (8) we obtain that the component u1 of u in the direction of b1 is (see Fig. 8): u1 ¼ v1 þ P Q
ð26Þ
where P ¼ ðL=2Þϕ_ cos θ and Q ¼ Rðϕ_ sin θ þ ψ_ Þ. The friction force associated to u1 is Fd1 ¼ μk Nu1 =j uj , opposite to the direction of u1 ¼ u1 b1 , thus if u1 o 0, then Fd1 has the direction of b1 , which is opposite to the direction of the spin velocity ψ_ and the same that the precession velocity ϕ_ (see Fig. 8). Therefore, the spin velocity will decrease while the precession velocity will increase. We have made the above considerations for the case of the sliding cylinder, but they apply also to the case of the rolling cylinder. In the case of the rolling cylinder the friction force Fd1 is governed by Eq. (15). We show in Fig. 9(e) and (f) the evolution of these friction forces that we call Fs1 in the sliding case and Fr1 in the rolling case. For the actual case, only thick lines of both graphics must be considered. Note that the friction force for the case of rolling is one order of magnitude lower than for the sliding; for this reason we have made two separated graphics. We can observe, comparing the graphics (a), (b), (e) and (f) in Fig. 9, that when Fd1 is positive ψ_ decreases and ϕ_ increases, whereas when Fd1 is negative ψ_ increases and ϕ_ decreases. In our case, the initial spin velocity was ψ_ ð0Þ ¼ 280 rad=s, the precession velocity was ϕ_ ð0Þ ¼ 80 rad=s, and the inclination angle was very low, θð0Þ ¼ 0:01 rad. For these values we have P CðL=2Þϕ_ ð0Þ and Q C Rψ_ ð0Þ, so that for v C 0 we get u C 0:04 m=s, and then, the 1
1
friction force Fd1 has the direction of b1 (see Fig. 8), and from the above argument, the spin velocity ψ_ decreases and the precession velocity ϕ_ increases, see Fig. 9(a) and (b).
Here, as in the spinning top (see [1]), if we hurry the precession, the cylinder rises, whereas if we retard the precession, the cylinder falls. In fact, to see this in our case, consider the component of the friction force in the direction of b2 , Fd2 . This force, that provides the necessary normal acceleration, produces a torque M1 (see Fig. 8) that tends to rise the cylinder, whereas the reaction normal force N produces a torque that tends to make it falls. In Fig. 9(d) we have shown the evolution of the total torque with respect to G in the direction of b1 . Comparing the graphics (a), (c) and (d) in Fig. 9 we see that when ϕ_ increases M1 is positive and θ rises, and the opposite happens when ϕ_ decreases. After the initial rising of the cylinder, the increasing of ϕ_ and the decreasing of ψ_ make that after some given value of θ, P 4 Q , and then, the signs of u1 and F d1 are changed, causing that ψ_ increases and ϕ_ decreases. Simultaneously, M1 becomes negative and so θ decreases, until the moment at which again Q 4 P and then the process repeats. In the rolling motion the analogous evolution of F d1 and M1 (see Fig. 9(f)) produces a similar behavior of ϕ_ , ψ_ , and θ. From the point of view of the energy evolution, the cylinder rises so its potential energy Ep increases (see Fig. 9(g)) at the expense of its kinetic energy of rotation, Ecr (see Fig. 9(h)). The decreases of Ecr correspond essentially to the decreases of the energy of spin rotation. The above analysis shows how friction acts as a self-adjusting force and produces an oscillating tendency to return to the steady motion.
4. Conclusion This paper presents a theoretical and numerical analysis of the motion of an elongated cylinder with a single point of contact on a horizontal plane under the action of gravity. Transitions between rolling and sliding, both cases with dissipation, have been taken into account. It has also been considered the airflow effect using the Kutta–Joukowski theorem. The results of the numerical integration of the obtained differential equations for the case of an AA battery show that, after a transitory state in which the inclination angle rises with large oscillations, the cylinder goes to a state of quasi-stationary oscillations with θ decreasing slowly. In the transitory state interval, the cylinder slips most of the time while, in the quasi-stationary interval, there are two regimes of motion: in the first one the cylinder is rolling during several time subintervals separated by very short sliding times; in the second, the cylinder only slips until the final collision with the horizontal plane where θ ¼ 0. The results of the simulations with and without airflow effect have been compared, and it has been concluded that this effect is mainly important for large rotation velocities and therefore mainly in the transitory time. It has also shown that there exists a region for the inclination angle where it is not possible the quasi-stationary oscillation state. Time evolution of the precessional velocity, the spin velocity, the friction force, and the energy, as well as the motion of the center of mass have also been characterized. We also give some simple qualitative explanations of the motion of the sliding and rolling cylinder. All these results are qualitatively in agreement with the observed real motion.
Acknowledgments This work was partially supported by the Junta de Andalucia Grants RNM-1399 (C. Criado), FQM-192 (C. Criado) and FMQ-213 (N. Alamo), and by the Ministerio de Economía y Competitividad Grant MTM2013-41768-P (N. Alamo). We also thank the reviewers for pointing out some important remarks that have improved the paper.
C. Criado, N. Alamo / International Journal of Non-Linear Mechanics 82 (2016) 24–36
35
Appendix A. Derivation of the equations of the sliding cylinder From Eq. (8) we have u1 ¼ u b1 ¼ v b1 A, thus differentiating this expression and using Newton's second law (1) we obtain 1 1 _ u_ 1 ¼ v_ b1 þ v b_ 1 A_ ¼ Fd b1 þ FK b1 þ v ϕ_ b2 A: m m
ðA:1Þ
By using the values of Fd , FK , and N, and after some calculation, we get the differential equation _ 1 þ a15 ω _ 2 þ a16 ω _ 3 ¼ B1 ; u_ 1 þa14 ω
ðA:2Þ
where 8 a14 ¼ μ1 D1 > > qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > > 2 2 > > < μi ¼ μk ui = u1 þ u2 ; i ¼ 1; 2 a15 ¼ R > > > > > a16 ¼ L=2 > > : B ¼ μ ðg D ω2 Þ þ ðu D ω Þω sec θ ðHL=mÞðω ω tan θÞ½μ cos θðu þAÞ þ u sin θ ðL=2Þω : 1 2 1 2 2 1 3 2 3 1 2 1 1 1 Similarly, by differentiating u2 in Eq. (8) and using Eq. (1), we obtain the differential equation _ 1 ¼ B2 ; u_ 2 þa24 ω
ðA:3Þ
where ( a24 ¼ μ2 D1 D2
B2 ¼ μ2 ðg D2 ω21 Þ ðu1 þ AÞω3 sec θ þ ω21 D1 HLðω2 ω3 tan θÞðu1 þ AÞðμ2 cos θ sin θÞ: The third differential equation is
θ_ ¼ B3 ; where B3 ¼ ω1 . To obtain the remaining differential equations we will use the torque equation. After some calculation in Eq. (25) we get " # " # L3 L L3 _ _ MG ¼ NðD1 þ μ2 D2 Þ þ H ψ ω3 c1 þ Nðμ1 R μp s sin θÞc2 N μ1 þ μp s cos θ þ H ψ ω1 c3 : 2 12 12
ðA:4Þ
ðA:5Þ
On the other hand, by differentiating the expression for the angular momentum, Eq. (6), we get: _ 1 I 2 ω2 ω3 þ I 1 ω23 tan θÞc1 þ I 2 ω _ 2 c2 þ ð I 1 ω1 ω3 tan θ þ I 2 ω1 ω2 þ I 3 ω _ 3 Þc3 : J_ G ¼ ðI 1 ω
ðA:6Þ
Now, comparison between Eqs. (A.5) and (A.6) leads to the following three differential equations: _ 1 ¼ B4 ; a44 ω
ðA:7Þ
where ( a44 ¼ I 1 þ mD1 ðD1 þ μ2 D2 Þ; B4 ¼ mðg D2 ω21 ÞðD1 þ μ2 D2 Þ þ I 2 ω2 ω3 I 1 ω23 tan θ HLðω2 ω3 tan θÞ½ðu1 þ AÞðD1 þ μ2 D2 Þ cos θ ðL2 =12Þω3 ; _ 1 þ a55 ω _ 2 ¼ B5 ; a54 ω
ðA:8Þ
where 8 a ¼ mD1 D3 ; D3 ¼ μ1 R mup s sin θ; > < 54 a55 ¼ I 2 > : B ¼ mðg D ω2 ÞD þHLðω ω tan θÞðu þ AÞ cos θD ; 5 2 1 3 2 3 1 3 _ 1 þ a66 ω _ 3 ¼ B6 ; a64 ω
ðA:9Þ
where 8 a ¼ mD1 D4 ; D4 ¼ μ1 ðL=2Þ þ μp s cos θ; > < 64 a66 ¼ I 1 > : B ¼ mðg D ω2 ÞD I ω ω þI ω ω tan θ HLðω ω tan θÞ½ðL2 =12Þω þ ðu þ AÞD cos θ: 6 2 1 4 2 1 2 1 1 3 2 3 1 1 4 The system of differential equations formed by the six equations (A.2)–(A.4) and (A.7)–(A.9) can be written in the matrix form of Eq. (14).
Appendix B. Derivation of the equations of the rolling cylinder Substituting in Eq. (25) the expression of Fd given by Eqs. (15) and (16), and using the torque equation, we obtain the following three differential equations: _ 1 ¼ B04 ; a044 ω
ðB:1Þ
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C. Criado, N. Alamo / International Journal of Non-Linear Mechanics 82 (2016) 24–36
where 8 < a0 ¼ I 1 þ mððL=2Þ2 þ R2 Þ; 44 : B04 ¼ mgD1 þ mD2 Aω3 sec θ þI 2 ω2 ω3 I 1 ω23 tan θ HLðω2 ω3 tan θÞ½AððL=2Þ þ R sin 2θÞ ðL2 =12Þω3 ; _ 1 þ a055 ω _ 2 þ a056 ω _ 3 ¼ B05 ; a054 ω
ðB:2Þ
where 8 a0 ¼ μp msD1 sin θ; > > > 54 > > < a0 ¼ I þ mR2 2 55 0 > > a56 ¼ mRðL=2Þ > > > : B05 ¼ mRD2 ω1 ω3 sec θ μp msðg D2 ω21 Þ sin θ þ HLðω2 ω3 tan θÞðω1 RðL=2Þ μp s sin θ cos θÞ; _ 1 þ a065 ω _ 2 þ a066 ω _ 3 ¼ B06 ; a064 ω
ðB:3Þ
where 8 0 a ¼ μp msD1 cos θ; > > > 64 > > < a065 ¼ mRL=2 a066 ¼ I 1 þ mL2 =4 > > > > > : B06 ¼ mðL=2ÞD2 ω1 ω3 sec θ μp msðg D2 ω21 Þ cos θ I 2 ω1 ω2 þ I 1 ω1 ω3 tan θ HLðω2 ω3 tan θÞðω1 ðL2 =3Þ þ μp s A cos 2 θÞ: The above three differential equations together with the equation θ_ ¼ ω1 ¼ B03 can be written in the matrix form of Eq. (17).
Appendix C. Supplementary material Supplementary data associated with this paper can be found in the online version at http://dx.doi.org/10.1016/j.ijnonlinmec.2016.02. 006.
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