Oscillation modes of a rod8 Introduction A cylindrical rod suitably prepared may be made to oscillate around many points and axes, thereby revealing a surprising variety of physical phenomena, ranging from a regular physical pendulum through bifilar twisting motion and torsional oscillations, all the way to possible chaotic behavior in a double pendulum [1–3]. The construction of the various suspension and support points, the period measuring device, and the theoretical calculations of the appropriate periods all require ingenuity and increasing depths of understanding of the rigid body dynamics involved, and provide plenty of scope for playing around with the physics. Introduction 49 Theoretical ideas 49 Experimental suggestions 52 Theoretical ideas Let us suppose that we have prepared some means of supporting the rod in a series of positions illustrated by the diagrams that follow, Figs 8.1–8.12. In each case we shall describe the modes of oscillation and provide the formula for its period – proving them is in some cases a non-trivial exercise and may be considered part of the project. Proofs are given in the more complex cases. Figure 8.1 shows a rod of radius R, length L, and mass M having a series of holes drilled symmetrically on either side of the center of mass (com). The holes are of radius r h . A pin of radius rp < rh can be used to suspend the rod from any of the holes, or strings can be attached to the rod through them. Vertical pendulum R L com Figure 8.1 Drilled rod. The rod can oscillate in a vertical plane by suspending it from a pin of radius r p through one of the holes, as in Fig. 8.2. Figure 8.3 shows that the top of the hole will roll on the pin without slipping, keeping the point of contact at rest. If the distance from the com to the top of the hole is d, and if one neglects the finite diameter of the pin, the period for small oscillations is given by T0 = 2π ! (L 2 /12) + d 2 . gd T as a function of d will exhibit a well-known minimum. θ (8.1) Figure 8.2 Vertical oscillations. or of different lengths. The rod oscillates about an axis parallel to its long axis. making an angle θ with the rod at the com of the rod – see Fig.3 Finite radius. (8. (d) Inclined bifilar pendulum.6. For large amplitude oscillations the period will once again be given by an elliptic integral.4) (c) Torsional oscillations about a vertical axis through the center of mass (see Fig. The period is given by " T = 2π (0.5 (H is measured from the point of suspension to the rod axis). (8.4 Horizontal pendulum. Horizontal pendulum Figure 8.5 Bifilar suspension. 8. 8. in which case the period will be given by an elliptic integral [1] which can be evaluated numerically for any amplitude. like a swing.5R2 + H 2 )/gH. Axis for “swing” oscillations H Axis for torsional oscillations Figure 8. to take into account the finite radius of the pin. in its equilibrium plane. 8. (a) In and out of its equilibrium plane.7. secondly. The axis of rotation is parallel to the rod.3.2) Bifilar suspension Here. parallel to the rod length. Let the system execute torsional oscillations about a vertical axis. and vice versa. Figure 8. shown in Fig. passing through the suspension points of the strings. while the left end of the rod moves out of the paper the right end moves into the paper. The period for small oscillations will be ! H ! L/d T = 2π √ .6) .5).3) (b) Laterally. The period is given by " T = 2π (H – R)/g.50 Physics Project Lab A good exercise is to extend this formula in two ways: First. Its period is given by " T = 2π 3R/2g. Several motions are possible.5) g 12 H H θ Figure 8. to be evaluated numerically. Pins. the rod is suspended horizontally by two vertical parallel strings which may be of equal length H . 8. In this case. The moment of inertia about the vertical axis is given by I = M{L 2 /12 + R2 /4}cos2 θ + M{R2 /2}sin2 θ . where H ! = H – R. as in Fig.6 Inclined pendulum. (8. as in Fig. (8. 8. 8. (8.4. as shown in Fig. attached to the ends of the rod. to large amplitudes. Consider first equal length strings. are supported horizontally. 8. as in Fig. 2 2 # % $ $ 1 2 2 = Mg H1 + H2 – H1 – x1 – H2 – x2 . These parameters are related by xcm = (x1 + x2 )/2 and θ = (x1 – x2 )/2d. the result of (8. in which ω02 = g/Heff . The method of deriving the normal mode frequencies is similar to that for case (e) above.8) The Lagrangian is L = Ek – U .9) and (8. the center of mass moves by xcm .7) H2 H1 d Figure 8. 2 Ek = U = Mgycm (8. From (8. For swing-like oscillations.9.10) x2 d x1 θ xcm Figure 8. as in Fig. Let d be the distance of the points of attachments of the strings to the center of mass. (8. the period is expected to be the result of (8.8 Coordinates for unequal suspensions. this time the dynamical variables are θ1 and θ2 . 8.9) θ¨ = –ω02 λ(θ – εxcm /d). say θ < π /6. ω02 = g/H .11) and the periods T± = 2π /ω± . 2(1 – d/s) where s = [L 2 /12 + d 2 ]/d. 8. (8. and the frequencies are given by 2 ω± & ' $ ω02 2 = (H /s + 1) ± (H /s + 1) – 4(H /s)(1 – d/s) .3) is still valid.51 Oscillation modes of a rod For small angles. θ1 d θ2 xcm (8.7 Unequal suspensions. H Double pendulum Here the rod is hung from a string passing through one of its holes. λ = md 2 /Icm .9 Double pendulum. In a torsional displacement of the rod the attachment points of the string move out of the equilibrium plane by x1 and x2 . as in Fig. H1 H2 (8.8. ε = (H1 – H2 )/2Heff . In the small-angle approximation one obtains the coupled Euler–Lagrange equations for the two variables θ and xcm : x¨ cm = –ω02 (xcm – εdθ ).10) one gets the normal mode frequencies & ' $ 1 2 ω± = ω02 (1 + λ) ± 4(1 – λ)2 + λε2 2 (8. while the rod swings by an angle θ . (e) Horizontal rod suspended from unequal-length strings.7. . Heff = 2H1 H2 /(H1 + H2 ).12) Figure 8. The kinetic and potential energies of the system in terms of these parameters are given by 1 1 Icm θ˙ 2 + Mx2cm .5) multiplied by cos θ. String and rod can either swing together or in opposite directions. 52 Physics Project Lab Chaotic pendulum Support String Wire As shown in Fig. The period is the well-known expression " T = 2π 2HI /Gπ r 4 .11 Torsional pendulums. swinging unpredictably in one direction or the other.13) where G is the shear modulus. 8. and a distance s from the point of attachment of the string. Torsional pendulum Rod The torque in a wire of length H and radius r when twisted through an angle θ is given by Figure 8.11.12.14) The moment of inertia I depends on whether the rod is suspended horizontally or vertically. Releasing the upper part of the pendulum from a large angle will result in chaotic oscillations of the rod. shown for both cases in Fig. while a second point on the rod. τ =– Gπ r 4 θ.15) Experimental suggestions • Periods can be timed manually. • For the vertical pendulum. ! HI = 2π Mgds ! H (L 2 /12 + d 2 ) . (8. but preferably with suitably positioned photogates. H s d Suspend the rod at one end by a string. as in Fig. the string moves through an angle θ such that H θ = sϕ. a steel wire passing through a hole near the com and suspended by very short strings to a support allows the rod to swing freely about its axis of support. gds (8. 2H (8. where the tension in the string is F = Mgd/s. 8. then. the torque exerted by the string is given by τ = Fθs = Fs2 ϕ/H .10 Chaotic pendulum. . In the small-angle approximation. Horizontal pendulum Figure 8. Let the rod swing through an angle ϕ about P. Numerical solution of the Euler–Lagrange equation for the system will show extreme sensitivity to the initial conditions.10. for large angle oscillations it is best to use the outermost hole in the rod. out of the plane. Hence the period is given by P T = 2π Figure 8. P. 8. rests on a support which is a horizontal distance d from the com.12 Horizontal pendulum. 63.. • For the double pendulum. 54... 2.. Teach. <http://www... many variants are possible.. J..com>. albertiswindow... 53 . • • Experiment with different suspension lengths.. 382 (1992). release the rod after rotating it through a small angle about a string... 3... Zahopoulos....... R.... • If you have a device that can follow and record two. Nelson and M.. Chaos in the corridor. n. Olson..Oscillation modes of a rod • In case (e) for bifilar suspension......... Phys. REFERENCES 1....or three-dimensional motion. 112 (1995).. Silevich. 30. 4.. the normal modes are obtained by starting with the string and rod displaced in the same direction.... Phys...... 112 (1986)... Phys..A.....d.... A..... A...... Am.. the displacement versus time of the pendulums may be sampled and compared with numerical solutions of the differential equations of motion...... Alberti’s window motion visualizer... Cromer.. and M...... such as Alberti’s windows [4] (which uses two coupled video cameras).. J. Am. Many oscillations of a rigid rod. Invent oscillations that have not been described here. Cromer. The pendulum: Rich physics from a simple system.. ......G. or in the opposite direction. Alberti’s Window....... C.....
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