# Physics in the Laundromat

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2 Figure 1: Model of the spin cycle of a washing machine. The drum and symmetrical part of the load have mass M. The center of the drum is at (r, θ) and is connected to the origin by a zero-length spring of constant k. An unbalanced load of mass m lies at distance a from the center of the drum and at angle φ with respect to a fixed direction in the laundromat. The washer motor turns the drum with angular velocity φ = Ω. force proportional to its velocity. The frictional torque that opposes the forced rotation Ω does not, however, affect the motion. The equations of motion of this system of two degrees of freedom, r and θ, are readily deduced from the Newtonian approach: where the position of mass m is M r + m r m = kr γṙ, (1) r m = (r + a cos(φ θ))ˆr + a sin(φ θ)ˆθ, (2) and γ is the coefficient of the velocity-dependent (translational) friction on the shaft of the drum. The equation of motion associated with coordinate r is and that with coordinate θ is r = r θ 2 + bω 2 cos(φ θ) ω 2 0r Γṙ, (3) 0 = r θ + 2ṙ θ bω 2 sin(φ θ) + Γr θ, (4) where we have introduced the notation k ω 0 = m + M. (5) 2

3 for the natural frequency of vibration of the washing machine, b = for the distance of the center of mass from the shaft and Γ = m a, (6) m + M γ m + M. (7) These equations can be interpreted in a frame rotating with angular velocity θ. Equation (3) tells us that the total mass times the radial acceleration of mass M equals the spring force plus the radial component of the centrifugal force and friction. Equation (4) indicates that the azimuthal coordinate forces plus friction sum to zero; the term 2ṙ θ is the Coriolis acceleration. The equations of motion with the neglect of friction are also readily deduced from the Lagrangian: L = 1 2 (m+m)(ṙ2 +r 2 θ2 ) maṙω sin(φ θ)+mar θω cos(φ θ)+ 1 2 (I +ma2 )Ω kr2, (8) where I is the moment of inertia of the drum plus symmetric part of the load. The rotational kinetic energy is constant by assumption, so the moment of inertia does not appear further in the analysis. 3 Steady Motion We first discuss steady motion in which ṙ = 0, r = 0 and θ = 0. The shaft of the drum moves in a circle of radius r 0 and the mass m is at constant azimuth φ 0 = φ θ relative to the azimuth of the shaft. Then eq. (3) tells us r 0 = bω2 cos φ 0 ω 2 0 Ω 2, (9) while eq. (4) indicates Together, r 0 = bω sin φ 0. (10) Γ cos φ 0 = ω 2 0 Ω 2 (ω 2 0 Ω 2 ) 2 + Γ 2 Ω 2, sin φ 0 = ΓΩ (ω 2 0 Ω 2 ) 2 + Γ 2 Ω 2, (11) and r 0 = bω 2 (ω 2 0 Ω 2 ) 2 + Γ 2 Ω 2. (12) For a balanced load (m = 0) distance b is zero, so the equilibrium displacement is zero also. 3

4 For low spin (Ω ω 0 ) an unbalanced load finds itself at relative azimuth φ 0 0, while near resonance (Ω ω 0 ) the azimuth is π/2, and for high spin (Ω ω) the azimuth approaches π. In the latter case the system is a kind of inverted pendulum. The center of mass of the system is at distance r cm = bω 2 0 (ω 2 0 Ω 2 ) 2 + Γ 2 Ω 2. (13) Thus the center of mass approaches the origin as the spin Ω becomes large, even though the shaft is at radius r 0 b. The system can be called self-centering as the spin Ω increases, once it successfully passes through the resonance region. 4 Stability Is the desirable self-centering motion found above stable against small perturbations? If angle θ were locked at φ φ 0, i.e., if only radial oscillations were permitted, and Ω > ω 0 answer would be no! To see this we refer to eq. (3), which for the locked hypothesis reads r = (Ω 2 ω 2 0)r + bω 2 cos φ 0 Γṙ. (14) For oscillatory radial motion the coefficient of the term in r must be negative. Hence the locked motion would be stable only for low spin, Ω < ω 0. However, we will find that the motion is stable when both radial and azimuthal oscillations are considered. The linked system of masses m and M forms a kind of double pendulum. The motion in which φ θ + π that arises when the drive frequency Ω exceeds the resonant frequency ω 0 is an example of a stable inverted pendulum. To demonstrate this we perform a perturbation analysis, seeking solutions of the form r = r 0 (1 + ɛ), θ = φ φ 0 + δ, (15) where the perturbations are desired to be small and oscillatory with frequency ω: ɛ = ɛ 0 e iωt, and δ = δ 0 e iωt with ɛ 0, δ 0 1. (16) The constants ɛ 0, δ 0 and ω are complex in general, and, of course, the physical motion is described by the real parts of (15). Both the real and imaginary parts of ω should be positive; the real part is the frequency of oscillation and the imaginary part is the damping decay constant. In the first approximation we now have cos(φ θ) = cos φ 0 + δ sin φ 0, and sin(φ θ) = sin φ 0 δ cos φ 0. (17) Then, using (15-17) in (3) and keeping terms only of first order of smallness, we find ω 2 ɛ 0 = Ω 2 + 2iΩδ 0 + bω2 sin φ 0 r 0 δ 0 ω 2 0ɛ 0 iωγɛ 0. (18) 4

5 With eq. (10) this tells us Similarly, eq. (4) leads to Γω + 2iωΩ ɛ 0 = ω 2 ω0 2 + Ω 2 iωγ δ 0. (19) which together with eq. (9) tells us 0 = ω 2 δ 0 + 2iΩωɛ 0 + bω2 cos φ 0 r 0 δ 0 + ΓΩ 0 ɛ 0 + iωγδ 0, (20) δ 0 = Equations (19) and (21) are consistent only if which leads to the quadratic equation ΓΩ + 2iΩω ω 2 ω Ω 2 iωγ ɛ 0. (21) ΓΩ + 2iΩω ω 2 ω Ω 2 iωγ = ±i, (22) ω 2 2ω(±Ω iγ/2) ω Ω 2 ± iγω = 0. (23) The roots of this with positive real parts are ω0 2 (Γ/2) 2 ± Ω + iγ/2, Ω < ω0 2 (Γ/2) 2, ω = Ω ± ω0 2 (Γ/2) 2 + iγ/2, Ω > ω0 2 (Γ/2) 2. (24) In the above we have assumed that the damping is weak enough that ω 0 > Γ/2. Then perturbations die out with characteristic time 2/Γ which is greater than the natural period of oscillation, 1/ω 0. Thus stable motion exists for all values of the spin Ω. Refering to eq. (20), we note that the key coupling between the radial and azimuthal perturbations (ɛ and δ) is provided by the Coriolis force. As the spin frequency Ω approaches the resonant frequency ω 0 the lower frequency of the perturbed motion goes to zero. If the amplitude of the perturbation is large it will be noticeable throughout the laundromat. For high spin, eqs. (21) and (24) yield the relation δ 0 = 1 iγ iɛ 0 ( Γ 2 2Ω 8Ω Ω± ω 2 0 (Γ/2)2 ) iɛ 0, (25) which indicates that the radial and azimuthal perturbations are 90 out of phase. The angular velocity of the motion of the center of the drum is ( ) θ = Ω + ɛ 0 sin Ω ± ω0 2 (Γ/2) 2, (26) 5

6 which is Ω on average. This implies that the perturbed motion of the shaft of the drum is still a circle of radius r 0 to first approximation. However, since the frequency ω of the perturbation differs from the average rotation frequency Ω, the orbit of the center of the drum is not closed but precesses at angular frequency ω 0, as sketched in Fig. 2. In the limit that Ω ω 0 the motion of the center of the drum is essentially a circle of radius r 0 displaced by distance ɛ 0 r 0 from the origin, as shown in Fig. 3. In practice this displacement can be quite noticeable, as the reader can confirm on his or her next trip to the laundromat. Figure 2: The steady motion of the axis of the drum is at angular velocity Ω in a circle of radius r 0 about the origin. The perturbed orbit is nearly circular, but precesses with angular velocity ω 0 and lies in the annulus r 0 (1 ɛ 0 ) < r < r 0 (1 + ɛ 0 ). Figure 3: If the drive frequency Ω is large compared to the resonant frequency ω 0 the perturbed motion is a circle of radius r 0 whose center is displaced by ɛr 0. 6

7 The author wishes to thank A.B. Pippard for a gracious reading of the manuscript, and for the suggestion to add damping to the analysis. I would also like to salute the (unknown to me) inventor of the washing machine, who likely solved this problem without the benefit of a formal analysis. 5 References [1] L. Landau and A. Kitaigorodsky, Physics for Everyone (Mir Publisher, Moscow, 1978), pp [2] A.B. Pippard, The Physics of Vibration, Vol. 1 (Cambridge U. Press, London, 1978), pp

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