Introduction and BackgroundFrictionless BeadFrictionless ExamplesFriction Added to the EquationFriction ExamplesSupercritical Pitchfork BifurcationIntroduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 1 of 16Go BackFull ScreenCloseQuitA Bead on a Rotating HoopRyan Seng & Michael MeeksMay 16, 2008AbstractThis paper models a bead moving along a rotating hoop using differential equations.We show how to convert from a second order differential equation to a system of first orderdifferential equations, then graphically analyze the equations without friction and compareit to a model with friction. We’ll also show how to scale the variables to obtain a simplifieddimensionless equation.1. Introduction and BackgroundTypically used in first year physics, this model shows an example of a bifurcation in a me-chanical system. The model closely resembles a swinging pendulum. The main difference is anadditional motion, the rotation of the hoop, where the bead’s path along the hoop accountsfor the pendulum motion. Refer to figure 1 to help visualize the hoop in 3-space spinning at aconstant angular velocity about its vertical axis. The behavior of the bead will vary as it travelsalong the hoop, the dependent factor being the hoop’s angular velocity. As the angular velocityincreases, the bead’s equilibrium points move up the sides of the hoop. The equation we’ll useis autonomous. First, we’ll give an example without friction to give a good foundation to theproblem, then an example with friction.Introduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 2 of 16Go BackFull ScreenCloseQuitFigure 1: Hoop Diagram2. Frictionless BeadThis example represents a conservative system where no energy is dissipated. Here, the beadis forced along a wire hoop by gravitational and centrifugal forces, without friction. The hooprotates at a constant angular velocity about the vertical axis.By convention, g equals the gravitational constant near the surface of earth, 9.8 m/s, m,the mass of the bead, r, the distance of the bead from the it’s rotational axis, ω, the (constant)angular velocity about the vertical axis, and we’ll set φ equal to the angle between the beadand the downward vertical direction. In this problem we restrict φ to −π < φ < π. First we’llconsider the downward gravitational force mg.To arrive at the equation we need, we will consider the motion of the bead, using Newton’ssecond law:XF = ma (1)Here, we’ll define the forces, and obtain the acceleration of the bead. To find the acceleration,recall from previous classes that φ = s/r, where φ is the angle created between vertical and themass, s the arclength, and r, the distance of the mass from it’s rotational axis (radius). Wesolve for s and get s = rφ. Differentiating this we get ds/dt = rdφ/dt, the velocity of the mass.Differentiating this to get acceleration, we find that a is equal to rd2φ/dt2. This is also knownIntroduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 3 of 16Go BackFull ScreenCloseQuitas the tangential acceleration. Substituting this into the equation we arrive at:XF = mrd2φdt2(2)Next we define the forces. Referring to Figure 2, we can determine these forces. We knowthat one is gravity and the other is the centrifugal force, a ’fictitious’ force. These componentsdefine the motion of the bead To understand why this force is called a fictitious force, think ofthe last time you went around a corner in your car. You had to use your hands to hold on tothe steering wheel as you entered the corner because of Newton’s second law, the tendency ofyou to keep moving in a straight line. The car turned, so you had to grab the steering wheel toturn with the car. In essence, the ’fictitious’ force is the bead’s inability to ”grab the wheel”.Back to defining the forces. The gravitational force is mg and the sideways centrifugal force ismρω2. Breaking down the vector mg into two perpendicular vectors components, we get thatthe vectors mg sin φ + mg cos φ are equal to the vector mg. The second force, the mρω2vector,is broken down into two perpendicular vector components where the vector mρω2is equal tothe vectors mρω2sin φ + mρω2cos φ. These forces are perpendicular to the tangent line at anypoint. Substituting these forces into the equation we arrive at:−mg sin φ + mρω2cos φ = mrd2φdt2(3)Referring to figure 3, we let ρ equal r sin φ.−mg sin φ + mrω2sin φ cos φ = mrd2φdt2(4)Next we’ll make the problem dimensionless by introducing a new variable called τ . We’ll setτ equal to t/T, where t is in units of time, and T is a variable to be chosen later. Taking thederivative of τ with respect to time we get dτ/dt = 1/T .Taking the derivative of φ with respect to time we getdφdt=dφdτdτdt⇒dφdτ1T(5)Introduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 4 of 16Go BackFull ScreenCloseQuitφφφmg sin φρmρω2cos φmρω2mgrbφ0mρω2sin φmg cos φFigure 2: ForcesIntroduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 5 of 16Go BackFull ScreenCloseQuitφρrFigure 3: ρ = r sin φIntroduction and . . .Frictionless BeadFrictionless ExamplesFriction Added to the . . .Friction ExamplesSupercritical Pitchfork . . .Home PageTitle PageJJ IIJ IPage 6 of 16Go BackFull ScreenCloseQuitThen, take the second derivative of theta with respect to timed2φdt2=ddtdφdt⇒ddt1Tdφdτ⇒ddτ1Tdφdτdτdt⇒1Td2φdτ21T⇒1T2d2φdτ2(6)Substituting these into our equation, we arrive at:−mg sin φ + mrω2sin φ cos φ = mr1T2d2φdτ2(7)Next, divide equation (7) through by mg to get− sin φ +rω2gsin φ cos φ =rgT2d2φdτ2(8)To finish making the equation dimensionless, set T equal to b/mg, to arrive at− sin φ +rω2gsin φ cos φ =m2grb2d2φdτ2(9)Simplifying these terms, we set γ equal to rω2/g, and ε equal to m2gr/b2. The resultexpresses the original 5 parameters as two more easily analyzed