105Phys 325 Lecture 12 Tuesday March 3, 2015inverted pendulum, physical pendulum, damped oscillations, initial conditionsThe upside down pendulum. (Done also in discussion section)Its analysis is very similar to that of the regular pendulum:T = (1/2) m ( L dθ/dt )2PE = - mgL(1 – cosθ ) (such that PE is zero at the top and diminishes as θ increases)So E = T +PE = (1/2) m ( L dθ/dt )2 - mgL(1 – cosθ )By invoking dE/dt = 0, we find that the ODE ism L2 d2θ/dt2 - mgLsinθ = 0It is the same ODE as before, but with g replaced by –g.------------For small θ, we writem L2 d2θ/dt2 – mgL θ = 0which resembles what we had before, except that the effective stiffness is now negative. The sign of the effective stiffness verymuch affects the character of the solutions, which now are exponentials not sinusoids. With σ defined as [ g/L ]1/2 , thesolutions areθ (t) = Α cosh σ t + B sinh σ t (instead of A cosωnt + B sin ωnt)or C cosh( σt – φ )or, if you preferθ (t) = D1 exp σ t + D2 exp-σ tThe relations between the constants may be determined by various hyperbolic trigidentities. The constants themselves may be determined by referring to initialconditions.The more important conclusion , though, is that the motion is exponentiallygrowing at late times. The equilibrium position θ = 0 at the top is unstable.Query: This result says that θ goes faster and faster, exponentially in time,eventually whirling around at the hundreds of miles per hour and obviously notconserving energy. What's wrong?=========Now let us try to stabilize the upside down pendulum bysupporting it with a torsional spring that provides a restoring torque proportional to the anglethrough which it is bent. Γ = −κ θ. Here is a diagram with the relevant torques and forcesindicated. Net torque is MgLsinθ clockwise plus κθ counterclockwise. Torque = I α reads: ML2&&θ= −κθ+ MgL sinθWhen linearized around the equilibrium at θ=0 it becomes106 ML2&&θ+ (κ− MgL)θ= 0The natural frequency is the square root of the ratio of the coefficients ωn= keff/ meff= (κ− MgL) / mL2which is real if κ > MgL and motion is oscillatory . It is imaginary ( i.e the motion exponentialin time, ie. unstable ) if κ < MgLThe qualitative character of the motiondepends on whether or not the torsionalstiffness κ is great enough.Lesson: Whenever you derive agoverning ODE, you should check thesigns of its meff and keff. If either isnegative, or could be with a certainchoice of parameters, then the predictedmotion is unstable. If you did notexpect instability, then check your derivation. You could have made some sign errors.Aside: What actually happens if κ < MgL such that the equilibrium at the top isunstable ?It turns out that there is, if κ < MgL , another equilibrium ( at the solution toκθ= MgL sinθ) This equilibrium is stable. You'd get stable oscillations around thisother equilibrium, at a frequency that turns out to be not equal to √g/L – see next week'sHW)The "physical pendulum"We again write an expression for the total energyin terms of what θ is doing and what θ is.T = (1/2) IP (dθ/dt)2PE = mgh = mg l ( 1-cosθ)where h =l ( 1-cosθ) is distance above the lowestpoint the CM can be at. Alternatively choose thepivot point to be the reference height and getPE = -mg l cosθ Then E = (1/2) IP (dθ/dt)2 + mg l ( 1-cosθ)107and dE/dt is (using chain rule to find the time derivative of cos θ and (dθ/dt)2 )dE/dt = (dθ/dt) { IP (d2θ/dt2) + mg l sinθ }Our ODE is obtained from dE/dt =0: IP (d2θ/dt2) + mg l sinθ = 0(which could also have been gotten from torque = I α ). The eqn looks just like it did for themassless rod and point mass, except for different coefficients.As before, we linearize this by assuming θ close to θeq which is 0, i.e, θ <<1. This gives IP (d2θ/dt2) + mg l θ = 0aside: sanity check: are both coefficients positive? Answer: Yes.From which we recognize that that Meff = IP and Keff=mgl. We immediately concludeθ (t) = Α cos ωn t + B sin ωn twhere ωn = SQRT( Ratio Coefficients) = [ mgl / IP ]1/2Or we could write θ (t) = C cos (ωn t –φ)The only difference is that now we have a new expression for ωn :ωn = [ mg l / IP ]1/2For the uniform rod of length L pivoted at its end, l = L/2 and IP = m L2 /3, so ωn = [3g /2L]1/2Initial conditions serve to determine the so-far unknown coefficients A and BIf, as in all the above examples, we solve the ODE to get x, or θ or whatever, in the formx(t) = Α cos ωn t + B sin ωn tWe would note that at time zero x = xo = A, and dx/dt = vo = B ωn.Thus A and B are determined in terms of the initial x and dx/dt----------If we had our general solution in terms of C and φ, we'd have writtenx(t) = C cos (ωn t -φ)from which we would note xo = C cos(φ) and vo = -Cωn sinφ and could then solve for C and φ.----108It sometimes happens that subsidiary conditions are not initial. For example, what if we weretold that x = ξ at time τ and that total energy were E? We could still solve for A and B ( or Cand φ) but the algebra would be harder ( and there might be more than one solution, or none.)What can we do to extend this Simple Harmonic Oscillator ( m&&x + kx = 0 ) model?Three ways: Introduce energy dissipating elementsAnalyze effects of nonlinearity ( non-small deviations from equilibrium )External forces – how does system respond?We'll spend most of our effort on the analytically more tractable 1st and 3d ways.Damped oscillationsThe above analysis seems to have predicted that oscillations continue forever. Can we modifythe analysis to predict differently? In practice E diminishes, so we ought write….dE/dt = power dissipatedand invoke some model for how it dissipates. Acommon model for dissipation is linear viscousdamping in which there is a force proportional tovelocity. This is represented in the diagram by adevice like those used in screen doors and shockabsorbers in which a fluid is forced through narrowpores. Force = - c v. Thus Power which is forcetimes velocity is -cv2. c is a coefficient ofresistance.Thus ddtE =ddt(12m&X2+12k(X − L)2) = −c&X2Or, &X(m&&X + k(X − L)) = −c&X2Or, m&&X + c&X + k(X − L) = 0Again we shift the coordinate origin so thatx(t) = X-L represents deviation from equilibrium, and conclude m&&x + c&x + kx = 0This is a linear ODE for x(t). As we ought, we