29Phys 325 Lecture 4. Thursday Jan 29, 2015Bead on a whirling rod continued……Some kinematics: ( See p 33 of T+M, example 1.8. Or see section 1.7 of Taylor )Position vector: rr = r(t )ˆrVelocity vector: rv = drr / dt =&r(t )ˆr + r(t )dˆr / dtAs it sometimes happens, our unit vectors are changing in time as the rod rotates. Hence thesecond term is not zero. A little bit of geometry establishes that our unit vectors are changingsolely due to changes in φ: dˆr / dt =&φ(t )ˆφ; dˆφ/ dt = −&φ(t )ˆrAlternatively, these can be established by first noting thatˆr =ˆi cosφ+ˆj sinφ;ˆφ= −ˆi sinφ+ˆj cosφ ,and then taking time derivatives. e.g. using chain rule: ˆ&r = −ˆi sinφ&φ+ˆj cosφ&φ=ˆφ&φSee if you can derive dˆφ/ dt = −&φ(t )ˆr this way.Thus rv =&r(t )ˆr + r(t )&φˆφa result which may seem a little obvious. The velocity has two orthogonal components, dr/dt inthe outward direction, and r(t) dφ/dt in the direction of increasing φ . The latter term may befamiliar from Phys 211 in the form Rω.Acceleration vector:Taking d/dt of the velocity vector, and recalling the above expressions for how the unit vectors arechanging, we find after a bit of algebra: ra =ddt(&r(t )ˆr + r(t )&φˆφ) =&&r(t )ˆr +&r(t )ˆ&r +&r(t )&φˆφ+ r(t )&&φˆφ+ r(t )&φˆ&φ= (&&r(t ) − r(t )&φ2)ˆr + (r(t )&&φ+ 2&r&φ)ˆφThe r-component of this may look familiar; note the contribution of the centripetal term -r ω2.The φ component is perhaps less obvious. It is worthwhile to examine all four of these terms toconfirm that they are dimensionally correct.We now return to the bead on the rod. F = ma becomes30 N (t)ˆφ= m(&&r(t ) − r(t )2&φ2)ˆr + m(r(t )&&φ+ 2&r&φ)ˆφThere are two components of this equation. I will discuss the ˆφ component below. The ˆr-component tells us 0 =&&r(t) − r(t)&φ2; or&&r(t)= r(t )ω2which is a homogeneous linear second order constant coefficient ODE for r(t). Let us solve it byfollowing the (hopefully familiar?) suggestion to try exponential solutions whenever you have alinear constant coefficient homogeneous ODE : Try r(t) = exp(λt)Substituting this we find that λ = ±ω. There are therefore two values of λ that permit exp(λt) to bea solution. Either one gives us a solution of the differential eqn. Because the eqn is linear andhomogeneous, then any linear combination of solutions is also a solution. We write r(t) = A exp(ωt) + B exp(-ωt)This satisfies the ODE for any choice of constants A and B. (check that!) The constants ofintegration A and B can be determined by invoking initial conditions. For example, if at t= 0, r =ro, and dr/dt = vr = 0, then ro= A + B 0 = ω (A-B).Thus A = B = ro/2 andr(t) = (1/2) ro [ exp(ωt) + exp(-ωt) ] = ro cosh(ωt)The particle moves out on the rod in an exponential spiral.The plot is the position of the bead (the dots are for equally spaced intervals of time.) Once itcomes off the rod (not shown) it would of course move in a straight line, since there would then beno forces on it.31The ˆφcomponent of F= ma is perhaps less interesting; it tells us only what N must be to keep theparticle on the rod N (t ) = m{r(t )&&φ+ 2&r&φ} = 2m&rωyou will need this for the HW in which there is a friction force proportional to N---Query : This particle goes faster and faster as it moves out. Why isn't it conserving energy?Answer: N is doing work on the particle! N dot v = (2m &rω ) (r ω) = m ω2 dr2/dt ≠ 0. Somethingis doing work on the system and a little reflection indicates that there must be a motor that ismaintaining the constant speed ω.-------In HW2 you are to consider a mass sliding on africtionless horizontal table and being pulled inwardsby a cord ( of as yet unknown tension T) at a constantradial speed. This is a 2-d polar coordinate problemfor which you will need both F and a in polarcoordinates. r(t) is specified, so the interestingquestion is the differential equation for φ.------Here is a problem using 3-d spherical coordinates, of abead on a spinning wire hoop.A hoop of radius R spins on its vertical axis at a fixed rate Ω as illustrated32with a bead of mass m free to slide along the hoop. We wish to determine the ODE that governsθ(t) This problem is a natural for spherical coordinates. The bead has spherical coordinates:r(t) = R constant. It has azimuthal coordinate φ = Ω t specified. Its other spherical coordinate isθ(t) which is to be determined. Spherical coordinates often take θ to be defined as angle from thenorth pole, but it is easier to do it from the south pole here, as the south pole is an obvious stableequilibrium point when the hoop is not spinning; the bead sits at the bottom.You may find it useful to think of these coordinates as they are used in geography: r is radius ofthe earth plus altitude, φ is longitude; θ is latitude above the south pole.The force on the bead is some as yet unknown normal force Nφ(t) that acts in the φ direction, plussome other unknown normal force Nr(t) that acts radially, plus gravity -mg{ˆr cosθ−ˆθsinθ}Therefore the total force is rF = mg{ˆr cosθ−ˆθsinθ} + Nrˆr + Nφˆφ[ The quantities N may end up negative so don't worry about the signs in front of them ].We also need to do the kinematics: How do the three components of acceleration ( and velocity)depend on how fast the coordinates are changing? Oddly, this is not in the text, but I find itonline. (We'll need it a couple more times in the near future.) Acceleration is obtained bymethods similar to those used above for polar coordinates (but much more complicated) :The r and φ components of this will tell us about the normal forces N that are usually of lessinterest. The θ component gives us a differential equation for the motion along the hoop: Fθ= −mgsinθ= maθ= m{r&&θ+ 2&θ&r − r cosθsinθ&φ2}We recall r = R; dr/dt = 0, dφ/dt = Ω, d2φ/dt2 = 0, and conclude: &&θ= cosθsinθΩ2− (g / R)sinθa differential equation for θ(t). ( There are other arguably more elegant ways to derive thisgoverning differential equation for θ; we will see some of them later in the course. )---How to solve it?We notice that there are potential points of equilibrium at which &&θ is zero. Two such areat θ = 0 (the bottom) and θ = π (the top) .33Another would be θ such that cosθ= (g / RΩ2) which has a root only if the hoop isspinning