38Lecture 5 Phys 325 Tuesday Feb 3, 2015Most of the above systems were for particles in one-dimension. That is, they had but onedynamically changing coordinate – usually called x(t) = a Cartesian coordinate, but sometimesit was the distance of a bead as is slides out along a spinning rod r(t), or the angle θ(t) of a beadas it slides along a hoop.If often happens that a system that initially appears rather complicated really has only onedegree of freedom – and our job is to find the equation that governs its dynamics. The firstpart of that job is to choose a dynamical coordinate.Here are two examples for systems (composed of more than one particle but with only onedegree of freedom) in which we skip using F=ma in favor of a direct statement of energyconservation.example 1) An Atwood machine – two masses on arope over a pulley.We choose a coordinate x representing the distance ofM below the pulley center. We notice that thevelocity of M is then dx/dt ( downwards) and that mhas the same speed, but upwards. The total kineticenergy is then (ignoring that of the pulley -let us take itto be massless )T = (1/2)(M+m)(dx/dt)2The total potential energy is ( where L is a constantrelated to the length of the rope )U = - Mgx - mg(L-x) = constant + (m-M)gxThen conservation of energy tells us ( the rope is doing no work )E = (1/2)(M+m)(dx/dt)2 + (m-M)gx is aconstant.One nice thing about such a formulation, is that itdoesn't introduce -or have to solve for- the (relatively)uninteresting constraint force of rope tension. Nordoes it introduce the concepts of acceleration or force.example 2) A block rocking on a hill with enoughfriction to prevent sliding39Total energy is conserved ( friction does no work if there is no sliding)The system's potential energy is mgh ( m is the mass of the block)U = mg [ (r+b)cosθ + rθ sin θ ]Here are some plots of U for different values of bNewton’s Laws and particle motion in Three Dimensions - Conservative and CentralForcesWe now turn to a general discussion of the motion of a particle in two or more dimensions,systems for which we will want to use at least two apriori unknown dynamical coordinates (forexample x(t) and y(t), or r(t) and θ(t)). We already know the basic formula: rF(rr ,rv,t) = m d2rr / dt2= mraIn three dimensions, and in Cartesian coordinates, this vector equation represents threeWithout ever finding the differentialequation that governs θ(t), ordetermining expressions for kineticenergy, we can use these plots of U(θ) todetermine its qualitative behavior, inparticular we show that b > r means thatthe equilibrium at the top is unstable andb < r means that it is stable.40simultaneous equations: Fx(rr,rv,t) = Fx(x , y, z,vx,vy,vz,t) = m d2x / dt2Fy(rr,rv,t) = Fy(x , y, z,vx,vy,vz,t) = m d2y / dt2Fz(rr,rv,t) = Fz(x , y, z,vx,vy,vz,t) = m d2z / dt2In cylindrical coordinates, it reads Fr(rr,rv,t) = Fr( r,φ, z, vr,vφ,vz,t) = mar= m {&&r − r&φ2)Fφ(rr,rv,t) = Fφ( r,φ, z, vr,vφ,vz,t) = maφ= m {r&&φ+ 2&r&φ}Fz(rr,rv,t) = Fz( r,φ, z, vr,vφ,vz,t) = maz= m&&zIf the forces separate in a convenient fashion, such that for example Fx depends only on x andvx and t, then the equations are decoupled and solving them amounts to solving three 1dproblems. Generally speaking, that doesn’t happen (and in cylindrical coordinates theexpressions for acceleration couple the equations regardless of how the forces may depend onvarious variables) and solutions can become quite complicated. There are few general resultsand we usually require numerical solutions.There are many cases, however, for which we can get good insight into the solution, or evencomplete solutions, by being sufficiently clever. The most prominent of these are the cases ofconservative force fields and central force fields. Each of these provides a great simplificationby giving us constants of the motion, ie., first integrals of the dynamics.Conservative (and nonconservative) ForcesWe have already used energy conservation in one dimension where we invented a potentialenergy function for a force that was purely a function of x :dU(x)/dx = - F(x)U = - ∫ F(x) dxWe showed if that is the case then the total energy E = T + U, is a constant of the motion. (that is, it is a constant during the motion; it is not an absolute constant, as it depends on initialconditions) This concept of a constant of the motion will recur throughout the course. Thisobservation allowed us to do a first integral of the ODE, and gave us speed as a function ofposition. It did not tell us x(t) which would have required a second integral of the ODE.Conservation of total energy is a powerful problem solving concept in 3-d also, though thereare numerous differences from 1-d. In 2 d and 3-d the analogy to F = -dU/dx is rF(rr ) = −r∇U(rr )41In Cartesian coordinates it becomesFx(x, y, z) = −∂∂xU(x, y, z); Fy(x, y, z) = −∂∂yU(x, y, z); Fz(x, y, z) = −∂∂zU(x, y, z)It looks different in cylindrical polar coordinates:Fr(r,φ, z) = −∂∂rU(r,φ, z); Fφ(r,φ, z) = −∂r∂φU(r,φ, z); Fz(x, y, z) = −∂∂zU(r,φ, z)In 1-d we can integrate any F(x) to get –U(x). Not so in 3-d. Not every F(r) is the gradientof some potential function – U(r) . Taking a curl of rF(rr ) = −r∇U(rr ) we notice ∇ ×rF(rr ) = −∇ × ∇U(rr ) = 0Thus if F can be written as the gradient of some -U, then it must necessarily be that F's curlvanishes. It turns out that this is a sufficient condition also.Do remember that not every F(r) has a vanishing curl. Force fields F(r) that havenonvanishing curls r∇ ×rF(rr ) ≠ 0cannot be written as rF(rr ) = −r∇U(rr ). Such are termed non-conservative force fields.Force fields F(r) that have vanishing curls r∇ ×rF(rr ) = 0can be written as rF(rr ) = −r∇U(rr ).(though finding U might require some work.) Such are termed conservative force fields. Theyallow us to do a first integral of the motion and write the particle's total energy E = T + U as aconstant of the motion, equal to its initial value.To prove this we consider the time derivative of the kinetic energy of a particle. It is ddtT =ddt12mv2=ddt12mrv ⋅rv =rv ⋅ mddtrv =rv ⋅rFThis is the familiar work-energy theorem. It says (what we already knew) that the kineticenergy changes like the rate at which the force is doing work. This is true whether the forcesare conservative or not.The time derivative of any scalar