10Lecture 2. 22 Jan 2015 solutions for position and velocity versus time, given prescriptionsfor F, e.g F(t), and F(r) or F(v) in 1-dSolution to these many special cases: (a-e)We are seeking the motion r(t) subsequent to specified initial conditions, and given someprescription for how F varies with position and velocity and time. The process involves thesolution of an ODE. The ODE is second order, so we'll need two (vector) constants of integration.Case a) rF = 0:We solve by doing two indefinite integrations, getting two constants of integration rvoandrro: ra = 0 ⇒ drv / dt = 0 ⇒rv = constant =rvorr (t) =rv∫(t)dt =rvot +rroThe constants of integration are recognized as corresponding to the velocity and position at timezero.Case b) rF = constant =rFo ( for which case (a) was a special case) ra =rFo/ m ⇒Now integrate each side (indefinite integral) with respect to time: rv(t) = (rFo/ m∫)dt =rFot / m +rC1The constant of integration rC1 may be recognized as the velocity rvoo it has at time t = 0. If youknew instead its velocity rvi at some other time, ti, you would identify the constant of integrationby solving rv(t1) =rFot1/ m +rC1. You'd get rC1=rvi−rFoti/ mNow integrate again (indefinitely) with respect to time rr (t) =∫(rFot / m +rC1)dt =rFot2/ 2m +rC1t +rC2The new constant of integration rC2may be identified with the position rro at time zero. Ifinstead you knew the position at some other time, you could use that to solve for rC2.This case should be familiar from Phys 211 where you so often saw the formulax = (a/2)t2 + vo t + xoDo keep in mind that this simple formula applies only to the one-dimensional case with F = constantand initial conditions specified at time zero.11case(b) continued. You could alternatively choose to do the integrals as definite integrals. Justmake sure that the limits on the integrals on the right and the left side of the equation correspond.For example, if you start with ra = drv / dt =rFo/ mSlap on an integral sign and a factor of dt on both sides to get v1v 2∫drv = (rFo/ m ) dtt1t 2∫(NB note the limits!)with the understanding the times t1 and t2 correspond respectively to the velocities v1 and v2.The result is rv2−rv1= (rFo/ m ){t2− t1}There are now no constants of integration to consider, and initial conditions (at time t1) enternaturally into the expressions.We now replace v2 and t2 with more generic v and t to get: rv = (rFo/ m ){t − t1} +rv1and integrate again with respect to time using corresponding limits on the two sides t1t∫rvdt =t1t∫((rFo/ m ){t − t1} +rv1)dtwhich after evaluating will give position as a function of time in terms of the velocity v1 at time t1and the position r1 at time t1. Again no constants of integration appear- because the integralswere definite.Case c) rF =rf (t) We need merely integrate twice with respect to time. (we'll use indefiniteintegrals and constants of integration here but we could do it with definite integrals if wepreferred) Cases a and b are special cases of (c). ra(t) =rf (t) / m ⇒drv = (rf (t) / m)dt∫drv =∫(rf (t) / m)dtrv(t) =rF(t) / m +rC1∫drr =∫(rF(t) / m +rC1)dtrr (t) =rF(t) / m +rC1t +rC2where F(t) is the anti-derivative of f (t) and F (t) is the anti-derivative of F(t).The vector constants of integration C1 and C2 may be solved for by invoking additionalknowledge, for example the velocity and position at some particular time. Nota Bene: Thevalues of C1 and C2 are related to the initial velocity and positions, but are not always equal tothem! ( See Miniquiz 2)12Case d) rF =rF(rr ) The force depends on where the particle is. Cases a and b can also beconsidered special cases of this. We will for now restrict to 1-dimension (x).In 1-d it takes the form: m dv/dt = F(x), where v = dx/dt.NB! You cannot profit by slapping an integral sign on both sides of m dv/dt = F(x) andintegrating with respect to t…. because the right hand side isn't (yet) known as a function of time.m (dv / dt ) dt∫= mv + C1= F(x)dt∫ . This expression is correct, and you can evaluate the leftside, but you cannot evaluate the right side ( unless you already know x(t) ), so it isn't good formuch.Trick: Multiply by v:m v dv/dt = v F(x) = F(x) dx/dtnow multiply by dt:mv dv = F(x) dxThis can be integrated: ∫mv dv = ∫F(x) dxIf we do a definite integral, from vo to v on the left and xo to x on the right, (with the understandingthat vo is the speed it has when it is at xo, and v is the speed it has when it is at x ) it becomes(m / 2)[v2− vo2] = F(x)dxxox∫which gives us speed v as a function of position x. This is a first integral of the second orderequation of motion: m d2x/dt2 = F(x). The next step would involve using the above to expressv = dx/dt as a function of x, and then solving the resulting differential equation for x(t). Butbefore doing that it is instructive to re-write the above in more physical terms:We define the kinetic energy as T ≡ (mv2/ 2)and notice that the Kinetic energy when it was at xo was To≡ (mvo2/ 2)Then the above first integral of the differential equation becomes the work-energy theoremT = To+ F(x)dxxox∫= To+ work doneby F over the interval from xoto xIf we define the potential energy U(x) as the anti-derivative of -F(x), then13dU/dx = - F(x) ( the function U(x), we note, is unique only up to some additive constant )then the above first integral of the differential equation becomesT = To− U(x) + U(xo) or T + U(x) = To+ U(xo)which is conservation of total energy E = T + U (x) = Eo= To+ U(xo)-----------The expressions derived for case (d) are essentially giving us v as a function of x. We may beable to do another integral so as to determine x as a function of time. Solve E = T + U (x) forthe velocity: ( recalling that we know the constant E = Eo in terms of the initial velocity andposition) Thenv = dx / dt = ±2m(E − U(x))If you slapped an integral sign on the above and integrated with respect to t,x = v dt∫= ±∫2m(E −U(x)) dtyou could do the integral on the left hand side, but not the integral on the right side.So the above is true, but of no utility.Instead, let us solve dx / dt = ± 2(E − U(x)) / m! for dt and integrate (again being careful that the lowerand upper limits correspond)dttot∫= t − to= ±dx2(E − U (x)) / mxox∫If this integral can be done analytically, we'll get time as a function of x. Not exactly what wewanted, but