Goldstein Chapter 1 Exercises

Transcription

1 Golstein Chapter 1 Exercises Michael Goo July 17, Exercises 11. Consier a uniform thin isk that rolls without slipping on a horizontal plane. A horizontal force is applie to the center of the isk an in a irection parallel to the plane of the isk. Answer: Derive Lagrange s equations an fin the generalize force. Discuss the motion if the force is not applie parallel to the plane of the isk. To fin Lagrangian s equations, we nee to first fin the Lagrangian. L = T V Therefore T = 1 2 mv2 = 1 2 m(rω)2 V = 0 Plug into the Lagrange equations: L = 1 2 m(rω)2 t ẋ x = Q 1 2 mr2 ω mr2 ω 2 = Q t (rω) x t m(rω) = Q m(r ω) = Q 1

2 If the motion is not applie parallel to the plane of the isk, then there might be some slipping, or another generalize coorinate woul have to be introuce, such as θ to escribe the y-axis motion. The velocity of the isk woul not just be in the x-irection as it is here. 12. The escape velocity of a particle on Earth is the minimum velocity require at Earth s surface in orer that that particle can escape from Earth s gravitational fiel. Neglecting the resistance of the atmosphere, the system is conservative. From the conservation theorme for potential plus kinetic energy show that the escape veolcity for Earth, ingnoring the presence of the Moon, is 11.2 km/s. Answer: GMm = 1 r 2 mv2 GM = 1 r 2 v2 Lets plug in the numbers to this simple problem: ( ) ( ) ( ) = 1 2 v2 This gives v = m/s which is 11.2 km/s. 13. Rockets are propelle by the momentum reaction of the exhaust gases expelle from the tail. Since these gases arise from the raction of the fuels carrie in the rocket, the mass of the rocket is not constant, but ecreases as the fuel is expene. Show that the equation of motion for a rocket projecte vertically upwar in a uniform gravitational fiel, neglecting atmospheric friction, is: m v m = v t t mg where m is the mass of the rocket an v is the velocity of the escaping gases relative to the rocket. Integrate this equation to obtain v as a function of m, assuming a constant time rate of loss of mass. Show, for a rocket starting initally from rest, with v equal to 2.1 km/s an a mass loss per secon equal to 1/60th of the intial mass, that in orer to reach the escape velocity the ratio of the wight of the fuel to the weight of the empty rocket must be almost 300! Answer: This problem can be tricky if you re not very careful with the notation. But here is the best way to o it. Defining m e equal to the empty rocket mass, m f is the total fuel mass, m 0 is the intitial rocket mass, that is, m e + m f, an m t = m0 60 as the loss rate of mass, an finally the goal is to fin the ratio of 2

3 m f /m e to be about 300. The total force is just ma, as in Newton s secon law. The total force on the rocket will be equal to the force ue to the gas escaping minus the weight of the rocket: ma = t [ mv ] mg m v m = v t t mg The rate of lost mass is negative. The velocity is in the negative irection, so, with the two negative signs the term becomes positive. Use this: Solve: v m m t = v t m v m m = v m t t mg v m m t = v m m t g v m = v m + 60g m 0 Notice that the two negative signs cancelle out to give us a positive far right term. Integrating, v = v 60g m + m m m 0 v = v me m 0 m me m + 60g m m 0 m 0 v = v ln m e m g m 0 (m e m 0 ) v = v ln m e m e + m f + 60g m e m e m f m e + m f v = v ln m e + m f m e m f 60g m e + m f Now watch this, I m going to use my magic wan of approximation. This is when I say that because I know that the ratio is so big, I can ignore the empty 3

4 rocket mass as compare to the fuel mass. m e << m f. Let me remin you, we are looking for this ratio as well. The ratio of the fuel mass to empty rocket, m f /m e. v = v ln m e + m f m e m f 60g m e + m f v = v ln m f m e 60g m f m f v + 60g v = ln m f m e exp[ v + 60g v ] = m f m e Plug in 11,200 m/s for v, 9.8 for g, an 2100 m/s for v. m f m e = 274 An, by the way, if Golstein han t just converte 6800 ft/s from his secon eition to 2.1 km/s in his thir eition without checking his answer, he woul have notice that 2.07 km/s which is a more accurate approximation, yiels a ratio of 296. This is more like the number 300 he was looking for. 14. Two points of mass m are joine by a rigi weightless ro of length l, the center of which is constraine to move on a circle of raius a. Express the kinetic energy in generalize coorinates. Answer: T 1 + T 2 = T Where T 1 equals the kinetic energy of the center of mass, an T 2 is the kinetic energy about the center of mass. Keep these two parts seperate! Solve for T 1 first, its the easiest: T 1 = 1 2 Mv2 cm = 1 2 (2m)(a ψ) 2 = ma 2 ψ 2 Solve for T 2, realizing that the rigi ro is not restricte to just the X-Y plane. Don t forget the Z-axis! T 2 = 1 2 Mv2 = mv 2 Solve for v 2 about the center of mass. The angle φ will be the angle in the x-y plane, while the angle θ will be the angle from the z-axis. 4

5 If θ = 90 o an φ = 0 o then x = l/2 so: x = l sin θ cos φ 2 If θ = 90 o an φ = 90 o then y = l/2 so: If θ = 0 o, then z = l/2 so: y = l sin θ sin φ 2 Fin v 2 : z = l 2 cos θ ẋ 2 + ẏ 2 + ż 2 = v 2 ẋ = l 2 (cos φ cos θ θ sin θ sin φ φ) ẏ = 1 2 (sin φ cos θ θ + sin θ cos φ φ) Carefully square each: ż = l sin θ θ 2 ẋ 2 = l2 4 cos2 φ cos 2 θ θ 2 2 l 2 sin θ sin φ φ l 2 cos φ cos θ θ + l2 4 sin2 θ sin 2 φ φ 2 ẏ 2 = l2 4 sin2 φ cos 2 θ θ l 2 sin θ cos φ φ l 2 sin φ cos θ θ + l2 4 sin2 θ cos 2 φ φ 2 ż 2 = l2 4 sin2 θ θ 2 Now a, striking out the mile terms: ẋ 2 +ẏ 2 +ż 2 = l2 4 [cos2 φ cos 2 θ θ 2 +sin 2 θ sin 2 φ φ 2 +sin 2 φ cos 2 θ θ 2 +sin 2 θ cos 2 φ φ 2 +sin 2 θ θ 2 ] Pull the first an thir terms insie the brackets together, an pull the secon an fourth terms together as well: v 2 = l2 4 [cos2 θ θ 2 (cos 2 φ + sin 2 φ) + sin 2 θ φ 2 (sin 2 φ + cos 2 φ) + sin 2 θ θ 2 ] 5

6 v 2 = l2 4 (cos2 θ θ 2 + sin 2 θ θ 2 + sin 2 θ φ 2 ) v 2 = l2 4 ( θ 2 + sin 2 θ φ 2 ) Now that we finally have v 2 we can plug this into T 2 T = T 1 + T 2 = ma 2 ψ 2 + m l2 4 ( θ 2 + sin 2 θ φ 2 ) It was important to emphasize that T 1 is the kinetic energy of the total mass aroun the center of the circle while T 2 is the kinetic energy of the masses about the center of mass. Hope that helpe. 15. A point particle moves in space uner the influence of a force erivable from a generalize potential of the form U(r, v) = V (r) + σ L where r is the raius vector from a fixe point, L is the angular momentum about that point, an σ is a fixe vector in space. 1. Fin the components of the force on the particle in both Cartesian an spherical poloar coorinates, on the basis of Lagrangian s equations with a generalize potential 2. Show that the components in the two coorinate systems are relate to each other as in the equation shown below of generalize force 3. Obtain the equations of motion in spherical polar coorinates Q j = i F i r i q j Answer: This one is a fairly teious problem mathematically. First lets fin the components of the force in Cartesian coorinates. Convert U(r, v) into Cartesian an then plug the expression into the Lagrange-Euler equation. Q j = [V ( x t q 2 + y 2 + z 2 )+σ (r p)] [V ( x j q 2 + y 2 + z 2 )+σ (r p)] j Q j = [σ [(xî+yĵ+zˆk) (p x î+p y ĵ+p zˆk)] [V ( x t v j x 2 + y 2 + z 2 )+σ (r p)] j 6

7 Q j = [σ [(yp z zp y )î+(zp x xp z )ĵ+(xp y p x y)ˆk] [V ( x t v j x 2 + y 2 + z 2 )+σ (r p)] j Q j = [mσ x (yv z zv y )+mσ y (zv x xv z )+mσ z (xv y v x y)] [V ( x t v j x 2 + y 2 + z 2 )+σ (r p)] j Where we know that mσ x (yv z zv y ) + mσ y (zv x xv z ) + mσ z (xv y v x y) = σ (r p) So lets solve for just one component first an let the other ones follow by example: Q x = t (mσ yz mσ z y) x [V ( x 2 + y 2 + z 2 )+mσ x (yv z zv y )+mσ y (zv x xv z )+mσ z (xv y v x y)] Q x = m(σ y v z σ z v y ) [V ( x 2 + y 2 + z 2 )(x 2 +y 2 +z 2 ) 1 2 x mσy v z +mσ z v y ] Q x = 2m(σ y v z σ z v y ) V x r If you o the same for the y an z components, they are: Q y = 2m(σ z v x σ x v z ) V y r Thus the generalize force is: Q z = 2m(σ x v y σ y v x ) V z r F = 2m(σ v) V r r Now its time to play with spherical coorinates. The trick to this is setting up the coorinate system so that σ is along the z axis. Thus the ot prouct simplifies an L is only the z-component. U = V (r) + mσ(xẏ yẋ) With spherical coorinate efinitions: Solving for (xẏ yẋ) x = r sin θ cos φ y = r sin θ sin φ z = r cos θ 7

8 ẋ = r( sin θ sin φ φ + cos φ cos θ θ) + ṙ sin θ cos φ ẏ = r(sin θ cos φ φ + sin φ cos θ θ) + ṙ sin θ sin φ Thus xẏ yẋ is = r sin θ cos φ[r(sin θ cos φ φ + sin φ cos θ θ) + ṙ sin θ sin φ] r sin θ sin φ[r( sin θ sin φ φ + cos φ cos θ θ) + ṙ sin θ cos φ]. Note that the ṙ terms rop out as well as the θ terms. Thus xẏ yẋ = r 2 sin 2 θ cos 2 φ φ + r 2 sin 2 θ sin 2 φ φ xẏ yẋ = r 2 sin 2 θ φ U = V (r) + mσr 2 sin 2 θ φ Plugging this in to Lagrangian s equations yiels: For Q r : For Q θ : For Q φ : Q r = U r + t ( U ṙ ) Q r = V r 2mσr sin2 θ φ + t (0) Q r = V r 2mσr sin2 θ φ Q θ = 2mσr 2 sin θ φ cos θ Q φ = t (mσr2 sin 2 θ) Q φ = mσ(r 2 2 sin θ cos θ θ + sin 2 θ2rṙ) Q φ = 2mσr 2 sin θ cos θ θ + 2mσrṙ sin 2 θ For part b, we have to show the components of the two coorinate systems are relate to each other via Q j = i F i r i q j Lets take φ for an example, Q φ = F r φ = F x x φ + F y y φ + F z z φ 8

9 Q φ = Q x ( r sin θ sin φ) + Q y (r sin θ cos φ) + Q z (0) Q φ = [2m(σ y v z σ z v y ) V x r ]( r sin θ sin φ)+[2m(σ zv x σ x v z ) V y ](r sin θ cos φ)+0 r Because in both coorinate systems we will have σ pointing in only the z irection, then the x an y σ s isappear: Q φ = [2m( σ z v y ) V x r ]( r sin θ sin φ) + [2m(σ zv x ) V y ](r sin θ cos φ) r Pull out the V terms, plug in x an y, see how V terms cancel Q φ = V (x sin θ sin φ y sin θ cos φ) 2mr sin θσ[v y sin φ + v x cos φ] Q φ = V (r sin 2 θ cos φ sin φ r sin 2 θ sin φ cos φ) 2mr sin θσ[v y sin φ + v x cos φ] Plug in v y an v x : Q φ = 2mr sin θσ[v y sin φ + v x cos φ] Q φ = 2mr sin θσ[sin φ(r sin θ cos φ φ + r sin φ cos θ θ + ṙ sin θ sin φ) + cos φ( r sin θ sin φ φ + r cos φ cos θ θ + ṙ sin θ cos φ)]. Gather sin 2 s an cos 2 s: Q φ = 2mσr sin θ[r sin 2 φ cos θ θ + ṙ sin θ sin 2 φ +r cos 2 φ cos θ θ + ṙ sin θ cos 2 φ]. Q φ = 2mσr sin θ[r cos θ θ + ṙ sin θ] This checks with the erivation in part a for Q φ. This shows that inee the components in the two coorinate systems are relate to each other as Q j = i F i r i q j Any of the other components coul be equally compare in the same proceure. I chose Q φ because I felt it was easiest to write up. For part c, to obtain the equations of motion, we nee to fin the generalize kinetic energy. From this we ll use Lagrange s equations to solve for each component of the force. With both erivations, the components erive from the generalize potential, an the components erive from kinetic energy, they will be set equal to each other. In spherical coorinates, v is: 9

10 v = ṙˆr + r θˆθ + r sin θ φ ˆφ The kinetic energy in spherical polar coorinates is then: For the r component: T = 1 2 mv2 = 1 2 m(ṙ2 + r 2 θ2 + r 2 sin 2 θ φ 2 ) t ( T ṙ ) T r = Q r t (mṙ) mr θ 2 mr sin 2 θ φ 2 = Q r From part a, Set them equal: m r mr θ 2 mr sin 2 θ φ 2 = Q r Q r = V 2mσr sin 2 θ φ m r mr θ 2 mr sin 2 θ φ 2 = Q r = V 2mσr sin 2 θ φ m r mr θ 2 mr sin 2 θ φ 2 + V + 2mσr sin 2 θ φ = 0 For the θ component: m r mr θ 2 + mr sin 2 θ φ(2σ φ) + V = 0 t ( T T ) θ θ = Q θ t (mr2 θ) mr 2 sin θ φ 2 cos θ = Q θ From part a, mr 2 θ + 2mrṙ θ mr 2 sin θ φ 2 cos θ = Q θ Set the two equal: Q θ = 2mσr 2 sin θ cos θ φ mr 2 θ + 2mrṙ θ mr 2 sin θ φ 2 cos θ + 2mσr 2 sin θ cos θ φ = 0 mr 2 θ + 2mrṙ θ + mr 2 sin θ cos θ φ(2σ φ) = 0 10

11 For the last component, φ we have: t ( T T ) φ φ = Q φ t (mr2 sin 2 θ φ) 0 = Q φ mr 2 t (sin2 θ φ) + 2mrṙ sin 2 θ φ = Q φ mr 2 sin 2 θ φ + 2mr 2 sin θ cos θ θ φ + 2mrṙ sin 2 θ φ = Q φ From part a, Q φ = 2mσr 2 sin θ cos θ θ + 2mσrṙ sin 2 θ Set the two equal: mr 2 sin 2 θ φ+2mr 2 sin θ cos θ θ φ+2mrṙ sin 2 θ φ 2mσr 2 sin θ cos θ θ 2mσrṙ sin 2 θ = 0 mr 2 sin 2 θ φ + 2mr 2 sin θ cos θ θ( φ σ) + 2mrṙ sin 2 θ( φ σ) = 0 That s it, here are all of the equations of motion together in one place: m r mr θ 2 + mr sin 2 θ φ(2σ φ) + V = 0 mr 2 θ + 2mrṙ θ + mr 2 sin θ cos θ φ(2σ φ) = 0 mr 2 sin 2 θ φ + 2mr 2 sin θ cos θ θ( φ σ) + 2mrṙ sin 2 θ( φ σ) = A particle moves in a plane uner the influence of a force, acting towar a center of force, whose magnitue is F = 1 r 2 (1 ṙ2 2 rr c 2 ) where r is the istance of the particle to the center of force. Fin the generalize potential that will result in such a force, an from that the Lagrangian for the motion in a plane. The expression for F represents the force between two charges in Weber s electroynamics. Answer: This one takes some guess work an careful hanling of signs. To get from force to potential we will have to take a erivative of a likely potential. Note that if you expan the force it looks like this: 11

12 We know that F = 1 r 2 ṙ2 c 2 r r c 2 r U r + U t ṙ = F So lets focus on the time erivative for now. If we want a r we woul have to take the erivative of a ṙ. Let pick something that looks close, say 2ṙ c 2 r : t ( 2ṙ c 2 r ) = 2ṙ c 2 ( ṙ r 2 ) + 2 r c 2 r = 2ṙ2 c 2 r r c 2 r Excellent! This has our thir term we were looking for. Make this stay the same when you take the partial with respect to ṙ. ṙ 2 ṙ c 2 r = 2ṙ c 2 r So we know that the potential we are guessing at, has the term ṙ2 c 2 r in it. Lets a to it what woul make the first term of the force if you took the negative partial with respect to r, see if it works out. That is, So might work. Checking: We have an r 1 r = 1 r 2 U = 1 r + ṙ2 c 2 r U r + U t ṙ = F U r = 1 r 2 ṙ2 c 2 r 2 thus U t ṙ = t 2ṙ c 2 r = 2ṙ c 2 ( ṙ r 2 ) + 1 (2 r r c 2 ) = 2ṙ2 c 2 r r c 2 r U r + U t ṙ = 1 r 2 + ṙ2 c 2 r 2 2ṙ2 c 2 r r c 2 r U r + U t ṙ = 1 r 2 ṙ2 c 2 r r c 2 r 12

13 This is inee the force unexpane, F = 1 r 2 (1 ṙ2 2 rr c 2 ) = 1 r 2 ṙ2 c 2 r r c 2 r Thus our potential, U = 1 r + ṙ2 c 2 r works. To fin the Lagrangian use L = T U. In a plane, with spherical coorinates, the kinetic energy is Thus T = 1 2 m(ṙ2 + r 2 θ2 ) L = 1 2 m(ṙ2 + ṙ 2 θ2 ) 1 ṙ2 (1 + r c 2 ) 17. A nucleus, originally at rest, ecays raioactively by emitting an electron of momentum 1.73 MeV/c, an at right angles to the irection of the electron a neutrino with momentum 1.00 MeV/c. The MeV, million electron volt, is a unit of energy use in moern physics equal to J. Corresponingly, MeV/c is a unit of linear momentum equal to kg m/s. In what irection oes the nucleus recoil? What is its momentum in MeV/c? If the mass of the resiual nucleus is kg what is its kinetic energy, in electron volts? Answer: If you raw a iagram you ll see that the nucleus recoils in the opposite irection of the vector mae by the electron plus the neutrino emission. Place the neutrino at the x-axis, the electron on the y axis an use pythagorean s theorme to see the nucleus will recoil with a momentum of 2 Mev/c. The nucleus goes in the opposite irection of the vector that makes an angle θ = tan = 60 1 from the x axis. This is 240 from the x-axis. To fin the kinetic energy, you can convert the momentum to kg m/s, then convert the whole answer that is in joules to ev, T = p2 2m = [2( )] 2 1MeV J 106 ev 1MeV = 9.13eV 13

14 18. A Lagrangian for a particular physical system can be written as L = m 2 (aẋ2 + 2bẋẏ + cẏ 2 ) K 2 (ax2 + 2bxy + cy 2 ). where a, b, an c are arbitrary constants but subject to the conition that b 2 ac 0. What are the equations of motion? Examine particularly the two cases a = 0 = c an b = 0, c = a. What is the physical system escribe by the above Lagrangian? Show that the usual Lagrangian for this system as efine by Eq. (1.57 ): L (q, q, t) = L(q, q, t) + F t is relate to L by a point transformation (cf. Derivation 10). What is the significance of the conition on the value of b 2 ac? Answer: To fin the equations of motion, use the Euler-Lagrange equations. For x first: q = t q Thus x = ( Kax Kby) = K(ax + by) = m(aẋ + bẏ) ẋ = m(aẍ + bÿ) t ẋ K(ax + by) = m(aẍ + bÿ) Now for y: y = ( Kby Kcy) = K(bx + cy) Thus = m(bẋ + cẏ) ẋ = m(bẍ + cÿ) t ẋ K(bx + cy) = m(bẍ + cÿ) 14

15 Therefore our equations of motion are: K(ax + by) = m(aẍ + bÿ) K(bx + cy) = m(bẍ + cÿ) Examining the particular cases, we fin: If a = 0 = c then: If b = 0, c = a then: Kx = mẍ Ky = mÿ Kx = mẍ Ky = mÿ The physical system is harmonic oscillation of a particle of mass m in two imensions. If you make a substitution to go to a ifferent coorinate system this is easier to see. Then u = ax + by v = bx + cy Ku = mü Kv = m v The system can now be more easily seen as two inepenent but ientical simple harmonic oscillators, after a point transformation was mae. When the conition b 2 ac 0 is violate, then we have b = ac, an L simplifies to this: L = m 2 ( aẋ + cẏ) 2 K 2 ( ax + cy) 2 Note that this is now a one imensional problem. So the conition keeps the Lagrangian in two imensions, or you can say that the transformation matrix ( ) a b is singluar because b 2 ac 0 Note that ( ) ( ) ( u a b x = v b c y b So if this conition hols then we can reuce the Lagrangian by a point transformation. 19. Obtain the Lagrange equations of motion for spherical penulum, i.e., a mass point suspene by a rigi weightless ro. c ). 15

16 Answer: The kinetic energy is foun the same way as in exercise 14, an the potential energy is foun by using the origin to be at zero potential. T = 1 2 ml2 ( θ 2 + sin 2 θ φ 2 ) If θ is the angle from the positive z-axis, then at θ = 90 the ro is aligne along the x-y plane, with zero potential. Because cos(90) = 0 we shoul expect a cos in the potential. When the ro is aligne along the z-axis, its potential will be its height. V = mgl cos θ If θ = 0 then V = mgl. If θ = 180 then V = mgl. So the Lagrangian is L = T V. L = 1 2 ml2 ( θ 2 + sin 2 θ φ 2 ) mgl cos θ To fin the Lagrangian equations, they are the equations of motion: Solving these yiels: θ = t θ φ = t φ Thus an Thus θ = ml2 sin θ φ 2 cos θ + mgl sin θ t θ = ml2 θ ml 2 sin θ φ 2 cos θ + mgl sin θ ml 2 θ = 0 φ = 0 t φ = t (ml2 sin 2 θ φ) = ml 2 sin 2 θ φ + 2 φml 2 sin θ cos θ ml 2 sin 2 θ φ + 2 φml 2 sin θ cos θ = 0 16

17 Therefore the equations of motion are: ml 2 sin θ φ 2 cos θ + mgl sin θ ml 2 θ = 0 ml 2 sin 2 θ φ + 2 φml 2 sin θ cos θ = A particle of mass m moves in one imension such that it has the Lagrangian L = m2 ẋ mẋ 2 V (x) V 2 (x) where V is some ifferentiable function of x. Fin the equation of motion for x(t) an escribe the physical nature of the system on the basis of this system. Answer: I believe there are two errors in the 3r eition version of this question. Namely, there shoul be a negative sign infront of mẋ 2 V (x) an the V 2 (x) shoul be a V 2 (x). Assuming these are all the errors, the solution to this problem goes like this: L = m2 ẋ 4 mẋ 2 V (x) V 2 (x) 12 Fin the equations of motion from Euler-Lagrange formulation. Thus x = mẋ2 V (x) 2V (x)v (x) ẋ = m2 ẋ 3 + 2mẋV (x) 3 t ẋ = m2 ẋ 2 ẍ + 2mV (x)ẍ mẋ 2 V + 2V V + m 2 ẋ 2 ẍ + 2mV ẍ = 0 is our equation of motion. But we want to interpret it. So lets make it look like it has useful terms in it, like kinetic energy an force. This can be one by iviing by 2 an seperating out 1 2 mv2 an ma s. mẋ 2 2 V + V V + mẋ2 mẍ + mẍv = 0 2 Pull V terms together an mẍ terms together: Therefore: ( mẋ2 2 + V )V + mẍ( mẋ2 2 ( mẋ2 2 + V )(mẍ + V ) = 0 + V ) = 0 17

18 Now this looks like E E = 0 because E = mẋ2 2 + V (x). That woul mean t E2 = 2EE = 0 Which allows us to see that E 2 is a constant. If you look at t = 0 an the starting energy of the particle, then you will notice that if E = 0 at t = 0 then E = 0 for all other times. If E 0 at t = 0 then E 0 all other times while mẍ + V = Two mass points of mass m 1 an m 2 are connecte by a string passing through a hole in a smooth table so that m 1 rests on the table surface an m 2 hangs suspene. Assuming m 2 moves only in a vertical line, what are the generalize coorinates for the system? Write the Lagrange equations for the system an, if possible, iscuss the physical significance any of them might have. Reuce the problem to a single secon-orer ifferential equation an obtain a first integral of the equation. What is its physical significance? (Consier the motion only until m 1 reaches the hole.) Answer: The generalize coorinates for the system are θ, the angle m 1 moves roun on the table, an r the length of the string from the hole to m 1. The whole motion of the system can be escribe by just these coorinates. To write the Lagrangian, we will want the kinetic an potential energies. T = 1 2 m 2ṙ m 1(ṙ 2 + r 2 θ2 ) V = m 2 g(l r) The kinetic energy is just the aition of both masses, while V is obtaine so that V = mgl when r = 0 an so that V = 0 when r = l. L = T V = 1 2 (m 2 + m 1 )ṙ m 1r 2 θ2 + m 2 g(l r) To fin the Lagrangian equations or equations of motion, solve for each component: t θ = 0 θ = m 1r 2 θ θ = t (m 1r 2 θ) = 0 t θ = m 1r 2 θ + 2m1 rṙ θ = 0 18

19 Thus an t (m 1r 2 θ) = m1 r(r θ + 2 θṙ) = 0 Thus r = m 2g + m 1 r θ 2 ṙ = (m 2 + m 1 )ṙ t ṙ = (m 2 + m 1 ) r m 2 g m 1 r θ 2 + (m 2 + m 1 ) r = 0 Therefore our equations of motion are: t (m 1r 2 θ) = m1 r(r θ + 2 θṙ) = 0 m 2 g m 1 r θ 2 + (m 2 + m 1 ) r = 0 See that m 1 r 2 θ is constant. It is angular momentum. Now the Lagrangian can be put in terms of angular momentum. We have θ = l/m 1 r 2. The equation of motion L = 1 2 (m 1 + m 2 )ṙ 2 + l2 2m 1 r 2 m 2gr Becomes m 2 g m 1 r θ 2 + (m 2 + m 1 ) r = 0 (m 1 + m 2 ) r l2 m 1 r 3 + m 2g = 0 The problem has been reuce to a single secon-orer ifferential equation. The next step is a nice one to notice. If you take the erivative of our new Lagrangian you get our single secon-orer ifferential equation of motion. t (1 2 (m 1 + m 2 )ṙ 2 + l2 2m 1 r 2 m 2gr) = (m 1 + m 2 )ṙ r (m 1 + m 2 ) r l2 m 1 r 3 m 2g = 0 l2 m 1 r 3 ṙ m 2gṙ = 0 Thus the first integral of the equation is exactly the Lagrangian. As far as interpreting this, I will venture to say the the Lagrangian is constant, the system is close, the energy is converse, the linear an angular momentum are conserve. 19

20 22. Obtain the Lagrangian an equations of motion for the ouble penulum illustrate in Fig 1.4, where the lengths of the penula are l 1 an l 2 with corresponing masses m 1 an m 2. Answer: A the Lagrangian of the first mass to the Lagrangian of the secon mass. For the first mass: T 1 = 1 2 ml2 1 θ 2 1 Thus V 1 = m 1 gl 1 cos θ 1 L 1 = T 1 V 1 = 1 2 ml 1 θ mgl 1 cos θ 1 To fin the Lagrangian for the secon mass, use new coorinates: x 2 = l 1 sin θ 1 + l 2 sin θ 2 y 2 = l 1 cos θ 1 + l 2 cos θ 2 Then it becomes easier to see the kinetic an potential energies: T 2 = 1 2 m 2(ẋ ẏ2) 2 V 2 = m 2 gy 2 Take erivatives an then plug an chug: T 2 = 1 2 m 2(l 2 1 sin 2 θ 1 θ l 1 l 2 sin θ 1 sin θ 2 θ1 θ2 + l 2 2 sin 2 θ 2 θ2 2 +l 2 1 cos 2 θ 1 θ l 1 l 2 cos θ 1 cos θ 2 θ1 θ2 + l 2 2 cos 2 θ 2 θ2 2 ) an T 2 = 1 2 m 2(l 2 1 θ l 2 2 θ l 1 l 2 cos(θ 1 θ 2 ) θ 1 θ2 ) Thus V 2 = mg(l 1 cos θ 1 + l 2 cos θ 2 ) L 2 = T 2 V 2 = 1 2 m 2(l 2 1 θ l 2 2 θ l 1 l 2 cos(θ 1 θ 2 ) θ 1 θ2 ) + m 2 g(l 1 cos θ 1 + l 2 cos θ 2 ) A L 1 + L 2 = L, 20

21 L = 1 2 m 2(l 2 1 θ 2 1+l 2 2 θ 2 2+2l 1 l 2 cos(θ 1 θ 2 ) θ 1 θ2 )+m 2 g(l 1 cos θ 1 +l 2 cos θ 2 )+ 1 2 ml 1 θ 2 1+m 1 gl 1 cos θ 1 Simplify even though it still is pretty messy: L = 1 2 (m 1+m 2 )l 2 1 θ 2 1+m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 1 θ m 2l 2 2 θ 2 2+(m 1 +m 2 )gl 1 cos θ 1 +m 2 gl 2 cos θ 2 This is the Lagrangian for the ouble penulum. To fin the equations of motion, apply the usual Euler-Lagrangian equations an turn the crank: For θ 1 : θ 1 = m 2 l 1 l 2 sin(θ 1 θ 2 ) θ 1 θ2 (m 1 + m 2 )gl 1 sin θ 1 θ 1 = (m 1 + m 2 )l 2 2 θ 1 + m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 2 t [ θ ] = (m 1 + m 2 )l 2 θ m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 2 + θ 2 1 t [m 2l 1 l 2 cos(θ 1 θ 2 )] Let s solve this annoying erivative term: θ 2 t [m 2l 1 l 2 cos(θ 1 θ 2 )] = θ 2 m 2 l 2 l 1 t [cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ] Using a trig ientity, = θ 2 m 2 l 2 l 1 [ cos θ 1 sin θ 2 θ2 cos θ 2 sin θ 1 θ1 + sin θ 1 cos θ 2 θ2 + sin θ 2 cos θ 1 θ1 ] An then more trig ientities to put it back together, = θ 2 m 2 l 2 l 1 [ θ 2 sin(θ 1 θ 2 ) θ 1 sin(θ 1 θ 2 )] = θ 2 2m 2 l 2 l 1 sin(θ 1 θ 2 ) m 2 l 2 l 1 sin(θ 1 θ 2 ) θ 1 θ2 Plugging this term back into our Euler-Lagrangian formula, the secon term of this cancels its positive counterpart: θ + t [ θ 1 ] = (m 1 +m 2 )gl 1 sin θ 1 +(m 1 +m 2 )l 2 2 θ 1 +m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 2 + θ 2 2m 2 l 2 l 1 sin(θ 1 θ 2 ) Finally, cancel out a l 1 an set to zero for our first equation of motion: (m 1 +m 2 )g sin θ 1 +(m 1 +m 2 )l 1 θ1 +m 2 l 2 cos(θ 1 θ 2 ) θ 2 + θ 2 2m 2 l 2 sin(θ 1 θ 2 ) = 0 Now for θ 2 : 21

22 θ 2 = m 2 l 1 l 2 sin(θ 1 θ 2 ) θ 1 θ2 m 2 gl 2 sin θ 2 θ 2 = m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 1 + m 2 l 2 2 θ 2 t θ = m 2 l 2 θ θ 1 m 2 l 2 l 1 cos(θ 1 θ 2 ) + θ 1 [ 2 t (m 2l 2 l 1 cos(θ 1 θ 2 ))] Fortunately this is the same erivative term as before, so we can cut to the chase: = m 2 l 2 2 θ 2 + θ 1 m 2 l 2 l 1 cos(θ 1 θ 2 ) + m 2 l 1 l 2 θ1 [ θ 2 sin(θ 1 θ 2 ) θ 1 sin(θ 1 θ 2 )] Thus + θ 2 t θ = +m 2 gl 2 sin θ 2 +m 2 l 2 θ θ 1 m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 1m 2 2 l 1 l 2 sin(θ 1 θ 2 ) 2 Cancel out an l 2 this time, set to zero, an we have our secon equation of motion: m 2 g sin θ 2 + m 2 l 2 θ2 + θ 1 m 2 l 1 cos(θ 1 θ 2 ) θ 2 1m 2 l 1 sin(θ 1 θ 2 ) = 0 Both of the equations of motion together along with the Lagrangian: L = 1 2 (m 1+m 2 )l 2 1 θ 2 1+m 2 l 1 l 2 cos(θ 1 θ 2 ) θ 1 θ m 2l 2 2 θ 2 2+(m 1 +m 2 )gl 1 cos θ 1 +m 2 gl 2 cos θ 2 (m 1 +m 2 )g sin θ 1 +(m 1 +m 2 )l 1 θ1 +m 2 l 2 cos(θ 1 θ 2 ) θ 2 + θ 2 2m 2 l 2 sin(θ 1 θ 2 ) = 0 m 2 g sin θ 2 + m 2 l 2 θ2 + θ 1 m 2 l 1 cos(θ 1 θ 2 ) θ 2 1m 2 l 1 sin(θ 1 θ 2 ) = Obtain the equation of motion for a particle falling vertically uner the influence of gravity when frictional forces obtainable from a issipation function 1 2 kv2 are present. Integrate the equation to obtain the velocity as a function of time an show that the maximum possible velocity for a fall from rest is v + mg/k. Answer: Work in one imension, an use the most simple Lagrangian possible: With issipation function: L = 1 2 mż2 mgz 22

23 The Lagrangian formulation is now: F = 1 2 kż2 Plug an chug an get: t ż z + F ż = 0 m z mg + kż = 0 Note that at terminal velocity there is no total force acting on you, gravity matches force ue to friction, so m z = 0: mg = kż ż = mg k But lets integrate like the problem asks. Let f = ż mg the equation of motion: k an substitute into Note that f = z. Thus m z mg + kż = 0 m z k mg k + ż = 0 m z k + f = 0 Therefore mf k + f = 0 f f = k m ln f = k m t + C f = Ce k m t ż mg k = Ce k m t Plugging in the bounary conitions, that at t = 0, ż = 0, we solve for C Thus ż mg k mg k = C = mg k e k m t 23

24 an with t we have finally ż = mg k 24. A spring of rest length L a ( no tension ) is connecte to a support at one en an has a mass M attache at the other. Neglect the mass of the spring, the imension of the mass M, an assume that the motion is confine to a vertical plane. Also, assume that the spring only stretches without bening but it can swing in the plane. 1. Using the angular isplacement of the mass from the vertical an the length that the string has stretche from its rest length (hanging with the mass m), fin Lagrange s equations. 2. Solve these equations fro small stretching an angular isplacements. 3. Solve the equations in part (1) to the next orer in both stretching an angular isplacement. This part is amenable to han calculations. Using some reasonable assumptions about the spring constant, the mass, an the rest length, iscuss the motion. Is a resonance likely uner the assumptions state in the problem? 4. (For analytic computer programs.) Consier the spring to have a total mass m << M. Neglecting the bening of the spring, set up Lagrange s equations correctly to first orer in m an the angular an linear isplacements. 5. (For numerical computer analysis.) Make sets of reasonable assumptions of the constants in part (1) an make a single plot of the two coorinates as functions of time. Answer: This is a spring-penulum. It s kinetic energy is ue to translation only. T = 1 2 m(ṙ2 + (r θ) 2 ) The more general form of v is erive in problem 15 if this step was not clear. Just isregar φ irection. Here r signifies the total length of the spring, from support to mass at any time. As in problem 22, the potential has a term epenent on gravity, but it also has the potential of your normal spring. V = mgr cos θ k(r L a) 2 Note that the potential ue to gravity epens on the total length of the spring, while the potential ue to the spring is only epenent on the stretching from its natural length. Solving for the Lagrangian: 24

25 L = T V = 1 2 m(ṙ2 + (r θ) 2 ) + mgr cos θ 1 2 k(r L a) 2 Lets solve for Lagrange s equations now. For r: r = mg cos θ + mr θ 2 k(r L a ) t ṙ = m r For θ: = mgr sin θ θ t θ = t (mr2 θ) = mr 2 θ + 2mrṙ θ Bring all the pieces together to form the equations of motion: t ṙ r = m r mr θ 2 + k(r L a ) mg cos θ = 0 t θ θ = mr2 θ + 2mrṙ θ + mgr sin θ = 0 For part b, we are to solve these equations for small stretching an angular isplacements. Simplify the equations above by canceling out m s, r s an substituting θ for sin θ, an 1 for cos θ. r r θ 2 + k m (r L a) g = 0 θ + 2ṙ r θ + g r θ = 0 Solve the first equation, for r, with the initial conition that θ 0 = 0, θ 0 = 0, r 0 = 0 an ṙ 0 = 0: r = L a + mg k Solve the secon equation, for θ, with the same initial conitions: θ = 0 This is the solution of the Lagrangian equations that make the generalize force ientically zero. To solve the next orer, change variables to measure eviation from equilibrium. x = r (L a + mg k ), θ Substitute the variables, keep only terms to 1st orer in x an θ an the solution is: 25

26 ẍ = k m x θ g = L a + m k g θ In terms of the original coorinates r an θ, the solutions to these are: r = L a + mg k k + A cos( m t + φ) kg θ = B cos( kl a + mg t + φ ) The phase angles, φ an φ, an amplitues A an B are constants of integration an fixe by the initial conitions. Resonance is very unlikely with this system. The spring penulum is known for its nonlinearity an stuies in chaos theory. 26