Tutorial Exercises: Incorporating constraints

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1 Tutorial Exercises: Incorporating constraints 1. A siple pendulu of length l ass is suspended fro a pivot of ass M that is free to slide on a frictionless wire frae in the shape of a parabola y = ax. The pendulu oves in the plane of the frae. (a) Write down the cartesian coordinates of both asses in ters of x θ. M θ l (b) Calculate the tie derivatives of the cartesian coordinates. (c) Write down the kinetic potential energies using x θ as generalised coordinates. (d) Write down the Lagrangian using the approxiation that x, θ their derivatives are sall, solve the corresponding linear Lagrange equations. 1. Solution: (a) The cartesian coordinates of ass M are The cartesian coordinates of ass are x 1 = x y 1 = ax x = x + l sin θ (b) The tie derivatives of these coordinates are (c) The kinetic energy is y = ax l cos θ ẋ 1 = ẋ ẏ 1 = axẋ ẋ = ẋ + l cos θ θ ẏ = axẋ + l sin θ θ T = 1 M(ẋ 1+ẏ 1)+ 1 (ẋ +ẏ ) = 1 (M +)(1+4a x )ẋ + 1 l θ +lẋ θ(cos θ+ax sin θ) the potential energy is the Lagrangian is then L = T V. V = Mgy 1 + gy = (M + )gax gl cos θ 9

2 (d) The Lagrangian including only the quadratic ters is L = 1 (M + )ẋ + 1 l θ + lẋ θ (M + )gax 1 glθ The Euler-Lagrange equations are which can be siplified to (M + )ẍ + l θ + (M + )gax = 0 l θ + lẍ + glθ = 0 ẍ + M + l θ + gax = 0 ẍ + l θ + gθ = 0 These are linear coupled ODEs, the solutions will be trigonoetric so we try solutions of the for x = A exp(iωt) Substituting gives ( ω + ag)a The deterinant of this syste gives Thus, (ag ω )(g lω ) ω = y = B exp(iωt) M + lω B = 0 ω A + (g lω )B = 0 M + lω4 = M M + lω4 g(1 + al)ω + ag = 0 g(1 + al) ± g (1 + al) 8ag l M M+ l M M+. A pendulu of length l with ass at the end wraps around a circular obstacle of radius a as it swings (see diagra). The point R is fixed, the point Q is where the string first loses contact with the circle, the point P is location of the ass. Both Q P change with tie. R Q O θ a You can assue that the string reains taut so that the line P Q is always tangent to the circle. In case you have forgotten your basic geoetry, the angle between the tangent P Q the radius OQ is 90 degrees. P 30

3 (a) Start by using cartesian coordinates for both Q P. What is the length of the part of the string between R Q? Thus, what are the coordinates of Q? (b) Then deterine the length of the part of the string between Q P, thus, the coordinates of P. (c) Write down the kinetic potential energies in ters of the cartesian coordinates of P. (d) Rewrite everything in ters of θ θ thus write down the Lagrangian. (e) Find approxiate solutions of the dynaics of the pendulu by exping the Lagrangian up to quadratic order near the equilibriu point. (f) If the string starts out horizontally, how far round the circle does the point Q ove? Express your answer in ters of l a assue that l is sufficiently large.. Solution: (a) The length of the part of the string between R Q is aθ. The coordinates of Q are (a sin θ, a cos θ). (b) The length of the part of the string between Q P is l aθ. The coordinates of P are x = a sin θ + (l aθ) cos θ The tie derivatives are y = a cos θ (l aθ) sin θ ẋ = a cos θ θ a cos θ θ (l aθ) sin θ θ = (l aθ) sin θ θ (c) Thus ẏ = a sin θ θ + a sin θ θ (l aθ) cos θ θ = (l aθ) cos θ θ T = 1 (ẋ + ẏ ) V = gy (d) Thus T = 1 (l aθ) θ V = ga cos θ g(l aθ) sin θ L = 1 (l aθ) θ ga cos θ + g(l aθ) sin θ (e) The general solution for this echanical syste is rather difficult to find. Particular solutions like equilibria are always easier to find. There is an equilibriu with the string hanging vertically down when θ = π/. Linearising near an equilibriu give approxiate solutions. Approxiating near the equilibriu point just found gives L = 1 (l aπ/) θ + 1 g(l aπ/)(θ π/) Which is obviously just the equation for a linearised pendulu of length l aπ/. In general this syste is hard to solve. Writing down the energy which is a first integral we can separate the variables are led to a coplicated integral. 31

4 (f) If the syste starts fro rest with θ = 0 then the initial V = ga. Thus the ass will swing around until V = ga again. This occurs at ga = ga cos θ g(l aθ) sin θ thus cannot be solved in closed for but it siplifies to l cos θ + θ sin θ 1 = a sin θ showing that the answer only depends on the ratio l/a. The shortest length for which this is possible is l/a = 1 + 3π/ 5.7, then θ = 3π/. For larger l/a > 1 + 3π/ one has to find θ as the root of the equation in the interval [π, 3π/] for l the corresponding θ π. 3. A particle of ass slides freely fro rest down a sooth plane inclined at an angle α to the horizontal. The plane is a wedge of ass M which can slide freely in the horizontal direction. y s h α x (a) Construct the Lagrangian in ters of the distance s oved down the slope the distance x oved by the wedge. [Treat the wedge as a single particle of ass M.] (b) Derive solve the corresponding Lagrange equations. Hint: Rearrange your final set of DE s to give expressions for s ẍ first. 3. Solution: (a) Using the coordinate syste shown in the figure the coordinates of the particle are x 1 = x s cos α, y 1 = h s sin α. We have had to introduce another quantity h, the height of the initial position of the particle above the horizontal plane. Since it is not specified in the question, you ay guess that it has no real significance. The coordinates of the corner of the wedge are Thus the velocities are ẋ 1 = ẋ ṡ cos α, x = x, y = 0. ẏ 1 = ṡ sin α ẋ = ẋ, ẏ = 0. Note that we treat the wedge as a particle here. 3

5 Then T = 1 (ẋ 1 + ẏ 1) + 1 M(ẋ + ẏ ) = 1 ( + M)ẋ + 1 ṡ ṡẋ cos α V = gy 1 + Mgy = g(h s sin α) L = 1 ( + M)ẋ + 1 ṡ ṡẋ cos α + gs sin α. Note firstly that the expression for the K.E. involves the cross ter ṡẋ because the s-axis the x-axis are not orthogonal, secondly that the constant ter in the P.E. has been oitted fro the Lagrangian, where it has no significance. (b) There are two Euler-Lagrange equations, ( ) d L ṡ d ( L ẋ L s = 0 or d (ṡ ẋ cos α) g sin α = 0 ) L x = 0 or d (( + M)ẋ ṡ cos α) = 0. However, x is an ignorable co-ordinate so instead of the second Lagrange equations we can use a first integral instead ( + M)ẋ ṡ cos α = C 1 In either case, since we require the accelerations, the Lagrange equations should be exped, (or the first integral can differentiated once to give) s ẍ cos α = g sin α ( + M)ẍ s cos α = 0. The two accelerations are found by solving these siultaneously. This gives s = ( + M)g sin α sin α + M These are now trivial to solve give s = x =, ẍ = g sin α cos α sin α + M. ( + M)g sin α ( sin α + M) t + C t + C 3 g sin α cos α ( sin α + M) t + C 3 t + C 4 If everything starts fro rest when x = 0 s = 0 then all the constants are zero. The value of the first integral is C 1 = ( + M)C 3 cos αc. 4. An inextensible rope hangs over a pulley which has a oent of inertia I radius a. The pulley can rotate freely about its centre is rough so that the rope does not slip on the circuference. Masses M are attached at either end of the rope. If the syste is released fro rest, show that the acceleration of the asses is given by (M )ga I + (M + )a. 33

6 y 1 M y 4. Solution: Take coordinates fro the centre line of the pulley with positive y downwards. Since the length of the string reains constant, there is a constraint where l is a constant. Thus y 1 + πa + y = l, y = l πa y 1 ẏ = ẏ 1. There is another constraint provided by the no-slip condition which requires the speed of the circuference of the pulley to equal that of the rope, i.e. a θ = ẏ 1 = ẏ, where θ is the angular speed of the pulley as indicated. Now T = 1 Mẏ ẏ + 1 I θ = 1 Mẏ ẏ I ẏ 1 a = 1 (M + + Ia ) ẏ1 V = Mgy 1 gy = Mgy 1 g(l πa y 1 ). Hence L = 1 (M + + Ia ) ẏ 1 + (M )gy 1, ignoring constant ters. There is now a single Lagrange equation d ((M + + Ia ) ) ẏ 1 (M )g = 0, which gives ÿ 1 = (M )g M + + I/a, ÿ (M )g = M + + I/a. 5. A unifor solid cylinder, ass M radius a, rolls without slipping on a rough plane inclined at angle α to the horizontal. Analyse the proble in two diensions. As it rolls the cylinder rolls up a light string which passes over a fixed light pulley supports a freely hanging ass. Choose appropriate coordinates no-slip condition. Write down the Lagrangian solve to find the accelerations of the cylinder the ass. 34

7 y M s α 5. Solution: Let P be the point of attachent of the string on the cylinder. Let θ be the angle through the cylinder has rolled at tie t, when the centre of ass has oved a distance s down the plane. The constant length of the string provides one constraint, aθ + aπ + s + y + constant = l, where the constant is a function of the geoetry of the pulley. Thus a θ + ṡ + ẏ = 0. The other constraint is the no-slip condition, s = aθ or Now we can for The single Lagrange equation is then ṡ = a θ. T = 1 Mṡ + 1 I θ + 1 ẏ = 1 Mṡ + 1 ( 1 Ma = 1 ( ) 3 M + 4 ṡ V = Mgs sin α gy ) ( ṡ a ) + 1 ( ṡ) = Mgs sin α + gs + constant L = 1 ( ) 3 M + 4 ṡ + (M sin α )gs. d (( 3 M + 4 ) ) ṡ (M sin α )g = 0. Hence s = (M sin α )g 3 M + 4, (M sin α )g ÿ = s = 3 M

8 6. Physical Pendulu The so called atheatical pendulu (as treated in the lecture notes) is a point ass suspended on a assless rod. In the physical pendulu the rod has ass, or ore generally, the body of the pendulu has a suspension point, a centre of ass, soe oent of inertia I G about the centre of ass. Let the distance fro the suspension point the centre of ass of the pendulu body be l. Denote by θ the angle between the vertical the line connecting the suspension point the centre of ass. (a) The parallel axis theore states that if a body of ass M with oent of inertia I G (about the centre of ass) rotates about an axis that has distance d fro the centre of ass then it has oent of inertia I = I G + Md. Prove this theore by considering a body as coposed of point asses. (b) Show that the kinetic energy of the pendulu can be written as T = 1 Ml θ + 1 I G θ. 6. Solution: Physical Pendulu (a) About the centre of ass we have I G = i r i, where r i is the vector fro the origin, which coincides with the centre of ass. If the body is now rotating about a point s instead the oent of inertia with respect to that point is I s = i r i + s = i ( r i + r i s + s ) = i r i + M s = I G + Md since i r i = 0. (b) Since can be seen as a special case of the decoposition into otion of the centre of ass the rotation about the centre of ass. Alternatively, using the parallel axis theore this can also be read as the rotational energy 1 I s θ for rotation about the suspension point, which has distance l fro the centre of ass. 7. Double Pendulu Consider a double pendulu ade of two point asses. The first ass 1 is suspended on a assless rod a distance c 1 fro the point about which the rod is rotating. The first rod is rotating about the origin (x, y) = (0, 0). Denote the counterclockwise angle between the rod the downward vertical by θ 1. The second ass is suspended on a second assless rod a distance c fro the point about which the rod is rotating. The second rod that is rotating about the end for the first rod which is a distance l away fro its suspension point. Denote the counterclockwise angle between the second rod the downward vertical by θ. Gravity is on. (a) Write down the Lagrangian of this double pendulu by writing down the position (x 1, y 1 ) of ass 1 (x, y ) of ass as a function of the generalised coordinates θ 1 θ. (b) Write down the Lagrangian if instead of point asses we would have extended bodies of asses 1 with oents of inertia I1 G IG, respectively. Here the G indicates that these oents of inertia are defined with respect to the centre of ass (or centre of gravity). Fro now on work with this general Lagrangian (c) Can you choose paraeters c 1, c, l such that the potential energy vanishes? What does this ean? Give two constants of otion linear in angular velocities in this case. (d) After setting the potential to zero (e.g. by arranging the double pendulu so that it rotates in a horizontal plane), find a linear change of coordinates that introduces an ignorable coordinate. Choose the transforation atrix so that its deterinant is 1. 36

9 (e) In the ost general case the origin, the centre of ass of the first pendulu, the suspension point of the second pendulu are not collinear. Give the additional ters in the Lagrangian in this case. 7. Solution: Double Pendulu (a) Fro basic geoetry (x 1, y 1 ) = c 1 (sin θ 1, cos θ 1 ) (x, y ) = l(sin θ 1, cos θ 1 )+c (sin θ, cos θ ). The kinetic energy is the potential energy is L = 1 c θ 1 1+ T = 1 1(ẋ 1 + ẏ 1) + 1 (ẋ + ẏ ) V = 1 gy 1 + gy. Inserting the expression for (x i, y i ) collecting ters gives ( l θ 1 + ls cos(θ θ 1 ) θ 1 θ + c θ ) +g( 1 c 1 + l) cos θ 1 +g c cos θ. (b) Use the general observation that kinetic energy can be decoposed into contributions fro translation of the centre of ass rotation about the centre of ass. This eans to add the two ters 1 IG 1 θ IG θ to the kinetic energy, since we already have the contribution fro the otion of the centres of ass. Notice that even though the angles are defined with respect to the assless rods, they do give the rotational angle of the centres of ass as well. (c) Yes, c = 0 1 c 1 + l = 0. It eans that the second pendulu is suspended in its centre of ass, the first pendulu is balanced so that the centre of ass the suspension point of the second pendulu are on opposite sides of the origin (the suspension point of the first pendulu). The Lagrangian is L = 1 IG θ + 1 (IG 1 +c 1 1 +l ) θ 1, which describes two uncoupled free rotors. The corresponding angular oenta are I G θ = const (I G 1 + c l )θ 1 = const. Of course we would eliinate either l or c 1. (d) The kinetic energy depends on the difference of the angles θ 1 θ only, so we should choose this cobination as a new coordinate. A possible transforation whose atrix has deterinant one is u = θ 1, v = θ θ 1. The velocities are θ 1 = u θ = u + v. Then u is ignorable, L u = const. (e) This requires going back to the beginning replacing the ter l(sin θ 1, cos θ 1 ) in (x, y ) by the ore general l x (cos θ 1, sin θ 1 )+l y (sin θ 1, cos θ 1 ), where l = l y. In the kinetic energy this replaces l by l x + l y any other l by l y. The additional cross ter in the kinetic energy is l x c sin(θ θ 1 ), in the potential the additional ter is gl x sin θ 1. 37