The Lagrangian Method vs. other methods (COMPARATIVE EXAMPLE)
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1 The Lagrangian ethod vs. other ethods () This aterial written by Jozef HANC, Technical University, Kosice, Slovakia For Edwin Taylor s website 6 January 003 The ai of this aterial is a deonstration of power of Lagrange s equations for solving physics probles based on coparing the solution of an eaple by five different ethods: Vectorial ethods Using Newton s laws (inertial frae of reference) Using Newton s laws with fictitious forces (accelerated frae of reference) Using D Alebert s principle (The etended principle of virtual work) Scalar ethods Using Conservation laws Using Lagrange s equations (The Lagrangian ethod) Here is our proble: OVING PLANE: A block of ass is held otionless on a frictionless plane of ass and angle of inclination. The plane rests on a frictionless horizontal surface. The block is released (see Fig.). What is the horizontal acceleration of the plane? SOLUTIONS: Fig. VECTORIAL ETHODS: I. ETHOD: Newton s laws (in an inertial frae of reference) : g N Fig. Diagra with forces applying on the oving block showing the notation: N agnitude of the noral force fro the plane g agnitude of gravity horizontal The proble was taken fro section.7 in orin [Ref.], ch.pdf, p.8 This solution according to orin [Ref. ], ch.pdf, p.
2 N F Fig. 3 Diagra of the oving plane shows notation: N agnitude the noral force fro the block g agnitude of gravity F agnitude of the noral force fro the base g horizontal Equations of otion of the block horizontally ly Equations of otion of the plane horizontally N sin = a (.) g N cos = a (.) y N sin = A (.3) ly g N cos F = A (.4) + y Kineatical equation resulting fro the otion of the block: ( a ayt / ay = A ) t / a A = tan (.5) We have five equations with five unknowns. To find the quantity A, it is sufficient to resolve equations (.), (.), (.3) and (.5): ay N sin = a g N cos = ay N sin = A = tan a A The solution for A has to be done by a ore tie consuing procedure: gtan A = (.6) ( + tan ) + tan II. ETHOD: Newton s laws with fictitious forces: FORCES acting on the BLOCK (oving in an accelerated frae of reference) g N F trans Fig. 4 Diagra of the oving block in accelerated frae deterined by the plane. N agnitude of the noral force fro the plane F trans agnitude of the translation fictitious force equal to A and with opposite as A g agnitude of gravity horizontal
3 3 Equations of otion of the block: horizontally gsin + ( A ) cos = a (.) ly g cos N ( A )sin (.) Equations of otion of the plane are identical with those written in Section, because we consider the plane in the sae inertial frae of reference. So: horizontally N sin = A (.3) ly g N cos F = A (.4) + y We have four equations with four unknowns. To find the quantity A, it is sufficient to resolve equations (.) and (.3) g cos N + A sin N sin = A The solution for A is given by a quick way: gsincos A = (.5) + sin Using known trigonoetric forulas, we get the sae result as (.6). III. ETHOD: D Alebert s principle ( The etended principle of virtual work ) SHORT SUARY: D Alebert s Principle 3 : The total virtual work of the effective forces, is zero for all reversible variations (virtual displaceents), which satisfy the given kineatical conditions. REARKS: ) This principle uses the concept of the virtual displaceent, which eans, not actual sall displaceent, but it involves a possible, but purely atheatical eperient, so it can be applied in a certain definite tie (even if such a displaceent would involve physically infinite velocities). At the instant the actual otion of the body does not enter into account. To distinguish between virtual displaceent fro actual one, we use label δ instead of or d, e.g. δ, δy, δr. ) The special nae the effective forces represents a coon nae for the given eternal or applied forces on a echanical syste, augented by the inertial forces (see fictitious force in Section ), but without the forces of constraint. 3) Since the forces of constraint do not need to be taken into account, D Alebert principle iediately gives the consequence: The virtual work of the forces of reaction is always zero for any virtual displaceent which is in harony with the given kineatic constrains. Or in other words we restrict ourselves to systes for which the net virtual work of the forces of constraint is zero 4.This assertion is easily understandable fro the Newtonian echanics. If e.g. a particle is constrained to ove on a surface, the force of constraint arises fro the reaction forces (Newton s third law) and is perpendicular to the surface, while virtual displaceent ust be tangent to it, and hence the virtual work of the force vanishes. However, in general case it is not so evident and becoes postulate called POSTULATE A. 3 The forulation of D Alebert s Principle according to Lanczos [Ref. ], page 90, Section IV. See also Ref. 3, Section.4 4 According to Goldstein [Ref.3]
4 4 horizontal g g Fig. 5 Diagra of ipressed forces (g, g) the given eternal forces acting on the echanical syste. The forces of constraint have no influence on a value of virtual work and can be neglected in a virtual-work calculation. Total virtual work is represented by su of virtual works of all ipressed forces together with the inertial forces on echanical syste: block + plane (the vectorial for): δw = ( g a ) δr + ( g A) δr (3.) where a and A are accelerations of the block and the plane, g+( A), g+( a) are corresponding net effective forces acting on the block and the plane, and δr, δr are virtual displaceents of the block and plane in harony with the given kineatical conditions. We can rewrite the equation (3.) in ore appropriate, analytical language. Using Cartesian coordinates: a=(a, a y ), A=(A, 0), g+( A)=(-A, g), g+( a)=(-a, g-a y ), and δr=(δ, δy), δr=(δx, 0), we get the total virtual work as a su of virtual works of coponents of the effective forces in and y-: δw = ( g a ) δy + ( a ) δ + ( A δx (3.) y ) All these displaceents ust be in harony with the kineatical conditions (the block ust slide down the frictionless plane). It gives the condition 5 : δy = ( δ δx ) tan (3.3) In other words all the virtual displaceents δ, δy, δx are not independent. At the sae tie the previous equation (3.3) provides the following relations between accelerations a, a y, A : a = a A ) tan (3.4) y ( Since one equation (3.3) connects the virtual displaceents, one of the displaceent, say δ, can by (3.3) be epressed 6 in ters of the other two δy, δx. Substitution of δ in ters of δy, δx into the virtual work (3.) with an rearrangeent leads to: δw = ( g tan a y tan a ) δy ( a + A ) δx (3.5) This reaining pair δy, δx ay be considered as quite independent displaceents without further restriction (Think about it!). Then the vanishing of the resultant work requires: g tan a y tan a (3.6a) a A (3.6b) + These equations (3.6) together with equation (3.4) provide the syste of three equations with three unknowns. Solving the equations, we obtain in short tie the sae result for A as (.6). 5 This condition can be obtained by the sae way as in Section IV, Figure 7 of this aterial. 6 We can use any virtual displaceent and epress it in ters of all reaining ones.
5 5 ADDITIONAL NOTE (This note can be skipped the first tie): The analytical treatent according to D Alebert s principle takes only the eternal and inertial forces into account. Since the inner forces which produce constraints need not be consider, D Alebert s ethod offers a significant siplification in coparison with Newton s laws. But it is interested to look at the forces of constraints (see Fig. 6). horizontal N N F Fig. 6 Diagra of all forces of constraints N, N and F. According to D Alebert s principle the forces of constraint have no influence on a value of virtual work and their work has zero value. If we assue the validity of Newton s third law, then agnitudes of forces N, N are equal: N = N = N. Siultaneously according to POSTULATE 7 A the resultant work of these forces with respect to δ, δy, δx ust be zero. atheatically: δwconstraint s = N sin δx + F 0 + N sin δ N cos δy or since N = N = N cos δy = sin δ sin δx (3.8) This equation (3.8) is nothing else as the kineatical condition (3.3): δy = ( δ δx ) tan that was derived by geoetrical arguents fro a figure considering otion of the block on the plane. Use of Newton s third law together with Postulate A yield easily the kineatical condition without considering otion of the block on the plane.. Consider only Newton s third law with the kineatical condition δy = ( δ δx ) tan. Virtual work of the forces of constraints equals: δwconstraints = N sin δx + N sin δ N cos δy = N.cos(tan δ tan δx δy) (3.9) Then kineatical condition (the epression in brackets) provides the zero value of virtual work. So in that case the POSTULATE A is deducible fro Newton s third law. However, in general 8 case the third law of otion, action equals reaction, is not wide enough to replace Postulate A. 3. Finally we can start fro POSTULATE A and the kineatical condition δy = ( δ δx ) tan and coe to Newton s third law. The resultant virtual work (3.7) of the forces of constraints iplies: N δy = ( δ δx ) tan N (3.0) Subtracting the last equation (3.0) fro the δy = ( δ δx ) tan, we obtain the condition: = N / N δ tan and hence the use of arbitrariness of δx yields a result: N = N = N: ( ) 0 X (3.7) 7 See Reark 3 in this section. 8 According to Lanczos [Ref.], p. 77, footnote.
6 6 SCALAR ETHODS IV. ETHOD: Conservation laws We use the center-of-ass frae of reference. For easy work with potential energy we will assue a coordinate syste connected to this frae reference, which has the sae horizontal, but opposite (obvious), as in previous cases 9. Let the plane have the horizontal coordinate l + (t), where the l represents an initial -coordinate of the plane in center-of ass frae reference and (t) translation fro this position. Siilarly l + (t) for the block. Denote v, a the velocity and acceleration of the plane and v, a the velocity and acceleration of the block.. Conservation of oentu in -: initial 0= v - oentu + v. Conservation of otion of center of ass (c..): (4.) initial position l + l ( l + ) + ( l + ) + = 0= = of c Conservation of energy: If the block translates through and the plane through, then it is easy to see fro Fig. 7: (4.) h 7.a 7.b 7.c Fig.7 If the block translates trough a horizontal displaceent and the plane through and we find the height fallen by the block, we do not need to consider the translations of the block and the plane siultaneously. The sae effect is obtained if we first translate horizontally the block through (7.a), then the plane through (7.b) and finally the block ly through h (7.c). that the height h fallen by the block is h = ( ) tan. Hence we can write: changein potential energy g ( )tan = v + v [ + ( v v ) tan ] (4.3) of theblock 9 It eans y increases in the ly upward.
7 7 v Substituting epressions v = and a relationship between position and v : = fro (4.) and (4.) to (4.3), we obtain ( v ) + ( v + ) tan g( + )tan = v + v (4.5) Because we have a need for acceleration a of the plane - tie derivative of velocity v - it is sufficient to take derivative of the equation (4.5) with respect to tie: ( ) v a + va ( + ) gv + )tan = va + tan ( (4.6) Dividing by v ( + ), we get acceleration a : g tan = a + a ( + ) tan or gtan a = (4.7) + ( + )tan ADDITIONAL NOTE: (Here we present a non-calculus way how to obtain the acceleration a fro (4.5)). Instead derivative we can use another general epressions fro kineatics, which is valid for uniforly accelerated otion with zero initial velocity. We can write for the plane: = l a t l + + a t = ; v = at a = v (4.8) where s is passing distance, a and v are agnitudes of acceleration and velocity. After rearranging (4.5), we get: gtan = v + ( + )tan So according to (4.8) the agnitude of the a has a for: gtan a = (4.0) + ( + )tan (4.9) V. ETHOD: Lagrange s equations 0 We use the sae notation as in Section IV. oreover label h 0 the initial height of the block. We know fro the previous section that the height fallen by the block is ( ) tan. Then:. Potential energy of the syste (block + plane): [ ( ) ] V = gh tan 0. Kinetic energy of the syste: T = v + [ + ( v v ) tan ] v (5.) (5.) 0 See the detailed description of the eplanation and use of the Lagrangian ethod in orin [Ref.], ch5.pdf
8 8 3. The Lagrangian of the syste: [ + v v ) tan ] gh [ ( ) tan ] L T V = v + v ( 0 (5.3) 4. The equations of otion: L L d dt d dt L v L v (5.4a) (5.4b) After substituting Lagrangian (5.3) to (5.4a) and (5.4b), we have: g tan = a + ( a a) tan (5.5a) + g tan = a ( a a) tan (5.5b) Adding equations (5.5), we get a + a or a (5.5a), we easily obtain: a = and substituting this result to the gtan a = (5.6) ( + tan ) + tan REFERENCES: [] David J. orin, echanics and Special Relativity, [] Cornelius Lanczos, The Variational principles of echanics, (Dover Publications, New York, 986) [3] Herbert Goldstein, Charles Poole, and John Safko, Classical echanics, third edition (Addison-Wesley, 00)
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