Lagrangian Dynamics
Dr. Subhash Kumar and Dr. Sukanta Deb Department of Physics, Acharya Narendra Dev College (University of Delhi) Govindpuri, Kalkaji, New Delhi 110019
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Learning Objectives: • From a modern perspective, the Newtonian mechanics is unsatisfactory on various levels - it is cumbersome and complex. It is diifucult to deal with problems which involve extended objects rather than point particles. Moreover it obscures certain salient features of the dynamics. As a result certain concepts such as chaos theory took over 200 years to unfold after the formulation of Newtonian mechanics. • The purpose of this unit is to resolve these issues by presenting a new perspectives on Newton’s ideas. It includes description of the advancements that took place during the 150 years after Newton when the laws of motion were reformulated using more powerful techniques and ideas developed by some of the giants of Mathematical Physics viz., Euler, Lagrange, Hamilton and Jacobi. This provides a practical advantage and allows us to solve certain complicated problems with relative ease. • Introduction to the concepts of generalised coordinates, degree of freedom, constraints and generalised forces. • Classification of the dynamical system at several levels - basis of constraints and nature of potentials • Introduction to Lagrangian function and the Lagrangian equation of motion. • Solution to some problems using the Lagrangian equation for conservative systems.
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Introduction The subject of motion of material bodies have formed the work of some of the earlier scientists and philosophers in the field of physical science. These efforts have culminated into the development of a very vast and rich field which is now known as "Classical Dynamics" or "Classical Mechanics. Before we attempt to develop a basic framework of the classical mechanics it is necessary to define certain terms which are critical to formulate the basic contours of this branch of mathematics/physics.
Some Definitions • Dynamical System - A system of particles is known as dynamical system. • Configuration - The set of positions of all the particles is known as the configuration of the dynamical system. • Constraints - If a particle moves in space, it requires three independent coordinates to specify its location in space. Then, we say that the particle has 3 degree of freedom. On the other hand, if a particle moves on a plane surface, say a table-top, its motion is confined on a plane only. This implies that the particle now requires only two independent coordinates to specify its position on the table-top. When the particle is not allowed to move freely in three-dimension, we say that it is subject to constraints. it is clear that when constraints exists, the number of degree of freedom is reduced. For example, in case of the particle moving on a table-top, one constraints exists which can be expressed as z=0 for all times. In general, when a particle is constrained to move on a surface S : f (x, y, z) = 0, the coordinates are not independent to each other as the coordinate z can be expressed in terms of x and y. In turn this implies that the particle possess only two degree of freedom, instead of three. • Generalised Coordinates - Suppose that a particle or a system of N particles move subject to possible constraints, then there will be necessarily a minimum number of independent coordinates needed to specify the motion. These coordinates denoted by q1 , q2 , . . . , qn are called generalised coordinates and they may be distances, angles or quantities relating to them. Following are some examples of the generalised coordinates: (i) A simple pendulum of length l is a dynamical system. The angular displacement θ from the vertical is a generalised coordinate of this dynamical system. (ii) A particle on the surface of a sphere; in this case the polar coordinates θ and φ can describe the system and hence can be considered as its generalised coordinates. (iii) A lamina lying on a plane; generalised coordinates are x, y and θ. In this case (x, y) is the coordinate of the centroid and θ is the angle made by a line fixed in the plane. (iv) A rod lying on a plane surface; generalised coordinates are x, y and θ, where (x, y) are the coordinate of one end of the rod and θ is the angle between x-axis and the rod. We note here that the generalised coordinates need not necessarily have the dimension of length1 . The differential coefficients of qα (α = 1, 2, . . . , n) with respect to time ’t’ are termed as generalised velocities and are denoted by q˙α . • Degrees of freedom - The number of coordinates required to specify the position of the system of one or more particles is called the number of degrees of freedom of the system. The following examples specify the degrees of freedom for various dynamical systems. 1 Later,
we shall define generalised force , which again need not have the dimesion of forces, but the two are such that the scalar product of generalised coordinates and generalised force has the dimension of force.
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(a) A particle moving freely in space requires 3 coordinates, e.g. (x, y, z) to specify its position. Thus the number of degrees of freedom is 3. (b) A system consisting of N particles moving freely in space requires 3N coordinates to specify the position. Thus, number of degrees of freedom is 3N . (c) A rigid body which can move freely in space has 6 degree of freedom i.e., 6 coordinates are required to specify the position. Let 3 non collinear points of a rigid body be fixed in space, then the rigid body is also fixed in space. Let these points have coordinates (x1 , y1 , z1 ), (x2 , y2 , z2 ), (x3 , y3 , z3 ) respectively, a total of 9 since the body is rigid we must have (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 =constant (x2 − x3 )2 + (y2 − y3 )2 + (z2 − z3 )2 =constant (x1 − x3 )2 + (y1 − y3 )2 + (z1 − z3 )2 =constant Hence, 3 coordinates can be expressed in terms of the remaining six. Thus, six independent coordinates are needed to describe the motion i.e. there exists six degrees of freedom [1, 2, 3].
Transformation Equations Let rν = xν i+yν j+zν k be the position vector of ν th particle with respect to the cartesian coordinate system. The relationship of the generalised coordinates q1 , q2 , . . . , qn to the position coordinates are given by the transformation equations. xν = xν (q1 , q2 , . . . , qn ; t) yν = yν (q1 , q2 , . . . , qn ; t) zν = zν (q1 , q2 , . . . , qn ; t) where t denotes time. In vector notation, the above equations can be written as rν = rν (q1 , q2 , . . . , qn ; t).
(1)
The functions in the above equations are continuous and have continuous derivatives [3].
Classification of Dynamical System Holonomic and Non Holonomic System A dynamical system is called holonomic if it is possible to give arbitrary and independent variations to the generalised coordinates without violating constraints, otherwise it is called non holonomic. Let q1 , q2 , . . . , qn be n generalised coordinates of a dynamical system. Then for a holonomic system, we can change qr to qr + δqr without making any changes in the remaining n − 1 generalised coordinates [3, 4]. For example, in the case of rigid body, the constraint can be expressed as (ri − rj )2 = c2ij , where ri and rj are the position vectors of the ith and jth particle respectively and cij is a constant vector. This is clearly a holonomic constraint. Similarly, if the conditions of constraints can be expressed as equations connecting the coordinates of the particles (and the
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time) as f (r1 , r2 , . . . , t) = 0, then the constraints are said to be holonomic. A particle constrained to move along a curve or on a surface is another example of holonomic constraint. Constraints not expressible in this way are called non holonomic. For example, consider a particle placed on the surface of the sphere. The constraint on the particle can be expressed as r2 −a2 ≥ 0, where a is the radius of the sphere. Such constraints are termed as non holonomic.
Scleronomic and Rheonomic System Constraints are further classified according as they are independent of time or contain time explicitly. While the former are known as scleronomic, the later are called as rheonomous constraints. In other words, a scleronomic system is one which has ’fixed’ constraints, where as rheonomic system has ’moving’ constraints.
Conservative and non conservative system If the forces acting on the system are derivable from a potential function (or potential energy) V , then the system is called conservative otherwise it is non conservative.
Generalised Force in Holonomic System Consider a dynamical system having N particles of mass mi (i = 1, 2, . . . , N ) having position vectors ri with respect to the origin O. We suppose that the particles undergo a virtual displacement2. Let the i-th particle of mass mi at position ri at time t undergoes a virtual displacement to position ri + δri . Suppose Fi and ′ ′ Fi are the external and internal forces acting on the i-th particle, then the virtual work done on the i-th particle is (Fi +Fi )·δri . Hence, the total work done on all the particles of the system is δW =
N X
′
(Fi + Fi ) · δri
i=1
=
N X
(Fi · δri ) +
i=1
N X
′
(Fi · δri ),
(2)
i=1
PN ′ where i=1 (Fi · δri ) is the total work done by the internal forces of the system, which we consider to be zero. Then from the above equation N N X X (xi δxi + yi δyi + zi δzi ), (3) (Fi · δri ) = δW = i=1
i=1
where F = (xi , yi , zi ) and δri = (δxi , δyi , δzi ). Let us suppose the system is holonomic and the generalised coordinate are (q1 , q2 , . . . , qn ) then ri = ri (q1 , q2 , . . . , qn , t) Therefore,
n
δri =
X ∂ri ∂ri ∂ri ∂ri δq1 + δq2 + · · · + δqn = δqk . ∂q1 ∂q2 ∂qn ∂qk
(4)
k=1
2 A virtual displacement is an arbitrary, instantaneous, infinitesimal displacement of a dynamical system independent of time and consistent with the constraints of the system. The work done due to virtual displacement is known as virtual work.
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Substituting Equation (4) into Equation (3), we have ! n X ∂ri Fi · δW = δqk ∂qk i=1 k=1 ! n N X X ∂ri δqk Fi · = ∂qk i=1 N X
(5)
k=1
=
n X
Qk δqk ,
k=1
where Qk =
N X
Fi ·
i=1
∂ri ∂qk
(6)
The coefficient Q1 , Q2 , . . . , Qn are called components of generalised force associated with the coordinates q1 , q2 , . . . , qn respectively. It is called generalised force because Qk is the coefficient of δqk in Equation (6). Further, if the system is conservative then, if V be the potential energy of the system in any configuration space, we have δW = −δV In this case, V = V (q1 , q2 , . . . , qn ). Thus we have, n X
k=1
Qk δqk = −
n X ∂V δqk ∂qk
(7)
k=1
Since the dynamical system is holonomic, dq1 , dq2 , . . . , dqn are independent and arbitrary. Thus from the above equation, we get ∂V Qk = − , (k = 1, 2, . . . , n). (8) ∂qk
Lagrange’s Equations for a Holonomic System The term Lagrangian was named after the Italian Enlightenment Era mathematician and astronomer Joseph-Louis Lagrange (1736 − 1813). In order to simplify formulas and ease calculations, Lagrange reformulated the Newtonian approach of problem solving in classical mechanics, which is referred to as the Lagrangian mechanics. He made remarkable contributions in the field of mathematical analysis, number theory, classical and celestial mechanics. His treatise on "Analytical Mechanics" offers one of the most comprehensive treatment of classical mechanics since Newton and forms a basis for the development of mathematical physics in the nineteenth century. Numerous concepts in mathematics and physics emanated from his scientific ideas and thoughts are named after him. For an instance, Euler-Lagrange equation, Lagrange bracket, Lagrange interpolation, Lagrange multiplier, Lagrange points are to name a few. (Source: Wikipedia, Photo Courtesy: Wikipedia).
Joseph-Louis Lagrange
Let q1 , q2 , . . . , qn be n generalised coordinates of a holonomic dynamical system with n degrees of freedom. Suppose, the number of particles in the system be N and mi be the mass of a typical particle and ri be its position vector at time t with respect to origin, so that ri = ri (q1 , q2 , . . . , qn , t), (i = 1, 2, . . . , N ) (9)
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therefore, r˙i =
dri ∂ri dq1 ∂ri dq2 ∂ri dqn ∂ri = + + ···+ + dt ∂q1 dt ∂q2 dt ∂qn dt ∂t ∂ri ∂ri ∂ri ∂ri q˙1 + q˙2 + · · · + q˙n + = . ∂q1 ∂q2 ∂qn ∂t
(10)
Differentiating both sides of Eq. (11) partially with respect to q˙k , we obtain ∂ri ∂ r˙i = , ∂ q˙k ∂qk Also
(k = 1, 2, . . . , n)
(11)
n
X ∂ 2 ri ∂ 2 ri ∂ r˙i = q˙k + ∂qj ∂qj ∂qk ∂qj ∂t
(12)
k=1
Again d dt
∂ri ∂qj
=
d dt
∂ri ∂qj
=
n X ∂ 2 ri ∂ 2 ri q˙k + ∂qk ∂qj ∂qj ∂t
(13)
k=1
Comparing Eqs. (12) and (13), we get
∂ r˙ i , (j = 1, 2, . . . , n) ∂qj
(14)
The kinetic energy T of the system at time t is given by N
T =
N
1X 1X mi r˙ 2i = mi r˙i · r˙i 2 i=1 2 i=1
therefore, from Eq.(11) we obtain N
and
N
X X ∂ r˙i ∂ri ∂T mi r˙i · mi r˙i · = = ∂ q˙k ∂ q˙k ∂ q˙k i=1 i=1 N
X ∂ r˙i ∂T mi r˙i · = ∂ q˙k ∂qk i=1
From Eqs. (14) and (15), we get
N
(15)
N
X d ∂T ∂ri X ∂ r˙i )= mi r¨i · + mi r˙i · ( dt ∂ q˙k ∂qk i=1 ∂qk i=1 =
N X
mi r¨i ·
i=1
∂T ∂ri + ∂qk ∂qk
Thus, we obtain N
X ∂ri ∂T d ∂T mi r¨i · )− = ( dt ∂ q˙k ∂qk ∂qk i=1
(16)
′
Let Fi be the applied force on the ith particle of mass mi and Fi be the force of constraint on it. Let it undergo a virtual displacement δri (i = 1, 2, . . . , N ), consistent with the constraints on the system, at time t. Then by d’Alembert’s principal3, we have N X ′ [(Fi + Fi ) − mi¨ ri ] · δri = 0 i=1
3 D’Alembert’s principal: A fundamental result in analytical mechanics. It was introduced in 1708 by Jacques Bernoulli to understand static equilibrium, P r) · δrk = 0 and further developed by D’Alembert in 1743 to solve dynamical problems [3]. The principle asserts for N particles N k=1 (Nk + Ck − mk ¨ where Nk and Ck represents the non-constraint and constraint force respectively on the kth particle
7 N X
(Fi − mi¨ ri ) · δri +
N X
′
Fi · δri = 0
(17)
i=1
i=1
Restricting ourselves to systems for which the virtual work of the forces of constraints vanishes. Hence the second term in the above equation vanishes. We obtain, N X (Fi − mi¨ ri ) · δri = 0 (18) i=1
Now
! n X ∂ri Fi · Fi · δri = δqk ∂qk i=1 i=1 k=1 ! n N X X ∂ri δqk Fi · = ∂qk i=1 N X
N X
k=1
=
n X
Qk δqk
(19)
k=1
where Qk =
N X
∂ri , ∂qk
Fi ·
i=1
(k = 1, 2, . . . , n)
are the components of generalised force associated with the coordinates qk . Next from Equation (16), we get ! n N N X X X ∂ri mi¨ ri · mi¨ ri · δri = δqk ∂qk i=1 i=1 k=1 ! n N X X ∂ri δqk mi¨ ri · = ∂qk k=1 i=1 n X d ∂T ∂T = − δqk dt ∂ q˙k ∂qk
(20)
k=1
Substituting Eqns. (19) and (20) in Equation (18), we obtain n X d ∂T ∂T − − Qk δqk = 0 dt ∂ q˙k ∂qk
(21)
k=1
Since the system is holonomic, δq1 , δq1 , . . . , δqn are independent and arbitrary, so it follows from the above equation that d ∂T ∂T − = Qk (22) dt ∂ q˙k ∂qk The above equation [Eq. (22)] is called Lagrange’s equation for a holonomic system. If the system be conservative then there exists a potential energy V = V (q1 , q1 , . . . , qn ), such that Qk = −
∂V , ∂qk
(k = 1, 2, . . . , n)
Then Eq. (22) gives d dt
∂T ∂ q˙k
−
∂T ∂V =− , ∂qk ∂qk
(23)
where k = 1, 2, . . . , n. We now define a scalar function L, the Lagrangian as L=T −V
(24)
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where V is independent of q˙1 , q˙2 , . . . , q˙n . Therefore ∂L ∂T = ∂ q˙k ∂ q˙k Also
∂L ∂T ∂V ∂(T − V ) = − = ∂qk ∂qk ∂qk ∂qk
Then Eq. (23) becomes d dt
∂L ∂ q˙k
=
∂L , ∂qk
(k = 1, 2, . . . , n).
(25)
These are Lagrange’s equation for a conservative, holonomic dynamical system. Lagrange’s equation can be put in the form of Eq. (25) even if the system is not conservative, provided the generalised forces are obtained from a function U = U (qj , q˙j ) if d ∂U ∂U (26) + Qj = − ∂qj dt ∂ q˙j In such cases, Eq. (25) still follows from Eq. (23) with the Lagrangian given by L=T −V U is called a ’generalised potential’ or velocity-dependent potential.
Example 1. Find the equation of motion for the one dimensional harmonic oscillator. Solution 1. For an one-dimensional harmonic oscillator, the kinetic (T ) and the potential energies (V ) are given by 1 T = mx˙ 2 2 1 V = kx2 , 2 where m denotes the mass of the particle and k represents the spring constant. The Lagrangian of the system is given by L =T − V 1 1 ⇒ L = mx˙ 2 − kx2 2 2 ∂L ∂L = − kx, = mx. ˙ ⇒ ∂x ∂ x˙ We know that the Lagrange’s equation for the given system is given by ∂L d ∂L − =0. dt ∂ x˙ ∂x d ⇒ (mx) ˙ + kx =0 dt ⇒ m¨ x + kx =0 This is the required equation of motion for the one-dimensional oscillator. Example 2. Find the equation of motion of a simple pendulum.
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Solution 2. For the case of a simple pedulum, the kinetic energy (T ) is given by 1 T = mv 2 2 2 1 ⇒ T = m(lθ˙ ) 2 1 ⇒ T = ml2 θ˙2 , 2 where m is the mass of the bob of the pendulum and l is the length of the string. The potential energy for the system is given by V =mgl(1 − cos θ). Therefore, the Lagrangian (L) of the simple pendulum is given by L =T − V 1 ⇒ L = ml2 θ˙2 − mgl(1 − cos θ) 2 ∂L ∂L ˙ = − mgl sin θ, ⇒ = ml2 θ. ∂θ ∂ θ˙ The Lagrange’s equation for the system is given by d dt ⇒
∂L ∂ θ˙
−
∂L =0. ∂θ
d ˙ + mgl sin θ =0 (ml2 θ) dt ⇒ ml2 θ¨ + mgl sin θ =0 g ⇒ θ¨ + sin θ =0 l
If the amplitude of the motion of taken to be small enough, i.e sin θ ≈ θ. Then the above equation of motion becomes g θ¨ + θ =0 l ¨ ⇒ θ + ω 2 θ =0, pg where ω = l is the angular velocity of the bob. The time period of the oscillation is given by P = Equation (27) gives the required equation of motion for the simple pendulum.
(27) 2π ω
= 2π
q
l g.
Example 3. A bead slides on a smooth rod which is rotating about one end in a vertical plane with uniform angular velocity velocity ω. Show that the equation of the system is given by r¨ = ω 2 r − g sin ωt. Solution 3. Let us consider the problem in plane polar coordinates (r, θ), where x = r cos θ and y = r sin θ. The kinetic energy of the bead is given by 1 T = m x˙ 2 + y˙ 2 2 1 2 ⇒ T = m r˙ + r2 θ˙2 2
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On the other hand, the potential energy of the bead is given by V =mgr sin θ. The Lagrangian (L) of the system is given by
⇒
L =T − V 1 L = m r˙ 2 + r2 θ˙2 − mgr sin θ 2
∂L ∂L = mg r, ˙ = −mgr cos θ + mrθ˙2 . ∂ r˙ ∂r
Substituting these values into the Lagrange’s equation, we have ∂L d ∂L − =0 dt ∂ r˙ ∂r ⇒ m¨ r − mrθ˙2 + mg sin θ =0 ⇒ r¨ =θ˙2 r − g sin θ. Example 4. Find the equation of motion of a compound pendulum.
Figure 2: A compound pendulum oscillating in a vertical plane through the point A above a fixed horizontal axis.
Solution 4. A rigid body capable of oscillating in a vertical plane above a fixed horizontal axis is called a compound pendulum. Let the vertical axis of the oscillation be xy. Let A be the point through which the axis of rotation passes. Let the point G denote the center of gravity. The distance between the point A and the centre of gravity (G) is called the length of the pendulum. Let the mas of the pendulum be m and the moment of inertia about the axis of rotation be I. Let the pendulum be given a samml angular displacement and then released to oscillate about the point A. Let the centre of gravity at the new position be G′ when the angular displacement is θ. The kinetic energy of the pendulum is given by T =
1 ˙2 Iθ . 2
The potential energy relative to the horizontal axis through A is V = −mgl cos θ.
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The Lagrangian of the system is give by L =T − V 1 ⇒ L = I θ˙2 + −mgl cos θ 2 ∂L ˙ ∂L = −mgl sin θ. =I θ, ⇒ ∂θ ∂ θ˙ Substituting these values into the Lagrange’s equation, we have d ∂L ∂L =0 − ˙ dt ∂ θ ∂θ ⇒ I θ¨ + mgl sin θ =0 mgl sin θ =0 I ⇒ θ¨ + ω 2 θ =0 (If the amplitude of oscillation is small sin θ ≈ θ),
⇒ θ¨ +
where r mgl mgl ⇒ω= . ω = I I 2
The time period of the pendulum is given by P = ⇒ T = 2π
s
2π ω
I . mgl
Example 5. A bead of mass m slides without friction along a wire having the shape of a parabola given by y = Ax2 . with the vertical axis in the earth’s gravitational field g (As shown in Fig. (3)). (i) Find the Lagrangian of the system taking the horizontal displacement as generalized coordinate. (ii) Write down the Lagrange’s equation of motion. Solution 5.
(i) The kinetic energy of the bead is given by 1 T = m(x˙ 2 + y˙ 2 ) 2 1 ⇒ T = m(1 + 4A2 x2 )x˙ 2 2
The potential energy is given by V =mgy ⇒ V =mgAx2 Therefore, the Lagrangian is given by L=
1 m(1 + 4A2 x2 )x˙ 2 − mgAx2 . 2
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Figure 3: A bead of mass m slides without friction along a wire having the shape of a parabola given by y = Ax2 with the vertical axis in the earth’s gravitational field g.
(ii) From the Lagrangian, we have ∂L = m(1 + 4A2 )x, ˙ ∂ x˙
∂L 1 = m(8A2 x)x˙ 2 − 2mgAx. ∂x 2
Therefore, the Lagrange’s equation of motion becomes d dt
∂L ∂ x˙
−
∂L =0 ∂x
1 ⇒ m(1 + 4A2 x2 )¨ x + m(8A2 x)x˙ 2 − m(8A2 x)x˙ 2 + 2mgax =0 2 ⇒ m(1 + 4A2 x2 )¨ x + m(4A2 x)x˙ 2 + 2mgAx =0 Example 6. Fig (4) shows the the sketch of Atwood’s machine. It consists of two two masses m1 and m2 which are suspended by an inextensible string over a frictionless mass pulley. Find the Lagrangian and hence obtain the equation of motion for the system.
Figure 4: Atwood’s machine
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Solution 6. The kinetic energy of the system is given by T =
1 1 m1 x˙1 2 + m2 x˙2 2 . 2 2
The potential energy of the system is V = −m1 gx1 − m2 gx2 , where V is taken to be zero at the center of the pulley. The system is subjected to the constraint x1 + x2 = l = constant. Therefore, the Lagragian of the system is given by L =T − V 1 1 ⇒ L = m1 x˙1 2 + m2 x˙2 2 + m1 gx1 + m2 gx2 2 2 Using x2 = l − x1 , we have the Lagrangian in terms of a single variable given by 1 1 L = m1 x˙1 2 + m2 x˙1 2 + m1 gx1 + m2 g(l − x1 ) 2 2 1 ⇒ L = (m1 + m2 )x˙1 2 + (m1 − m2 )gx1 + m2 gl. 2 Therefore, the Lagrange’s equation of motion for the system is given by ∂L d ∂L − =0 dt ∂ x˙1 ∂x1 d ⇒ (m1 + m2 )x˙1 − (m1 − m2 )g =0 dt m1 − m2 g. ⇒ x¨1 = m1 + m2 This is the required equation of motion for the system. Example 7. Find the Lagrange’s equation of motion for a particle moving under the influence of a central force field. Solution 7. We know that a particle moving under a central force field is a conservative force and the motion takes place in a plane. We will use plane polar coordinates (r, θ) to describe the motion of the particle. Let m be the mass of the particle. The kinetic energy of the particle is 1 T = m(x˙ 2 + y˙ 2 ) 2 1 ⇒ T = m(r˙ 2 + r2 θ˙2 ). 2 The potential energy of the particle is V (r). Hence the Lagrangian of the system is given by L =T − V 1 ⇒ L = m(r˙ 2 + r2 θ˙2 ) − V (r) 2 ∂V ∂L ∂L =mrθ˙2 − , = mr˙ ⇒ ∂r ∂r ∂ r˙ And, ∂L ∂L = mr2 θ˙ =0, ∂θ ∂ θ˙
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Therefore, the Lagrange’s equations of motion for the system are given by d ∂L ∂L d ∂L ∂L − =0, =0 − ˙ dt ∂ r˙ ∂r dt ∂ θ ∂θ ∂V d ˙ =0 ⇒ m¨ r − mrθ˙2 + =0, (mr2 θ) ∂r dt Example 8. A particle of mass m is constrained to move on a frictionless cylinder of radius R given by the equation ρ = R in cylindrical polar coordinates ρ, φ, z as shown in Fig. (5). The only force acting on the particle is F = −kr and directed towards the origin. Find the Lagrangian of the particle using z and φ as generalized coordinates and write down the equations of motion.
Figure 5: A particle of mass m confined to the surface of the cylinder ρ = R and subject to a Hook’s law force F = −kr.
Solution 8. Since the coordinate ρ of the particle is fixed at ρ = R, where R is the radius of the cylinder, the position of the particle can be described with two degrees of freedom specified by z and φ. Therefore, (z, φ) can be treated as generalized coordinates. The three components of the velocity are vρ = 0, vφ = Rφ˙ and vz = z. ˙ Therefore, the kinetic energy is
The potential energy is
1 T = mv 2 2 1 ⇒ T = m vρ2 + vφ2 + vz2 2 1 ⇒ T = m R2 φ˙ 2 + z˙ 2 2 1 ⇒ V = kr2 , 2
where r is the distance of the particle from the origin, given by r 2 = R2 + z 2 . Therefore, the potential energy becomes V =
1 k(R2 + z 2 ). 2
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Therefore, the Lagrangian of the system is given by L =T − V 1 1 ⇒ L = m R2 φ˙ 2 + z˙ 2 − k(R2 + z 2 ). 2 2
Since the system has two degrees of freedom (φ, z), there are two equations of motion corresponding to them. The Lagrange’s equation of motion for the z-coordinate is given by ∂L d ∂L − =0 dt ∂ z˙ ∂z ⇒ m¨ z = −kz The Lagrange’s equation of motion for the φ coordinate is given by ∂L d ∂L − =0 dt ∂ φ˙ ∂φ d ˙ − 0 =0 ⇒ (mR2 φ) dt ⇒ mR2 φ˙ =constant. Example 9. A particle of mass m is lying on a smooth plane. If the plane is raised to an inclination angle θ at a constant rate α (Given θ = 0 at t = 0) such that the particle moves down the plane. Find the resulting equation of motion for the particle (As shown in Fig. (6)).
Figure 6: The plane on which a particle lies is raised to an inclination angle θ at a constant rate α.
Solution 9. Let the bottom of the plane be taken as the origin. The kinetic energy is 1 T = m r˙ 2 + r2 θ˙2 − mgr sin θ 2 1 ⇒ T = m r˙ 2 + α2 r2 − mgr sin αt (∵ θ = αt; θ˙ = α). 2 Therefore, the Lagranage equation of motion becomes ∂L d ∂L − =0 dt ∂ r˙ ∂r
⇒ r¨ − α2 r = − g sin αt.
Example 10. Find the equation of motion of a simple pendulum of length b and a bob with mass m attached to a massless support moving horizontally with a constant acceleration a (As shown in Fig. (7)). Solution 10. The (x, y) components are given by 1 x = at2 − b sin θ 2 y = − b cos θ
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Figure 7: A pendulum of length b and a bob with mass m moving horizontally with a constant acceleration a.
Therefore, x˙ =at − bθ˙ cos θ y˙ =bθ sin θ The kinetic energy is
The potential energy is given by
1 T = m(x˙ 2 + y˙ 2 ) 2 1 2 2 ⇒ T = m a t − 2atbθ˙ cos θ + b2 θ˙2 2 V =mgy ⇒ V = − mgb cos θ
Hence, the Lagrangian of the given system is given by L =T − V 1 ⇒ L = m a2 t2 − 2atbθ˙ cos θ + b2 θ˙2 + mgb cos θ. 2 Therefore, the Lagrange’s equation of motion becomes d ∂L ∂L =0 − dt ∂ θ˙ ∂θ d ⇒ −matb cos θ + mb2 θ˙ + mb2 θ˙ =matbθ¨ sin θ − mgb sin θ dt a g ⇒ θ¨ + sin θ − cos θ = 0 b b
Summary • Classical Mechanics presents a new formalism of the Newtonion mechanics, which is more elegant and practical. It is more suited to the formalism of modern physics. • Dynamical systems may be classified on the nature of constraints and nature of potentials. • Minimum number of generalised coordinate required to describe the system is the degree of freedom of the system. • The concept of force, displacement and work is revisited and their definitions expanded into the concept of generalised force, virtual displacement and virtual work respectively. • The heart of Lagrangian mechanics is the Lagrangian function L = T − V , where T and V are the kinetic energy and potential energy of the system respectively. • In Lagrangian mechanics, the trajectory of a system of particles is obtained by solving the Lagrange equations of motion.
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Exercises 1. Obtain the Lagrangian and hence its equation of motion for a particle of mass m dropped from rest in a uniform gravitational field. The field is along the vertical axis which is taken as the z-axis. 2. Write down the Lagrangian for a projectile subjected to no air resistance in term of Cartesian coordinates (x, y, z), where z-axis is directed vertically upwards. Find the three Lagrange’s equations of motion. 3. Find the Lagrangian for a one-dimensional particle moving along the x axis and subjected to a force F = −kx, where k > 0. Hence obtain the Lagrange’s equation of motion for the system. 4. Let us consider a particle of mass m moving in two dimension having potential energy U (x, y) = 12 kr2 , where r = p x2 + y 2 . Find the Lagrangian using the two coordinates x and y and hence obtain the two Lagrange’s equations of motion. 5. Find the equation of motion of a simple pendulum of length b and a bob with mass m moving vertically upward with a constant acceleration a. 6. Find the equation of motion of a simple pendulum of length b and a bob with mass m moving vertically downward with a constant acceleration a. 7. Find the equation of motion of a simple pendulum of mass m suspended by a massless spring of unextended length b and spring constant k. 8. If the Lagrangian of a system is given by
1 −kt 2 e x˙ − p2 x2 . 2 Show that the Lagrange’s equation of motion is given by L=
x ¨ + k x˙ + p2 x = 0. 9. Find the equation of motions for a projectile in two dimensions. 10. A particle of mass m is attracted by a central force which varies inversely as the cube of its distance from the center. Obtain the Lagrange’s equation of motion for the system.
References [1] L.D. Landau and E.M. Lifshitz. Mechanics. Butterworth Heinemann. Butterworth-Heinemann, 1976. [2] S.T. Thornton and J.B. Marion. Classical Dynamics of Particles and Systems. Brooks/Cole, 2004. [3] H. Goldstein, C.P. Poole, and J.L. Safko. Classical Mechanics. Addison Wesley, 2002. [4] J.M. Finn. Classical Mechanics. Engineering series. Jones and Bartlett Publishers, 2009.
