Principles of Dynamics. Tom Charnock

Transcription

1 Principles of Dynamics Tom Charnock

2 Contents 1 Coordinates, Vectors and Matrices Cartesian Coordinates Polar Coordinates Cylindrical Polar Coordinates Spherical Polar Coordinates Matrices Identity Matrix Levi Civita Tensor Products Of Vectors Scalar (Dot) Product Vector (Cross) Product Product of Products Position, Velocity and Acceleration 6.1 Position Velocity Acceleration Operators Gradient Operator Gradient (Grad) Divergence (Div) Curl Classical Mechanics Orbits Kepler s Law of Planetary Motion Galileo s Principle of Inertia Isaac Newton s Principia Mathematica Newton s First Law Newton s Second Law Newton s Third Law Law of Universal Gravitation From Newton to Kepler Action Lagrangian Calculus of Variation Euler-Lagrange Equations Many Particle Systems 15 7 Hamilton s Principle Physical Lagrangian Euler-Lagrange Equations

3 8 Rigid Bodies Euler s Theorem Angular Momentum Moment of Inertia Tensor Principal Axis Perpendicular Axis Theorem Parallel Axis Theorem Kinetic Energy of a Rigid Body Gravitational Potential of a Rigid Body Compound Pendulum Chasles Theorem Double Pendulum 10 Constraints Newton s Constraints Holonomic Constraints Non-Holonomic Constraints Symmetries Constant Generalised Momentum Constant Pseudo Energy Hamilton s Equations The Hamiltonian Gyroscope Lagrangian Approach Euler s Angles Angular Velocity Hamiltonian Approach General Motion of the Gyroscope Stability of a Vertical Top

4 Chapter 1 Coordinates, Vectors and Matrices 1.1 Cartesian Coordinates Cartesian coordinates are given by a set of vectors which can be written as: (x, y, z) = ṽ = In index notation this can be written as: v 1 v v 3 v = = v 1 î + v ĵ + v 3ˆk 3 iv ê i i=1 x, y, z can be replaced with x 1, x, x 3 so that a generalisation can be made to arbitrary dimensions of index notation. 1. Polar Coordinates The symmetry of a system can often be solved more easily using a different coordinate system such focuses on the space around an object Cylindrical Polar Coordinates This coordinate system is useful for calculating fields around a line such as a wire and is denoted by (ϱ, θ, z) which gives the vector: v = v ϱ ˆϱ + v θ ˆθ + vz ẑ 1.. Spherical Polar Coordinates This coordinate system is useful for calculating fields around a point, such as an atom and us denoted by (r, θ, ϕ), so that the vector is: v = v rˆr + v θ ˆθ + vϕ ˆϕ 1.3 Matrices A matrix can be written in terms of index notation as M ij so that a matrix looks like: M = M 11 M 1 M 13 M 1 M M 3 M 31 M 3 M 33 A vector can be multiplied by a matrix to obtain another vector where the sum is written in index notation as: 3 u i = M ij v j j=1 4

5 1.3.1 Identity Matrix The identity matrix is the matrix which when multiplied gives by a vector gives the same vector I = This can be given by the Kronecker Delta δ ij where δ ij = 0 when i j and δ ij = 1 when i = j Levi Civita Tensor The Levi Civita Tensor can be used to indicate whether a matrix has cyclic permutations, anti-cyclic permutations or non-cyclic permutations. It is indicated by: ε 13 = ε 31 = ε 31 = 1 ε 13 = ε 13 = ε 31 = 1 Any other permutation of the matrix is given by ε ijk = Products Of Vectors Scalar (Dot) Product ũ ṽ = ũ ṽ cos θ Where ũ ṽ is the norm of the vectors and θ is the angle between the vector. Using this it can be shown that orthogonality is defined by two unit vectors pointing in the same direction: ũ ṽ = 1 This is because cos θ = 1 when θ is 0. In index notation this can be shown by: ê i ê j = δ ij So that when i = j then δ ij = 1 and when i j then δ ij = Vector (Cross) Product ũ ṽ = ũ ṽ sin θˆn Where ˆn is the unit vector orthogonal to both ũ and ṽ. The vector product only works in three dimensions. If ũ and ṽ are parallel then sin θ = 0 so that ũ ṽ = 0. This can be written as: ê i ê j = If ω = ũ ṽ then the components are given by: ω i = ijk 3 ε ijk ê k k=1 ε ijk u j v k Product of Products = u v 3 v u 3 u 3 v 1 v 1 u 3 ω u 1 v v 1 u 1 Some identities of the products of products are such that: ũ (ũ ũ) = 0 a c ) = (ã b ) = ã) (b c b (c a c ) = (ã (ã (b c )b b )c 5

6 Chapter Position, Velocity and Acceleration.1 Position Position can be given by a vector: (t) r = x 1 (t) x (t) = x 3 (t) 3 x i (t)ê i Distance is not the same as position because distance is a scalar quantity given by r(t) = r (t) = x1 (t) + x (t) + x 3 (t). i=1. Velocity Velocity is the rate of change of position and can be denoted: δr (t) ẋ 1 (t) 3 v = = ṙ (t) = ẋ (t) = ẋ i (t)ê i (t) δt ẋ 3 (t) i=1 Where ê is a basis vector fixed with time. The speed is a scalar quantity like the distance, which is related to the velocity by v(t) = ṽ(t) = ṙ where ṙ ṙ in general. ṙ = ẋ 1 + ẋ + ẋ 3 ṙ = x 1ẋ 1 + x ẋ + x 3 ẋ 3 x 1 + x + x 3 In circular motion the radius is given by the scalar value r, which remains constant and ṙ = 0. This does not imply that the speed has vanished because ṙ 0..3 Acceleration The acceleration is the rate of change of velocity or the rate of change of the rate of change of position. It is denoted by: a(t) = δṽ(t) = δ r (t) ẍ 1 (t) 3 δt δt = = ẍ (t) = ẍ i (t)ê i r ẍ 3 (t) i=1 The scalar acceleration is: r = ẍ 1 + ẍ + ẍ 3 6

7 Chapter 3 Operators Vector Calculus allows the rate of change in space to be calculated. 3.1 Gradient Operator The gradient operator is given by which in index notation can be written as where the latter is shortened terminology. 3. Gradient (Grad) For scalar valued functions of space f(x 1, x, x 3 ) the gradient (grad) is defined as: f x 1 f 3 f f = = ê i x i x f x 3 The operator maps the scalar value function into a vector function. 3.3 Divergence (Div) can also act on vector functions. This is called the divergence or div. i=1 F = F 1 x 1 + F x + F 3 x 3 = Divergence maps the vector quantity into a scalar function. 3.4 Curl The curl maps a vector to a vector such that: For any scalar function f = 0. F = 3 i=1 3 i=1 ( ) F k ε ijk ê i x j F i x i x i or even just x i, 7

8 Chapter 4 Classical Mechanics 4.1 Orbits Kepler s Law of Planetary Motion Kepler s First Law Planets move in elliptical orbits with the sun at one of the foci. Kepler s Second Law A line drawn between the sun and planet sweeps out equal areas in equal times. Figure 4.1: Orbit Around a Foci with Equal Areas in Equal Times Kepler s Third Law The square of the period is proportional to the cube of the radius of the semi-major axis: T r 3 This comes from: So that: v = πr Gm T = r ( ) 4π T = r 3 Gm 4. Galileo s Principle of Inertia Galileo stated that the masses of falling bodies were independent from the uniform acceleration of the body when resistance was negligible. This means that a body moving on a level surface would remain travelling in the same direction with the same constant speed unless it were disturbed. Although this was discovered there was no real theory or mathematics to prove it. 4.3 Isaac Newton s Principia Mathematica Isaac Newton developed three laws of motion and a law of universal gravitation which proposed the theory from which Kepler s laws can be derived. 8

9 4.3.1 Newton s First Law An object at rest will remain at rest or an object moving at constant speed will remain at constant speed if the net force is zero. This is the same as Galileo s principle of inertia Newton s Second Law The force applied on a body is equal to the rate of change of momentum. = F δp = ṗ δt When the mass remains constant then the velocity is equal to p written as: Newton s Third Law F = m δṽ δt = mã Every force occurs as one member of an action/reaction pair of forces Law of Universal Gravitation = mṽ and so the second law can be Newton s law of universal gravitation states that two masses can exert a force on each other at a distance of r. = F GMm ( ˆr ) Where the direction of the force is along the line joining the two masses. 4.4 From Newton to Kepler r Newton s laws can be uses to find Kepler s laws. Doing this uncovers the conservation of energy and the conservation of angular momentum. As = ˆr r r then the equation of motion from Newton s second law and law of universal gravitation can be written as: r = GM r 3 r By using scalar multiplication by ṙ on both sides of the equation then this becomes: This is because = r ṙ 1 δr δt and = r ṙ 1 1 ṙ = GM r δṙ δt. The equation found can be rearranged to obtain: 1 ṙ GM r Where E is the total energy of the system and is constant due to δe δt = 0. This is the conservation of energy where the two parts are the kinetic and potential energy of the system. Now by using vector multiplication on the original equation of motion by on both sides then it is found r that because = r r δ(ṙ r ) δt due to = 0 and because = 0 then it can be seen that: ṙ ṙ r r δ(ṙ r ) δt = E = δh δt = 0 This is the conservation of angular momentum. Without loss of generality, can point along the z axis due it being constant so that: h 0 = 0 h h 9

10 With this claim the motion of the planet s position is now confined to the x-y plane. x r = y 0 y ˆθ ˆr r θ x Figure 4.: Planar Polar Coordinates Now converting this into planar polar coordinates then x = r cos θ and y = r sin θ This makes: cos θ ˆr = sin θ, ˆθ sin θ = cos θ 0 0 The unit vectors ˆr and ˆθ are perpendicular so that ˆr ˆr = 0. The velocity can be calculated so that: ẋ ṙ cos θ r θ sin θ ṙ = ẏ = ṙ sin θ + r θ cos θ = ṙˆr + r θˆθ 0 0 Writing the angular momentum h = r ṙ in polar coordinates gives: h = (rˆr) (ṙˆr + r θˆθ) As rṙˆr ˆr = 0 and r θˆr ˆθ = 1 in the z direction then it can be seen that h = r θ, which is the angular momentum around z. The conservation of energy can also be written in polar coordinates in the form: ṙ = (ṙˆr + r θˆθ) (ṙˆr + r θˆθ) This expands to: ṙ = ṙ ˆr ˆr + rṙ θˆr ˆθ + r θ ˆθ ˆθ And because the middle of this is orthogonal then it is equal to zero. This leaves ṙ = ṙ + r θ which means the conservation of energy in polar form is: E = 1 (ṙ + r θ ) GM r The polar forms of the conservation of momentum and energy can be used to derive Kepler s Laws of Planetary Motion. As ṙ θ = δr δθ and θ = h r then: E = 1 ( ) [ ( ) ] h δr r + r GM δθ r Substituting in u = 1 r then: E = 1 (hu )6 [ ( 1 ) ] δu u + 1 δθ u GMu 10

11 This can be rearranged to get: ( ) ( δu = u GM ) δθ h + E ( GM h6 + h ) Now taking another substitution of ω = u GM h then it is seen that: δω E δθ = h + GM h ω In general, when integrating a k th variable then a trigonometric function should be used, so making E ω = αθ cos β gives where α = h + ( ) GM : h δω δθ = α sin β δβ δθ This can be placed into the conservation of energy equation in polar coordinates to give: ( ) δβ α sin β = α sin β δθ This means that β = ±(θ θ o ) where θ o is the constant of integration. When p = h the position can be written as: p r = 1 + ε cos(θ θ o ) GM and ε = αp then For a planet to reach there must be an angle θ where 1 + ε cos(θ θ o ) = 0. For ε < 1 there is no solution and so the system is bound and the planet is in orbit, so the energy is E < 0. If ε > 1 then there is a value of θ and the planet is unbounded, which is the case when the energy is E > 0. Kepler s First Law By rearranging the equation for position and placing in the values of x and y from polar coordinates gives: x + y = p εpx + ε x Which can be rearranged to obtain: (1 ε ) p This has the general form of an ellipse: This therefore has shown Kepler s First Law. Kepler s Second Law ( x + εp ) 1 ε + p 1 ε y = 1 (x + c) a + y b = 1 The second law can been found from using the rate of change of area. A small change in area is given by δa = 1 r δθ for small values of θ. The rate of change of the area is given by A = 1 r θ and using θ = h r the change in area in a given time is shown to be a constant of A = h. Kepler s Third Law Finally Kepler s Third Law can be found by using the total area of an ellipse: A tot = T 0 δa δt δt = h T The equation for the area of an ellipse is A tot = πab where b = a 1 ε. This leaves: T a 3 = 4π a(1 ε ) h 11

12 And using the values of ε, a and p stated earlier this leaves: Which is Kepler s Third Law. T = 4π GM a3 1

13 Chapter 5 Action If a path is denoted by x (t) then after a short time the path may change from x (t) to x (t + δt). 5.1 Lagrangian A Lagrangian is generally a function of L(t, x (t), ẋ (t)) and it can be used to calculated the action using: S = t t 1 Where the boundary conditions come from x (t 1 ) = x 1 and x (t ) = x. Nature always selects the path where the functional S[x] is minimised. This can be found using calculus of variation. 5. Calculus of Variation For F (t) the minima or maxima occurs at a time t = t and is given by F (t ) = 0. This definition of a derivative at a maxima or minima is: Lδt F (t + ε) F (t ) lim = 0 ε 0 ε When functionals, such as the action, are involved then this becomes: lim ε 0 S[x + εf(t)] S[x ] = 0 ε The boundary conditions on this are that x (t 1 ) = x 1, x (t ) = x and f(t 1 ) = f(t ) = 0. This can then be used to calculate the Euler-Lagrange equations of a system. 5.3 Euler-Lagrange Equations If the general Lagrangian is given by L(t, x i (t), ẋ i (t)) then the action is given by: S[x i ] = t t 1 L (t, x i, ẋ i ) δt If an arbitrary vector valued function f i (t) is introduced where the function can be minimised by f i (t 1 ) = f i (t ) = 0 then the Lagrangian can be minimised. S[x i + εf i ] = t Using Taylor Expansion on the Lagrangian gives: t 1 L(t, x i + εf i, ẋ i + ε f i )δt L(t, x i + εf 1, ẋ i + εf i ) = L(t, x i, ẋ i ) + εf i + ε x f i + O(ε ) i ẋ i 13

14 Now ignoring the orders of ε and using integration by parts on f i x i And because [ ] t t S[x i + εf i ] = S[x i ] + εf i + ε ẋ i t 1 t 1 [ ] t εf i ẋ i = 0 then finally minimising this gives: t 1 0 = t t 1 i=1 D i=1 D ( f i δ ) δt x i δt ẋ i And since f i (t) is arbitrary up to the boundary then for all i: δ = 0 x i δt ẋ i then the action becomes: δ f i f i x i δt δt ẋ i 14

15 x (3) (t) x () (t) x (1) (t) XA(t) Chapter 6 Many Particle Systems In a system with N particles in ( D dimensions the ) k th particle has the position x (k) i (t). This system is described by the Lagrangian L t, x (k) i (t), ẋ (k) i (t). Figure 6.1: Systems with Many Particles, or One Particle in Many Dimensions Many paths can be described as a single path in DN dimensions: x (1) 1. x (1) D x () (t) = X 1. x () Ḍ. x (N) D In index ( notation this ) is X A (t) where A goes from 1 DN. The Lagrangian in terms of DN dimensions is L t, X A (t), Ẋ A (t). From this the Euler-Lagrange equations are: x (k) i δ δt ẋ (k) i = 0 15

16 Chapter 7 Hamilton s Principle For a system described by a set of generalised coordinates q A, the correct path of motion q A (t) between an initial state q A (t o ) at time t o and the final state q A (t 1 ) at time t 1, corresponds to a stationary path of the action S = t 1 t o Lδt where L = L(t, q A, q A ) is the Lagrangian describing the system. 7.1 Physical Lagrangian The physical Lagrangian is in the form of L = T V where T is the kinetic energy and V is the potential energy. 7. Euler-Lagrange Equations The stationary path is given by the solution of the Euler-Lagrange equations: δ = 0 q A δt q A 16

17 Chapter 8 Rigid Bodies Rigid bodies are many particle systems which are held together by rigid bonds. The sum over the particles becomes an integral over the volume and the mass becomes the mass density at a position x i, which is given by ϱ(x i )δv. volume 8.1 Euler s Theorem ˆn e r r e (t + t) r e (t) O Figure 8.1: Rigid Body The general displacement of a rigid body with one fixed point, O, is a rotation about some axis, through O. The point is displaced by δr = r (t + δt) r (t) where δr is perpendicular to ˆñ. δr is also perpendicular to r (t) when the movement is infinitesimal rotation. This means that δr is parallel to ˆñ r (t). This means that: δr = ˆñ r sin θδϕ = (ˆñ r )δϕ When the rate of change of an infinitesimally small amount of time is taken into account then: ṙ = (ˆñ r ) ϕ = ( ϕˆñ) r This leads to the angular velocity ω = ϕˆñ so that ṙ = ω r. 8. Angular Momentum A rigid body as a system of particles of mass m k and position r (k) i the k th particle then the linear momentum is p (k) i 17 = m k ṙ (k) i relative to the fixed point. Considering (t) and the angular momentum is h (k) i =

18 [ ( r (k) i p (k) i = m k r (k) i ω (k) i )] r (k) i. Using vector product identities this can be shown to be: h (k) i = m k [ r (k) i ω (k) i (r (k) i ω (k) i ] )r (k) i The total angular momentum of the system can be found by summing over all the particles, and so for a rigid body the total angular momentum is given by: h i = ϱ(r i ) [ r i ] ω i (r i ω i )r i δv body This leads to h i = Iω i where I is the moment of inertia tensor. The total angular momentum of the system is: h i = ϱ(r i ) [ r ] δ ij r i r j ωj δv j So the moment of inertia tensor is: I ij = body body 8.3 Moment of Inertia Tensor ϱ(r i ) [ r δ ij r i r j ] δv In Cartesian coordinates the moment of inertia tensor is given by: [ ϱ y z ] δv ϱ [xy] δv ϱ [xz] δv I = [ ϱ [xy] δv ϱ x + z ] δv ϱ [yz] δv ϱ [xz] δv [ ϱ [yz] δv ϱ x + y ] δv The moment of inertia about an axis along the ˆñ direction is given by: I nn = ij I ij ˆn iˆn j Using the moment of inertia tensor the moment of inertia bout the x axis is selected from the matrix I xx = ϱ(y + z )δv. The moment of inertia tensor is both real and symmetric, which means that the axis can be chosen so that I is diagonal Principal Axis To find the principal axes the eigenvectors of the moment of inertia tensor must be computed. corresponding eigenvalues are the values of the principal moment of inertia. The 8.3. Perpendicular Axis Theorem The moment of inertia of a lamina about an axis perpendicular to the lamina is the sum of the moment of inertia about two perpendicular axes in the plane of the lamina. Figure 8.: Lamina in the x-y plane 18

19 ϱy δv ϱxyδv 0 I = ϱxyδv ϱx δv ϱ(x + y )δv So the moments of inertia in the different axes are I xx = ϱy δv and I yy = ϱx δv and I zz is given by: Parallel Axis Theorem I zz = I xx + I yy The moment of inertia about a parallel axis is I = I C M + Md where d is the perpendicular distance to the parallel axis. Figure 8.3: A Rigid Body Rotating about a Parallel Axis If the inertia of the centre of mass is about I xx and is shifted from y y d then: ϱ(y + z )δv ϱ((y d) + z )δv This gives the moment inertia as: I = I CM d ϱyδv + Md As ϱyδv is the y component of the centre of mass then is goes to zero if at the origin. It is also trivial that the integral of the density of the volume is the same as the total mass of the rigid body. This therefore leaves: I = I CM + Md 8.4 Kinetic Energy of a Rigid Body Treating a rigid body as a system of N particles with the k th particle having mass m k and position x (k) it is seen that: T l = 1 m (k) k ẋ = 1 ( ) m (k) kẋ (k) ω x Where ẋ (k) = ω x (k). Using scalar identities it can be seen that: T k = 1 [ ] ω x (k) (k) m k = ẋ 1 h k ω 19

20 And summing over these gives the total kinetic energy. T = 1 1 ω h Because h = Iω This shows that the total kinetic energy is the total angular momentum of the body. T = 1 I ij ω i ω j 8.5 Gravitational Potential of a Rigid Body ij For a single particle then the Gravitation Potential Energy of the particle (in the field of Earth) is V = mg x, where g = gẑ. Considering a system of N particles, then the Gravitational Potential Energy of the particle is V k = m k g (k). Using Newton s Third Law then it is known hat if one particle exerts a force F on another particle x then the other particle exerts a force of F on the first particle. This means that for a rigid body internal forces cancel out leaving the total Gravitational Potential Energy as: V = V k = g (k) m + k kx k The definition of the centre of mass is states that: Mx CM = k (k) m k x This means that for a rigid body the potential energy due to gravity of the body can be idealised as a particle with its entire mass located at the centre of mass x CM. 8.6 Compound Pendulum V = Mg x CM Figure 8.4: Compound Pendulum A compound pendulum is a shaped lamina which can swing freely about a point O. The moment of inertia about the axis perpendicular to O is given by I and the centre of mass of the pendulum is a fixed distance l from the origin. θ(t) is the angle made to the vertical at a time t. To find the Lagrangian of the system then the Kinetic Energy and the Potential Energy need to be found. The Kinetic Energy of a compound pendulum comes from the angular velocity, θ about the pivot point. Therefore in this case the Kinetic Energy is T = 1 I θ. The Potential Energy of a system with mass m is given by V = g x CM = mgl cos θ, This means that the Lagrangian is: L = T V = 1 I θ + mgl cos θ 0

21 And using the Euler-Lagrange equations it can be seen that: mgl sin θ θ = I This states that there are equilibrium points at both θ(t) = 0 and θ(t) = π. These points correspond to when the pendulum is below and above the vertical axis. Slightly perturbing the system will show if the points are stable or not. Using the small angle approximation of sin θ = θ the equation of motion becomes: θ = mglθ I This is the equation of motion for a simple harmonic oscillator with a frequency of ω = mgl I and the solution is θ(t) = A cos ωt + B sin ωt. The fact that this is oscillating about the point θ(t) = 0 indicates that the system is stable. For θ(t) = π then small fluctuations give θ = π + δθ so: This gives a phase change to δ θ = mgl I δ θ = mgl I sin (π + δθ) sin δθ. This means that the solution can be written as: δθ = Ae ωt + Be ωt This is an exponential growth and shows that the system is unstable. 8.7 Chasles Theorem So far it has been assumed that the pivot has been fixed, which is Euler s Theorem. The corollary of this is Chasles Theorem. It states that the most general displacement of a rigid body is a translation plus a rotation. Figure 8.5: Translation and Rotation of a Rigid Body Since the centre of mass is always fixed in the rigid body the motion of the body relative to the centre of mass can only be a rotation. A system of N particles with the k th particle having a mass of m k and position x (k) (t). The centre of mass position is x CM = 1 M k mkx (k) (t). As each particle has a position of S (k) = x (k) x CM then relative to the centre of mass k m ks (k) = 0. This can be used to calculate the total kinetic energy of the system. T = 1 (k) m k ẋ k This gives a equation with two parts. this first part is the Kinetic Energy of the total mass at the centre of mass and the second is the total Kinetic Energy of the particles relative to the centre of mass. T = m k Ṡ (k) 1 + m k ẋ CM k k 1

22 Chapter 9 Double Pendulum (t) l 1,m 1 (t) l,m Figure 9.1: Double Pendulum Two rods are uniform and can swing freely at a fixed point O. The Kinetic Energy is made up of T = T 1 + T. The first rod has a moment of inertia of I = 1 3 m 1l 1 and an angular velocity of θ and using T = 1 I θ then the Kinetic Energy of the first rod is: T 1 = 1 6 m 1l 1 θ The second rod is not attached to a fixed point and so its kinetic energy is given by T = T CM + T rel. T CM has a velocity of v and is given by T CM = 1 m v. The components of v can be found from the values of x CM and y CM. x CM = l 1 sin θ + 1 l sin ϕ and y CM = l 1 cos θ + 1 l cos ϕ. This therefore gives: T CM = 1 m (l 1 θ + 14 ) l ϕ + l 1 l θ ϕ cos(θ ϕ) T rel has an angular velocity ϕ with a moment of inertia I = 1 1 m l. T rel = 1 4 m l θ This gives the total kinetic energy of the system as: T = 1 6 m 1l1 θ + 1 (l 1 θ + 13 ) l ϕ + l 1 l θ ϕ cos(θ ϕ) The potential energy of the system due to gravity for the two systems is V = V 1 + V and as each rod has a potential energy of V 1 = 1 m 1gl 1 cos θ and V = m g(l 1 cos θ + 1 cos ϕ) so the Lagrangian of the system is: L = 1 6 (m 1 + 3m )l 1 θ m l ϕ + 1 m l 1 l θ ϕ cos(θ ϕ) + 1 (m 1 + m )gl 1 cos θ + 1 m gl cos ϕ

23 This can be solved by the Euler-Lagrange equations: This gives the equation for θ: And the equation for ϕ: θ δ δt θ = 0 ϕ δ δt ϕ = 0 0 = 1 m 1 l 1 l θ ϕ sin(θ ϕ) (m 1 + m )gl 1 sin θ (m 1 + 3m )l1 θ 1 ( m l 1 l ϕ cos(θ ϕ) ( θ ) ϕ ϕ ) sin(θ ϕ) 0 = 1 m 1 l 1 l θ ϕ sin(θ ϕ) m gl ϕ m l ϕ 1 m l 1 l ( θ cos(θ ϕ) ( θ θ ) ϕ) sin(θ ϕ) At the equilibrium points where θ(t) = 0 and ϕ(t) = 0, small fluctuations can be used to see if the system is stable or not. In this case θ and ϕ are small so that cos θ 1 θ + O(θ4 ), cos ϕ 1 ϕ + O(ϕ4 ) and cos(θ ϕ) 1 (θ ϕ) + O ( (θ ϕ) 4). By neglecting any terms beyond the quadratic order makes the Lagrangian: L 1 6 (m 1 +3m )l 1 θ m l ϕ + 1 m l θ ϕ 1 4 (m 1 +m )gl 1 θ 1 4 m gl ϕ + 1 (m 1 +m )gl m gl The last two parts of this equation are constant and so fall out of the Euler-Lagrange equations. The Euler-Lagrange equations can then be written in matrix form as: ( 1 3 (m 1 + 3m )l1 1 m ) ( ) ( l 1 l θ 1 1 m 1 l 1 l 3 m l + (m ) ( ) 1 + m )gl 1 0 θ 1 ϕ 0 m = 0 gl ϕ This is in the form k ψ + mpsi = 0. As the equations are coupled they are difficult to solve. The normal modes are combinations of θ and ϕ that oscillate with a well defined frequency. these frequencies are the normal frequencies which can be found if the above equation is written in the form ψ + Aψ = 0 where A = K 1 M. This can be done because det K 0 and the inverse is well defined. Two vectors can be made ũ 1 and ũ which are linearly independent eigenvectors of A with eigenvalues λ 1 and λ. Aũ 1 = λ 1 ũ 1, Aũ = λ 1 ũ If ψ = Θ(t)ũ 1 + Φ(t)ũ where Θ and Φ are the linear combinations of θ and Φ then: Θ(t) = Φ(t) = ( ũ ũ1 ( ũ 1 ũ ) ũ ) ψ ũ1 ũ ( ũ1 ũ ) ( ũ1 ũ ( ũ 1 ũ ) ũ1 ) ψ ũ1 ũ ( ũ1 ũ ) This can be found by dotting both sides of ψ = Θũ 1 + Φũ with ũ 1 and ũ and solving the two equations simultaneously. As these are linearly independent then: Θ = λ 1 Θ, Φ = λ Φ These are the normal modes of oscillation with corresponding normal frequencies ω Θ and ω Φ. Θ and Φ satisfy simple harmonic motion with a frequency of ω Θ = λ 1 and ω Φ = λ respectively. Finding the eigenvalues and eigenvectors of A gives the values ω Θ and ω Φ. Mũ = λkũ 3

24 Chapter 10 Constraints Some dynamical systems involve constraints. There are two ways of deal with constraints. This first uses Newton s Laws and the second uses the Lagrangian Newton s Constraints l m Figure 10.1: Mass on a Fixed Rope For a simple pendulum with mass m at the end of a rope with a fixed length l the constraint is defined by the length of the rope. It can be seen that the constraint for this system is x + y = l. By solving in the x and y directions the equations of motion can be found. Horizontally mẍ = T x l and vertically mÿ = mg T y l. By setting x = l sin θ and y = l cos θ then this constraint is immediately imposed such that ẋ = l θ cos θ, ẏ = l θ sin θ, ẍ = l θ cos θ l θ sin θ and ÿ = l θ sin θ l θ cos θ. The constraint on the tension is then T = ml θ + mg cos θ and θ = g l sin θ. Using vector multiplication with this and θ it can be seen that θ = g l cos θ + c and so the tension is given by T = 3mg cos θ + mlc. 10. Holonomic Constraints For a dynamical system where the position is at x A (t) then the holonomic constraint can be written F (x A, t) = 0. If there are N n constraints then F α (x A, t) = 0 where α = n + 1,..., N. This can be incorporated into a Lagrangian by introducing N n new variables, λ n. These are the Lagrange multipliers. If an unconstrained Lagrangian is L(t, x A, ẋ A ) then the constrained Lagrangian is: The Euler-Lagrange equations for λ β are: L = L(t, x A, ẋ A ) + N α=n+1 δ = 0 λ β δt λ β As the second part goes to zero then it can be seen that: λ α F α (x A, t) λ α λ β = δ αβ 4

25 And so the constraint is recovered for λ β. The Euler-Lagrange equations become: δ + x A δt ẋ A N α=n+1 λ α F α x A = 0 In the case of the simple pendulum then the Lagrange multiplier is λ = T l. This shows that with the Lagrange multiplier approach the nature of the constraint forces do not matter Non-Holonomic Constraints If the constraint is velocity dependent so that F (x a, ẋ a, t) then this can be absorbed into the Lagrangian formalism, and so can be solved. An example of a velocity dependent constraint is particles travelling into a magnetic field. 5

26 Chapter 11 Symmetries Symmetries are often connected to the action principle. The theories connected the symmetries of a system with physically conserved quantities such as conserved momentum and conserved pseudo-energy Constant Generalised Momentum If L(t, q A, q A ) is such that a particular generalised coordinate q j the Lagrangian does not depend upon it then: q j = 0 Now if the generalised momentum is p j = q A then the Euler-Lagrange equations imply that: δ = 0 δt q A This means that p j is constant. The coordinate, q j, is cyclic. If q j is a linear coordinate then it follows that p j is the linear momentum and so for an angular coordinate q j then p j is the angular momentum. 11. Constant Pseudo Energy The Lagrangian does not have to explicitly depend upon time so that L = L(q i, q i ). This means that = 0. In this case time is cyclic. This leads to: t For a general Lagrangian: E = N i=1 q q i L(q i q i ) = const δl δt = N t + q i + q i q i q i Now considering a function H = N i=1 q i q i L then the rate of change of this is: This leaves: Ḣ = N i=1 i=1 δ q i + q i q i δt H t = t δl q i δt When L is independent of time then H is constant and this defines the Hamiltonian. The Hamiltonian uses a Lagrangian with no explicit time dependence. This is related to the conservation of energy. It gives E = T + V which is the total energy of the system. This may not always be the case depending on the initial conditions. 6

27 11.3 Hamilton s Equations Hamilton s equations are an extension of the Lagrangian method. They use generalised momenta, p 1, p,..., p N, instead of generalised velocities q 1, q,..., q N, which is a useful technique when the generalised momenta are constants of motion. This makes the principle well suited to finding conserved quantities. The Euler-Lagrange equations are used to find the generalised momenta: p α = q α Then using N variables, every part of the system can be uniquely specified. Hamilton s approach is to solve p α = q α for q α (q, p). With the N new variables q and p serve equally well to specify the position and velocity of every particle The Hamiltonian Hamilton s First Equation As p β = q β H(q, p) = N p β q β (q, p) L(q, q(q, p)) β=1 H p α = q α + N β=1 then the last two terms cancel out and leave: Hamilton s Second Equation H = + q α q α pβ q β p α q β q β p α H p α = q α N β=1 p β q β q α q β q β q α Again the last two terms cancel and using q α = ṗ α then the second equation is: q α = ṗ α The Euler-Lagrange equations are a set of N nd order differential equations whereas Hamilton s equations are a set of N 1 st order differential equations. 7

28 Chapter 1 Gyroscope R! m Ω R cos ω m R θ The potential energy of the system is given by mgr cos θ and the kinetic energy can be calculated by finding the moment of inertia tensor. 1.1 Lagrangian Approach The principle axes are ê 1, ê and ê 3, which are orthonormal such that: I I ij = 0 I I 3 These axes are fixed to the rigid body and therefore must rotate with it. From this the angular velocity can be obtained so that ω = ω 1 ê 1 + ω ê + ω 3 ê 3. The angular momentum is then J = I ij ω = I 1 ω 1 ê 1 + I ω ê + I 3 ω 3 ê 3. This gives the kinetic energy as: T = 1 ( I1 ω 1 + I ω + I 3 ω 3) The coordinate system for this has not been set, and so although a generalised kinetic energy has been found, this cannot yet be used to find the dynamics of the system Euler s Angles Three angles are needed to specify the orientation of an object. Two of the angles are used to fix the ê 3 axis and then one is used to fix the rotation about ê 3. 8

29 k e (ê 3 e 0 ) j(ê 00 ) e e i e (ê 1 e 000 ) Figure 1.1: The axes are orientated along i, j and k. k e (ê 3 e 0 ) 0 ê e ' ' 00 ê 1 e Figure 1.: The axes are rotated by ϕ about k to give a still orthonormal set. ê 3 e 0 ê e 0 ê 1 e Figure 1.3: The axis are rotated by θ about the ê axis. ê 3 e ê e ê 1 e Figure 1.4: Finally rotate about the ê 3 axis by angle ψ Angular Velocity The first rotation in ϕ gives the angular velocity ϕˆk, the second rotation in θ gives the angular velocity thetaê and the final rotation in ψ gives the angular velocity ψê 3. This means that the angular velocity is: ω = ϕˆk + θê + ψê 3 This involves three different base axes and so an assumption can be made that, for a symmetrical gyroscope, I 1 = I. his means that rotating about ê 3 will remain in the principle axes. Any pair of orthonormal vectors can be used to define the ê 1 -ê plane. There is also a relationship between hatk and θ that can be seen from the rotation diagram. This is that ˆk = sin θê 1 + cos θê 3. This can be substituted into the angular velocity to give = ϕ sin θê ω 1 + θê + ( ψ + ϕ cos θ)ê 3. This means that the [ kinetic energy is T = 1 I 1 ϕ sin θ + θ ] ( + 1 I 3 ψ + ϕ cos θ) and so the Lagrangian is: L = 1 I 1 [ ψ sin θ + θ ] + 1 I 3 ( ψ + ϕ cos θ) mgr cos θ Both ϕ and ψ are cyclic coordinates and so the momenta are given by: ψ = ω 3 = ψ + ϕ cos θ ϕ = ω = I 1 ϕ sin θ + I 3 ω 3 In the special case when θ is constant then ϕ is constant and becomes Ω which is the precession frequency. 9

30 1. Hamiltonian Approach Using the Lagrangian for the gyroscope and the calculated values of the generalised moment gives: p ϕ = I 1 ϕ sin θ + I 3 ( ψ + ϕ cos θ) cos θ p θ = I 1 θ p ψ = I 3 ( ψ + ϕ cos θ) By rearranging the generalised momenta equation and substituting in the velocities into the Hamiltonian function then it is found to be: H = (p ϕ p ψ cos θ) I 1 sin θ + p θ I 1 + p ψ I 3 + MgR cos θ From this it can be seen that ψ and ϕ are cyclic such that p ψ and p ϕ are constant. This means that the system only has one degree of freedom, which is in the θ component. This gives the Hamiltonian finally in the form: Where U(θ) = (pϕ p ψ cos θ) I 1 sin θ + p ψ I 3 + MgR cos θ. H = p θ I 1 + U(θ) 1..1 General Motion of the Gyroscope Using Hamilton s equations then: θ = H p θ = p θ I 1 ṗ θ = H θ = U θ The Hamiltonian is independent of time and therefore H t = 0 and H = E is a constant. In the case where θ = 0 then p θ = 0 which means that U(θ) = E. When the motion is confined to the region U(θ) E the potential energy is a quadratic shape with a minimum at θ o. U( ) E 1 o Figure 1.5: Confined to the region U(θ) E It is clear that as U(θ), θ 0 or θ π. The minima occurs when δu δθ = 0. This minima corresponds to the steady precession. This is the equilibrium point, and so shows that the gyroscope will oscillate between θ = θ 1 and θ = θ. For the angular velocity about the vertical axis to vanish then cos θ vanish = pϕ p ψ. 30

31 1 Figure 1.6: θ 1 < θ vanish > θ so ϕ will never vanish and precession will continue to oscillate 1 Figure 1.7: θ 1 < cos 1 p ϕ p ψ < θ so ϕ 0 when θ θ vanish. This means that the precession is positive at the top of the loop, but negative near the bottom. Figure 1.8: θ 1 = cos 1 p ϕ p ψ = θ vanish which means the precession comes to rest at the top of each loop. 1.. Stability of a Vertical Top If the gyroscope is spinning with an angular velocity ω 3 with its axis vertical, so that θ = 0 then the potential energy U(0) must be finite. The only way this is possible is in the case that p ϕ = p ψ = I 3 ω 3. This is true because ψ = ω 3. The motion when θ = π also has a similar relation, but is always stable, and so, it is not as interesting a problem. Now using p ϕ = p ψ = I 3 ω 3, the potential energy is now: Where tan θ = 1 cos θ sin θ U(θ) = I 3 ω 3 I 1 tan θ + 1 I 3ω 3 + MgR cos θ. For small values of θ, tan θ and cos θ can be expanded to give: ( ) 1 U(θ) I 3ω3 + MgR + 1 ( I 3 ω ) 3 MgR θ 4I 1 This can be seen to stable when θ = 0. Also if the value of θ is positive then there is be a stable position, but for negative θ the equilibrium point is unstable. There is also a minimum value of ω 3 where the vertical top will be stable, and this is given by: ω 3 = 4MgRI 1 I 3 This shows that as long as ω 3 > ω o then the top will remain upright, but if it drops below ω o then the top will begin to wobble. If the vertical top has an energy given by E = p θ I 1 + U(θ) and θ = θ = 0 then: = ω o E = 1 I 3ω 3 + MgR 31

32 Using this then the condition the U(θ) = E is: I 3 ω 3 I 1 tan θ + 1 I 3ω 3 + MgR cos = 1 I 3ω 3 + MgR Which gives θ = θ 1 = 0 and cos θ = ω 3. If the top is set spinning slowly with ω ω0 3 < ω o and with its axis vertical then the gyroscope will oscillate between the vertical and θ. As θ increases, ω 4 decreases such that θ π when ω 3 0. When ω 3 = 0 then the top behaves like a compound pendulum and swings in a circle through the upward and downward verticals. 3