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Chapter VII Hamilton's Principle- Lagrangian & Hamiltonian Mechanics

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Chapter VII Hamilton's Principle- Lagrangian & Hamiltonian Mechanics Chapter VII Hamilton’s Principle- Lagrangian & Hamiltonian Mechanics Recommended problems: 7-1, 7-2, 7-3, 7-4, 7-6, 7-7, 7-10, 7-12, 7-13, 7-14, 7-15, 7-25, 7-26, 7-27, 7-29, 7-33, 7-34, 7-37, 7-39, . Hamilton’s Principle Of all the possible paths along which a dynamical system may move from one point to another within a specific time interval (consistent with any constraint), the actual path followed is that which minimizes the time interval of the difference between the kinetic and potential energies, i.e., t2 T U dt 0 (7.1) t1 This means that the integral of T-U must be an extremum. Defining the difference of T-U to be the Lagrangian as Lxi , xi;t Txi Uxi (7.2) Eq.(7.1) now reads t2 Lxi , xi;tdt 0 (7.3) t1 Comparing Eq.(7.3) and Eq.(.3) with f L, x t y x y x We get the Euler’s equations as L d L 0 (7.4) Lagrange Equation x dt x Example Find the equation of motion of the 1-Dimensional harmonic oscillator. Solution The Lagrangian of the system is L T U 1 mx2 1 kx2 2 2 L L kx & mx x x d k kx mx 0 kx mx 0 x x 0 dt m Example Find the equation of motion of the simple pendulum shown. Solution With respect to the top, the potential energy of the ball is U mgLcos Lcos And for the kinetic energy we have L T 1 mv2 1 mL22 2 2 L 1 mL22 mgLcos 2 m L L mgLsin & mL2 Using Lagrange equation we get d mgLsin mL2 0 mgLsin mL2 0 dt g sin 0 L Generalized Coordinates Generalized coordinates are any set of independent coordinates qi (not connected by any equations of constraint) that completely specifies the state of a system. The required number of generalized coordinates is equal to the system’s number of degrees of freedom. A single particle free to move in 3-dimensional space requires 3-coordinates to specify configuration and hence has a 3-degrees of freedom. n-free particles would require 3n coordinates and so on. For each constraint equation the number of generalized coordinated is decreased by one coordinate. This means that if there are m equations of constraints, then 3n-m coordinates are independent, and the system is said to posses s=3n-m degrees of freedom. The equations of constraints must be expressible of the form fk xi ,t 0 i 1,2,n k 1,2,m (6.5) Constraints that can be expressed in the form of Eq.(6.5) is called holonomic constraint, otherwise it is nonholonomic. Since the Lagrangian is a scalar function it is invariant to coordinates transformation. In terms of the generalized coordinates, the Lagrange’s equations can be written as L d L 0 j 1,2,s (6.6) Lagrange Equations q j dt q j There are s of these equations, and together with the m equations of constraints and the initial conditions, they completely describe the motion of the system. It is important to realize that the validity of Lagrange’s equations requires the following 2-conditions: 1- The forces acting on the system (apart from any constraint forces) must be conservative. 1- The Equations of constraints must be in the form of Eq.(6.5), i.e., the constraint must be holonomic.. Example Find a suitable set of generalized coordinates for a point particle moving on the surface of a hemisphere of radius R whose center is at the origin. Solution The particle is constrained to move on the surface, so we have x2 y2 z2 R2 x2 y2 z2 R2 0 If we choose the Cartesian coordinates our generalized coordinates will be 2 2 2 q1 x, q2 y z R x y Example Use the (x,y) coordinates system as shown in the figure to find the K.E. T, P.E. U, and the Lagrangian L for a simple pendulum moving in the x-y plane. Determine the transformation equations from the (x,y) coordinates to the coordinates . Find the equations of motion. Solution Using the rectangular coordinates we have T 1 m x2 y 2 U mgy 2 L T U 1 m x2 y 2 mgy 2 To transformation x&y into the coordinates we have x lsin & y l cos x lcos & y lsin L 1 m x2 y 2 mgy 1 ml22 mgl cos 2 2 To find the equation of motion we apply Lagrange’s equation (Eq.7.4). We have 2 L L dml mglsin ml2 mglsin 0 dt g sin 0 l y Example Consider the problem of a projectile motion under gravity in 2- dimensions. Find the equations of motion in both Cartesian and polar vo coordinates. x Solution Using the Cartesian coordinates we have, taking U=0 at y=0, T 1 m x2 y 2 U mgy 2 L T U 1 m x2 y 2 mgy 2 Here we have 2-generalized coordinates, x & y. So Lagrange’e equations give L d L L d L 0 & 0 x dt x y dt y d For the x-coordinate we have 0 mx 0 x 0 dt d mg my 0 y g dt In polar coordinates, the generalized coordinates are r & . Now we have T 1 m r2 r22 U mgrsin 2 L 1 m r2 r22 mgrsin 2 Lagrange’e equations give for the 2-generalized coordinates, r & gives L d L For the r-coordinate we have 0 r dt r d mr2 mg sin mr 0 r2 g sin r 0 dt L d L For the -coordinate we have 0 dt d mgr cos mr2 0 gr cos 2rr r2 0 dt Example A particle of mass m is constrained to move on the inside surface of a smooth cone of half-angle . Determine the generalized coordinates and the constraints. Find the equation of motions. Solution here we use the cylindrical coordinates r, , & z. The equation of constraint is z r cot So we 2-degrees of freedom and the generalized coordinates are r & . Now T 1 m r2 r22 z2 1 m r2 r22 r2 cot2 1 m r2 csc2 r22 2 2 2 U mgz mgrcot L 1 m r2 csc2 r22 mgr cot 2 L d L For the r-coordinate we have L d L 0 r dt r 0 dt d mr2 mg cot mrcsc2 0 r2 g cot rcsc2 0 dt For the -coordinate we have d 0 mr2 0 mr2 constant dt But L I mr2 constant So we recover the conservation of angular momentum about the axis of symmetry of the system. Example The point of support of a simple pendulum of length b moves on a massless rim of radius a rotating with constant angular velocity . Obtain the equation of motion of m. Solution Taking the origin of the coordinates be at the center of the rotating rim we get x acost bsin y asint bcos x asint bcos y a cost bsin The K.E & P.E are now T 1 m x2 y 2 1 m a22 b22 2basin t 2 2 U mgy mgasint bcos L 1 m a22 b22 2basin t mg asint bcos 2 The only generalized coordinate is . SO L mbacos t mgbsin L mb2 basin t d L 2 mb mba cos t dt Applying the Lagrange’e equation we get mb2 mba cos t mbacos t mgbsin mb2 mba2 cos t mgbsin a2 g cos t sin b b Note that for =0 we get g sin 0 b Which is the equation of motion of the simple pendulum. Example Find the frequency of a simple a pendulum placed in a rail-road car that has a constant acceleration a in the x-direction. l Solution We choose a fixed coordinate l system with x=0, and v=vo at t=0. The position of the mass m is m m x l sin v t 1 at 2 o 2 y l cos x l cos vo at y lsin The K.E & P.E are now 2 2 T 1 m x2 y 2 1 m v at lcos 1 m lsin 2 2 o 2 U mgy mgl cos L 1 m v at lcos 2 1 m lsin 2 mgl cos 2 o 2 The only generalized coordinate is again . SO L mlsinv at lcos ml22 sin cos mglsin o L mlsinv at mglsin o g a sin cos cos sin cos cos sin sin l l l l l l Applying the Lagrange’s equation L d L g a 0 sin cos (prove as HW ) l l dt If the mass doesn’t oscillate with respect to the train, then 0 g a a 0 sinl cosl tan l l l g Because the oscillations are small we can assume l is very small g a sin cos l l l l Expand the sin and cosine functions we get L ml cosv at lcos ml2sin2 o Using Taylor expansion for sin and cos to first order of we get g a sin cos cos sin l l l l l l 1 g sin acos g cos asin l l l l l The first bracket is already zero leaving g cos asin l l l a a g As tanl sin & cos g a2 g 2 a2 g 2 g 2 a2 g 2 a2 l g 2 a2 l g 2 a2 0 l g 2 a2 We have SHM with frequency 2 l Note that when the car is at rest (a=0) we get 2 g l Example A bead slides along a smooth wire bent in the shape of a parabola z=cr2.
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