Problem 1:
(a) Give the definition of a cyclic variable.
(b) How is it related to conservation laws and physical symmetry?
(c) Give two examples of systems described by Lagrangians with cyclic variables
and relate them to part b.
Solution:
 | Concepts, principles, relations
that apply to the problem: The Lagrangian, the Hamiltonian, conservation laws, cyclic variables |
| Why do they apply? The problem asks for the definition of a cyclic variable and its relationship to conservation laws. |
| How do they apply? (a) The Lagrangian is a function of the generalized coordinates and velocities and time. Any generalized coordinate, which does not appear explicitly in the Lagrangian, is said to be cyclic. ¶L/¶qi = 0 implies qi = cyclic. (b) The corresponding generalized momentum pi = ¶L/¶ 
is a constant of motion, it is conserved. A coordinate which is cyclic in the Lagrangian is also cyclic in
the Hamiltonian. The connection between the symmetry properties of the
physical system and constants of motion can be derived in terms of the
Hamiltonian.Example: H is symmetric about an axis. H is invariant under rotation about this axis, it does not depend on the angle of rotation about this axis. The angle is a cyclic coordinate, the corresponding angular momentum is conserved. |
| Details of the calculation: (i) A particle in a central potential:  .Lf = mr2 df/dt is constant. (ii) A particle subject to a constant force F = F0k.  .x and y are cyclic, px = mdx/dt and py = m dy/dt are constants of motion. |
Problem 2:
A particle of mass m is attached (so it can not slide) to the midpoint
of a weightless rod of length l. The ends of the rod are constrained to
move along the x and y axes. A uniform gravitational field acts
in the negative y-direction. Use q as a
generalized coordinate. Neglect friction.
(a) Write the Lagrangian and obtain the equation of motion.
(b) Solve the equation of motion for small q,
i.e. |q| << 1, assuming that, at t = 0,
q = q0 and dq/dt
= 0.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations,  |
| Why do they apply? All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. We are asked to write down and solve Lagrange's equation of motion. |
| How do they apply? (a) As long as the rod remains in contact with the wall, we have the constraints x = (l/2)sinq and y = (l/2)cosq for the coordinates of m. The Lagrangian L is given by L = T - V.     We need only one generalized coordinate and one generalized velocity. Lagrange's equation yields (1/4)ml2d2q/dt2 = (mglsinq)/2. |
| Details of the calculation: (b) Let q << 1, then sinq @ q and d2q/dt2 @ (2g/l)q. The general solution to this differential equation is  .The initial conditions are
q(0) =
q0, dq/dt|t=0 = 0. This solution is only valid as long as q is still small. |
Problem 3:
Consider the pendulum illustrated in the figure with l the length of the string, m the mass of the ball, Fg the gravitational force, and f the angular displacement. (You may assume the string to be of fixed length and negligible mass). Use Lagrange’s equation to derive an equation of motion that neglects terms of order f3 and higher.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations, |
| Why do they apply? All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. We are asked to find the equation of motion. |
| How do they apply? T = (1/2) m l2(df/dt)2, U = -mglcos(f), L = T - U. Here Fg = mg. We have one generalized coordinate, f, and one generalized velocity, df/dt. |
| Details of the calculation: The equation of motion therefore is ml2df2/dt2 + mglsin(f) = 0, or df2/dt2 + (g/l)sin(f) = 0. sinf = f - f3/3! + ... Neglecting terms of order f3 and higher we have df2/dt2 + (g/l)f = 0. |
Problem 4:
For the Atwood Machine shown, all pulleys are smooth and non-rotating and all cords are inextensible. Work out the accelerations of 3m and 5m.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations, | |||||||||||||||||||||||||||||||||||
| Why do they apply? All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. | |||||||||||||||||||||||||||||||||||
| How do they apply? Let the y-axis point downward. Let l1 be the hanging portion of the length of the red string, l2 the hanging portion of the length of the blue string, and l3 the hanging portion of the length of the green string. Let us find the kinetic and potential energy of the masses in terms of the generalized coordinates q1, q2, and q3. We can then write down the Lagrangian and the equations of motion, which will yield the accelerations of the masses. | |||||||||||||||||||||||||||||||||||
| Details of the calculation:
T = (21m/2)(dq1/dt)2+2m(dq2/dt)2+5m(dq3/dt)2+2m(dq1/dt)(dq2/dt)+2m(dq1/dt)(dq3/dt).
|
Problem 5:
A particle of mass m is described by the Lagrangian function
,
where l3 is the z-component of the angular momentum and
w is a constant angular frequency.
(a) Find the
equations of motion, write them in terms of the variables (x + iy) and z and
solve them.
(b) Construct the Hamiltonian function and find the kinematic and
canonical momenta. Show that the particle has only kinetic energy and that the
latter is conserved.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations: ; Canonical momenta:
 ;The Hamiltonian function:  |
| Why do they apply? The Lagrangian of the system is given, we are asked to solve the equations of motion and to construct the Hamiltonian function. |
| How do they apply? The generalized coordinates qi are the Cartesian coordinates. The coordinate z is cyclic. |
| Details of the calculation: (a)  From Lagrange's equations we have:
Let c = x + iy. Then d2c/dt2 + iwdc/dt = 0. c = Bexp(-iwt) + C. B and C are determined by the initial conditions.
H does not explicitly depend on time. |
Problem 6:
In the figure above the mass point m rotates about the point s. At the same time the thread holding the mass point is shortened continuously. Find the constants of motion.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations: |
| Why do they apply? All forces except the forces of constraint are derivable from a potential. To find the constants of motion we find the Lagrangian and look for cyclic coordinates. If the generalized coordinates and momenta we choose do not depend on time and the Lagrangian depends explicitly on time, then the energy is not conserved. |
| How do they apply? Let x and y be the Cartesian coordinates in the plane of the circle and r and f be the polar coordinates. r = R0 - ct is equation of constraint. The constraint is holonomic and time dependent. x = (R0 – ct)cosf, y = (R0 – ct)sinf. T = (1/2)m(vx2 + vy2) = (1/2)m(  2(R0 –
ct)2 + c2).L = T, we have only one generalized coordinate, f, which does not explicitly depend on time. ¶L/¶f = 0, f is cyclic. ¶L/¶ = m(R0
- ct)2
= constant,
since
(d/dt)(¶L/¶) = 0.The angular momentum about the vertical axis is conserved. H = H(t) = T(t). H explicitly depends on time. The energy is not a constant of motion. |
| Details of the calculation: This problem should remind you that holonomic constraint can be time dependent. As long as all the applied forces (not the force of constraints) are potential forces, Lagrange’s equations hold. |
Problem 7:
In a problem with one degree of freedom, a particle of mass m is subject to a force F(x,t) = F0t. The force is derivable from a potential.
(a) Find the potential energy of the particle and the
Lagrangian and Hamiltonian of the particle.
(b) Solve Hamilton's equations of motion.
Solution:
| Concepts, principles, relations
that apply to the problem: The Lagrangian and the Hamiltonian, Hamilton's equations of motion |
| Why do they apply? We are asked to find the Lagrangian and the Hamiltonian and to solve Hamilton's equations of motion. |
| How do they apply? (a) F(x,t) = -¶U/¶x = F0t, U(x,t) = -F0tx. (We can arbitrarily choose the zero of the potential energy.) L = T - U = (1/2)m  2
+ F0tx.
¶L/¶
= px = m.H =  px
- L = Px2/2m - F0tx. |
| Details of the calculation: (b) =
¶H/¶px,
 = -¶H/¶x,
dx/dt = px/m, dpx/dt = F0t, d2x/dt2
= F0t/m.dx/dt = F0t2/(2m) + v0, x = F0t3/(6m) + v0t + x0. |
Problem 8:
Consider the system shown below, in which two bodies of the same mass M
are connected by a massless string across a massless frictionless pulley. The
coefficient of friction for the mass on the incline is m
and the angle of the incline is 30o with respect to the
horizontal. Define the z-coordinate as shown. Assume downward motion
and m < 1.
(a) Determine the Lagrangian L(z, dz/dt) for the system.
(b) If at t = 0, z = z0, and v = v0, find z(t)
for all t.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's equations,  ,where L contains the potential of the conservative forces and Qj
represents the generalized forces not arising from a potential. |
| Why do they apply? In this problem conservative and non-conservative forces (gravity and friction) are present. |
| How do they apply? (a) We only need one generalized coordinate. Assume downward motion. Then T = M(dz/dt)2, U = Mgz + Mgzsin(30o) + constant = 1.5 Mgz + constant. L = T - U. Qz = mMgcos(30o). (b) Lagrange's equation yields 2Md2z/dt2 + 1.5Mg = mMgcos(30o). d2z/dt2 = (mcos(30o) - 1.5)g/2 = g'. Since m < 1 g' is negative. dz/dt = -v0 - |g'|t, z(t) = z0 - v0t - (1/2)|g'|t2. until the block is stopped by the pulley. |
| Details of the calculation: None |
Problem 9:
A heavy particle with mass m is placed on top of a vertical hoop. Calculate the
reaction of the hoop on the particle by means of the Lagrange undetermined
multipliers and Lagrange's equations. Find the height at which the
particle falls off.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations, ,
Lagrange multipliers ,
 . |
| Why do they apply? The problem requires us to use the method of Lagrange multipliers.  Imagine the particle to be constrained to move on the hoop. For a small q, the force of constraint points away from the origin. For a large q it points toward the origin. The angle for which the force of constraint is zero is the angle at which the particle falls off. To find the force of constraint as a function of q, we use the technique of Lagrange multipliers. |
| How do they apply? We have two coordinates, r and q, and one equation of constraint, r = R, which must be cast into the form l1S a1k dqk = 0, l1(a1r dr + a1q dq) = 0. The equation of constraint is r = R, dr = 0, therefore a1r = 1, a1q = 0. |
| Details of the calculation:
The equation of constraint is dr = 0. We have three equations and three unknowns,
To find an expression for
l1 we need to solve the differential equation
Try
Therefore B = 2g/R,
|
Problem 10:
Find the Lagrangian and the equations of motion of a board of mass
M and length
l, which sits inside a cylindrical pipe of radius
R. There is no friction between the board
and the sides of the pipe, and gravity is not to be
neglected. Assume that the board does not change its
z-coordinate, where
z is along the horizontal axis of symmetry
of the pipe. Determine the critical energy Ec
above which the board starts to rotate in the pipe.
Solution:
| Concepts, principles, relations
that apply to the problem: Lagrange's Equations: ;
Centripetal force: Fc = mv2/r |
| Why do they apply? All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. For the board to rotate in the pipe without loosing contact with the wall, we need that the centripetal force at the highest point is greater in magnitude than the force of gravity. |
| How do they apply? Set up a coordinate system as shown in the figure below.  For q < 90o, the board does not break contact with the wall. Then T = (1/2)M(hdq/dt)2 + (1/2)M(l2/12)(dq/dt)2 = TCM + Trot, U = -Mghcosq. Here h2 = R2 - l2/4. We have only one generalized coordinate, q, and one generalized velocity, dq/dt. |
| Details of the calculation:  The equation of motion is  For q > 90o, the board can break contact with the wall. For the board to rotate in the pipe we need Mh(dq/dt)2 > Mg at q = 180o, or (dq/dt)2 > g/h at q = 180o. E = T + U. At q = 180o we have E = (1/2)M(R2 - l2/6)(dq/dt)2 + Mgh. For the board to rotate in the pipe we need E > Ec = (1/2)M(R2 - l2/6)(g/(R2 - l2/4)1/2) + Mg(R2 - l2/4)1/2. We can also use Lagrange multipliers to find the force of constraint in the radial direction. As long as this force points towards the center of the pipe for all q, the board will rotate in the pipe.  ;h is now a generalized coordinate. Constraint: dh = 0. We have one equation of constraint, l1dh = 0. [Lagrange multipliers: l1(a1hdh + a1qdq) = 0, a1h = 1, a1q = 0.]  l1 is the force of constraint (Salkll = a1hl1) in the radial direction.  At q = 180o we have l1 = -Mh(dq/dt)2 + Mg. We want l1 to be negative for the board to rotate in the pipe. We therefore need Mh(dq/dt)2 > Mg, or (dq/dt)2 > g/h at q = 180o. |
.
.
.
(1)
(2) 
from (2)
from (1)
.
and equate
the solution with 
.
, then 
.
.
.