Homework 5, due (Friday Feb 28.) extended to Monday March 3
Reading: 7.8, 7.9; 5.35.4 (should have been assigned earlier); 6.16.3.
Hour Exam: On Friday March 7, 11 am we will have an hour exam on the material through last assignment, and problems through the present assignement (dealing with Lagragian mechanics).
Some reading to prepare for the exam:
 Problem 14 solved via F = ma, and in x, y coordinates. You check that an x, y Lagrangian gives the same anwer!
 Last year's first hour exam, with solution but without pictures. The first picture should be obvious, the second is shown in the popup*
 If you really want to know how bad it was in the old days: First hour exam of two years ago with solutions but no pictures. "Problem 15" to which this refers, is the same as our homework problem 17. The second problem of that exam is not relevant to our exam  no Hamiltonians (yet).
 A question about gravity. With some hacking, you can find the answer, too. Won't help a whole lot on the exam though
 "Some Lagrangian Problems". The answers are written as jpg's, so take a while to download on a slow connection (unlike the answers to the problems on our site!). Good exercise in finding kinetic energy, but every one of these problems can be done by conservation of energy, you don't really need Lagrange's equations.
 Other internet sites, but I have not found any good problems with more than one degree of freedom. Let me know if you do!
*Does anyone know how to change this into a regular link in a new window of that small size?
Problems:
16. Text, problem 710.
17. An embellishment of problem 16: A mass m is moving without friction on a smooth horizontal table, and is attached by a string of length L, which passes through a hole in the table, to a mass
M suspended beneath. M moves only in a vertical line.
(a) Find the Lagrangian for the system, using coordinates that exhibit its symmetries, and mention those symmetries.
(b) What is/are the conserved quantity/ies corresponding to the symmetries?
(c) Find the remaining equation of motion (which does not directly yield a conserved quantity).
(d) If m is given an initial velocity perpendicular to the string, find the velocity that will keep M stationary.
18. Example 7.6 from the text, but the railroad car have mass M and be free to move (not constrained to acceleration a) on the frictionless rail.
(a) Write the Lagrangian in terms of two generalized coordinates, chosen in such a way that one of them is cyclic (does not appear in the Lagrangian), find the conserved momentum conjugate to the cyclic coordinate, interpret physically, and write the equation of motion for the other coordinate.
(b) Eliminate the cyclic coordinate from the equations of motion and find the frequency of small oscillations for the other coordinate. Is the result reasonable in the limit M ® ¥ and M = m?
19. Text 727 (a) and (b) only. Thinking about symmetries of the problem, choose coordinates so you get the maximum number of conserved quantities.
20.Text 711 except last sentence: it acts like a pendulum in a unform gravitational field, even for large pendulum angles. which is more remarkable than reasonable.
Solutions
16. Text, problem 710.
This is one of those problems that can be done by conservation of energy, but assuming that the acceleration is a central item in the solution we use the Lagrangian. The kinetic energy for the two masses M and the string of mass m, all going with the same speed y^{·}t (where y = height of mass that hangs down, measured from table top  a negative quantity) is of course T = ^{1}/_{2}(2M + m) y^{·2}. The potential energy of the string is found from the center of mass (at y/2) of the fraction y/l of the string that hangs down, thus
L = ^{1}/_{2}(2M + m) 
. y

2

 Mgy + m(y/l)g(^{y}/_{2}) = ^{1}/_{2}(2M + m) 
. y

2

 Mgy + ^{1}/_{2}mgy^{2}/l 

leading to the single Lagrange equation
(2M + m) 
.. y

+ Mg  mg(y/l) = 0. (1) 

(a) for m = 0 we have a =  g/2, y =  (g/4)t^{2}. For m ¹ 0 Eq (1) is solved via the usual particular solution (y = (m/M)g which is unphysical by itself) plus general solution of the homogeneous equation, an "oscillator with negative spring constant", whose solutions are exp, cosh, and sinh. Thus one obtains the answer in the back of the book.
17.(a) Let the coordinates be r, the distance of m from the hole (so that y = (Lr) = rL)
and f, the angle that the string on the top makes with some fixed direction. Then the kinetic energy is T = ^{1}/_{2} My^{·2} + ^{1}/_{2}m(r^{·2} + r^{2}q^{·2}) and U = Mgy.
Put it together and use the constraint y = rL to find the Lagrangian L
L = ^{1}/_{2} M 
. r

2

+ ^{1}/_{2}m( 
. r

2

+ r^{2} 
. q

2

)  Mg(rL). 

The symmetries are: independence of q, independence of t.
(b) p_{q} = ¶L/¶q^{·} = mr^{2}q^{·}, H = T + U
(c)
requation: (M+m) 
.. r

 mr 
. q

2

+ Mg = 0 

(d) M is stationary if  .. r
 = 0, so  . f
 2
 = Mg/mr hence the required velocity is v = r  . f
 = 
 _______ Ö(M/m)gr

. 

18. (a) This system has translational symmetry, which will be exhibited if one coordinate X is the position of the RR car, and the other coordinate is position relative to the RR car. We use the book's q, then L is ^{1}/_{2}MX^{·2} + the first equation on p. 248 with (v_{0} + at) replaced by X^{·2}, so
L = ^{1}/_{2}(M+m) 
. X

2

+ ml 
. X


. q

cosq + ^{1}/_{2}ml^{2} 
. q

2

+ mglcosq 

Then
p_{X} = ^{¶L}/_{¶X·} = (M + m) 
. X

+ ml 
. q

cosq 

is the horizontal component of the total momentum (car + pendulum). It is constant because there is no external force on the system in the horizontal direction (frictionless rails).
qequation, after some cancellation: m(l^{2} 
.. q

+l 
.. X

cosq+ glsinq) = 0 

(b) For small angles X^{¨} =  mlq^{¨}/(M+m) hence ^{M}/_{(M+m)}q^{¨} = ^{g}/_{l}q.
For large M (RR car not moving) this gives the pendulum frequency of length l. For M=m we can think of M concentrated at the suspension, the CM is then halfway along the pendulum. Since it can stay fixed, the effective pendulum length is l/2, consistent with the above equation.
19. Text 727
To get a qindependent Lagrangian you pretty much have to choose x, y coordinates of the CM, r distance between masses, q angle of the line between them; and use the familiar decomposition of the kinetic energy. The distance of m_{1} from CM is (^{m2}/_{m1+m2})r and similar for m_{2}, so the moment of inertia of the systems about the CM is (^{m1m2}/_{m1+m2})r^{2} and we have for the lagrangian
2L = (M+m)( 
. x

2

+ 
. y

2

) + 
m_{1}m_{2} m_{1}+m_{2}

( 
. r

2

+ r^{2} 
. q

2

)  kr^{2} 

(a) The requation yields r^{¨}  rq^{·2} + kr = 0.
(b) Since x and y are cyclic their equations give p_{x} = const, p_{y} = const  conservation of total momentum.
Since q is also cyclic we have angular momentum conservation, p_{q} = (^{m1m2}/_{m1+m2})r^{2}q^{·} = const.
20. Text 711
From the geometry (figure to be supplied) we see that the particle is at angle f = wt + q/2 and radius r = 2Rcos(q/2) from the fixed point. To get v^{2} we use the polar coordinate expression in terms of r and f:
v^{2} = 
×
r

2

+ r^{2} 
×
q

2

= 4R^{2} 
æ ç
è

sin^{2} 
q
2


æ ç
è

2

ö ÷
ø

2

+ cos^{2} 
q
2


æ ç
è

w+ 
2

ö ÷
ø

2

ö ÷
ø

= R^{2} 
æ è

×
q

2

+ (2cosq+ 2)w^{2} 
ö ø


 (2) 
where we have used the double angle formula cos2a = 2cos^{2}a1 for a = q/2. Up to the factor ^{1}/_{2} m, Eq (2) is the Lagrangian.
Compared to the Lagrangian for a pendulum, ^{1}/_{2} m (R^{2}q^{·2} + gR(cosq1) it agrees for 2w^{2} = g/R except for a unimportant additive constant.