part 1: calculus of variationsastro3.sci.hokudai.ac.jp/~alex/teaching_materials/examplescm2.pdf ·...

Alex R. Pettitt Mechanics 2: Board work December 4, 2020

Document that contains worked examples of blackboard problems used in class.

Part 1: Calculus of variations

Problem 1. Ramp 1

The simple ramp problem, how long to fall down for a mass M , height h, incline of ✓? NL2set up parallel to surface (x down incline):

f

y

= N = mg cos ✓ , f

x

= mx = mg sin ✓ (1)

so, if starting from rest at x0

= 0, integrating f

x

gives:

x(t) =1

2g sin ✓ t

2 (2)

the time after travelling a distance x

1

= h/ sin ✓ gives:

t

f

=

s2h

g

1

sin ✓(3)

can also use energy: �T = ��U , so that vf

=p2gh. As accel is constant, using SUVAT:

x = 1

2

(vi

+ v

f

)tf

to get tf

.

Problem 2. Where my ramp at: the brachistochrone

Now our ramp is missing, we want to optimise it to make our journey as short as possible.Say we move along is some distance x of d, such that:

y(x = 0) = 0 (4)

y(x = d) = �h (5)

and what we are missing is the shape of the ramp; y(x) = f(x), which we had asy(x) = x tan ✓ before. We must make some assumptions such as the ramp is well behavedsuch that the block can’t get stuck before reaching x = h. This is actually quite simple, andwe can still use conservation of energy:

T + U = 0 (6)

1

2m(x2 + y

2) +mgy = 0 (7)

Notice though that with energy we are coupling the x and y coordinates, can’t treatseparately. We can then denote this in a di↵erent way to remove the time derivative ofy:

y =dy

dt

=dy

dx

dx

dt

= f

0(x)x (8)

1


subbing back in to remove the y dependance:

� 2gy = x

2(1 + f

0(x)2) (9)

1 = x

s1 + f

0(x)2

�2gf(x)(10)

if x = dx/dt, this re-arranges to give:

t

f

=

ˆdt =

ˆd

0

s1 + f

0(x)2

�2gf(x)dx (11)

which can give us the arbitrary time to go from 1 ! 2 for some path f(x).

Problem 3. Variations along some curve

Say we have some function, F = (dy/dx)2 (this is key!), where we have the simple case thaty(x) = x. Let’s try adding in ⌘(x) = sin x and finding the functional to minimise for ourpath spanning x = 0 and x = 2⇡. Our combined function is thus:

y(↵, x) = x+ ↵ sin x (12)

Note our new function still satisfies our constraints! We now need to find:

J(↵) =

ˆb

a

F{(y(↵, x), y0(↵, x), x} dx (13)

where our y(↵, x) = f(↵, x) needs to be found. So:

dy(↵, x)

dx

= 1 + ↵ cos x (14)

F =

✓dy

dx

◆2

= 1 + 2↵ cos x+ ↵

2 cos2 x (15)

so:

J(↵) =

ˆ2⇡

0

(1 + 2↵ cos x+ ↵

2 cos2 x) dx (16)

=

x+ 2↵ sin x+

1

2↵

2

x+1

2↵

2 sin x cos x

�2⇡

0

(17)

this gives J(↵) = 2⇡+↵

2

⇡. As you can see, this always moves away from the minima, J(0),for any J(↵), regardless of the sign of ↵. Also, the derivative wrt ↵ gives 0 = 2↵, so correctlyimplying that ↵ = 0 is required to minimise the functional.

2


Problem 4. Shortest path between two points in plane

We constructed this line integral earlier for the distance spanning two points in a plane;

L =

ˆx2

x1

ds =

ˆx2

x1

p1 + y

02dx (18)

we can therefore see our function in the integrand:

F (y, y0, x) = (1 + y

02)1/2 (19)

evaluating the derivatives:

@F

@y

= 0 and@F

@y

0 =y

0p1 + y

02(20)

so that the Euler equation gives:

0� d

dx

y

0p

1 + y

02

!= 0 (21)

integrating both sides gives a constant on the right, so that:

y

0 = C

p1 + y

02 (22)

so:y

02 = C

2

/(1� C

2) = m

2 (23)

so this gives dy/dx = m, some constant, i.e. a straight line! Magical. Integrating once moregives:

y(x) = mx+ c (24)

what an excellent use of our time that was.

3


Problem 5. Solving the brachistochrone

Now the J [y] we want to minimise is the time taken to travel; t[y]:

t

1!2

=

ˆdt =

ˆds

v

(25)

For any kind of ramp, we can equate potential to kinetic to obtain the velocity at the bottom.Saying we start from rest and defining 0 energy at this point (gravity is conservative) so:T + U = 1

2

mv

2

f

�mgy = 0, so that v =p2gy (y measured downwards). So:

t

1!2

=

ˆdt =

ˆds

v

=

ˆ pdx

2 + dy

2

p2gy

=1p2g

ˆ px

0(y)2 + 1py

dy (26)

where we used x

0(y) = dx/dy. Now, you can hopefully see this has the same form as (wherewe swapped x and y):

J =

ˆF{x(y), x0(y), y} dy (27)

so we can minimise by spotting our integrand of:

F (x, x0, y) =

px

0(y)2 + 1py

(28)

so putting that into Euler (with x and y again swapped):

@F

@x

� d

dy

@F

@x

0 = 0 (29)

the first term is 0 for our F , thus the second term is also equal to 0.

d

dy

@F

@x

0 =d

dy

x

0(y)py

px

0(y)2 + 1= 0 (30)

So when integrated wrt y, it must give a constant (@F/@x0 =const.), thus, integrating upand squaring (absorbing the -ve into the new constant):

x

02

y(1 + x

02)= const. =

1

2a(31)

where we have re-written our constant to make the final solution nicer! We can then re-arrange to give:

x

0 =dx

dy

=

ry

2a� y

(32)

so:

dx =

ˆ ry

2a� y

dy (33)

4


there is a substitution that makes this nice to solve, say: y = a(1 � cos ✓), and therefore:dy = d✓a sin ✓ so:

dx =

ˆ sa(1� cos ✓)

2a� (a� a cos ✓)dy =

ˆ r1� cos ✓

1 + cos ✓a sin ✓ d✓ (34)

=

ˆ r1� cos ✓

1 + cos ✓

r1� cos ✓

1� cos ✓a sin ✓ d✓ =

ˆ1� cos ✓p1� cos2 ✓

a sin ✓ d✓ (35)

= a

ˆ1� cos ✓p1� cos2 ✓

p1� cos2 ✓ d✓ = a

ˆ(1� cos ✓) d✓ = a(✓ � sin ✓) + x

0

(36)

this thus gives us some parametric solutions! We can set our initial position as (0,0) so that:

x = a(✓ � sin ✓) (37)

y = a(1� cos ✓) (38)

Problem 6. Two approaches

We are looking to minimise the area spanned by rotating tow points about the x or y axis.

• First, about y [this is a seminar Q, but I put it here for completeness]

A =

ˆx2

x1

dA =

ˆx2

x1

2⇡xds =

ˆx2

x1

2⇡x(dx2 + dy

2)1/2 (39)

= 2⇡

ˆx2

x1

x(1 + y

0 2)1/2 dx (40)

Our function to minimise is thus:

F (x, y, y0) = x(1 + y

0 2)1/2 (41)

So into Euler:@F

@y

= 0 (42)

@F

@y

0 =xy

0

(1 + y

0 2)1/2(43)

thus we get a constant:xy

0

(1 + y

0 2)1/2= const. = a (44)

Re-arranging for the gradient we have:

y

0 =a

(x2 � a

2)1/2(45)

y =

ˆa

(x2 � a

2)1/2dx (46)

5


This is a standard integral, and gives:

y(x) = a cosh�1(x/a) + b (47)

so:x(y) = a cosh((y � b)/a) (48)

This is the equation for a Catenary!

• Now, in x:

A =

ˆx2

x1

dA =

ˆx2

x1

2⇡yds =

ˆx2

x1

2⇡y(dx2 + dy

2)1/2 (49)

=

ˆx2

x1

2⇡y(1 + y

0 2)1/2 dx (50)

So our function to minimise is:

F (x, y, y0) = y(1 + y

0 2)1/2 (51)

We can now put into Euler again, giving:

@F

@y

=p1 + y

0 2 (52)

@F

@y

0 =yy

0p1 + y

0 2(53)

So putting together we get:

p1 + y

0 2 =d

dx

yy

0p1 + y

0 2

!(54)

Problem 7. The first integrals

Say we want to make the following stationary, i.e. the second integral in the previous questionwhich was the rotation about x (ignoring the 2⇡):

I =

ˆy

p1 + y

0 2dx =

ˆF (x, y, y0) dx,

@F

@y

� d

dx

@F

@y

0 = 0 (55)

if we were to blindly put this into Euler’s equation, it would be pretty horrid as we wouldhave to perform the first and second integrals. Instead, we will change variables:

p1 + y

0 2dx =

p1 + y

0 2x

0dy =

px

0 2 + 1 dy (56)

I =

ˆy

px

0 2 + 1 dy =

ˆF (y, x, x0) dy (57)

6


so that we will not need the second integral! We’ve gone from using (0, y, y0) dx with @/@y

to (y, 0, x0) dy with @/@x, so those mixed terms make things much easier to deal with. Butmake sure you keep track of your new Euler equation in terms of variables:

@F

@x

� d

dy

@F

@x

0 = 0 (58)

the first term would vanish (which it wouldn’t for the original integral with a Euler equationincluding a @/@y term) such that:

d

dy

@F

@x

0 =d

dy

✓y

x

0px

0 2 + 1

◆= 0 (59)

so:yx

0px

0 2 + 1= const. (60)

which is exactly the same as our rotation about y example, just with x $ y an

Problem 8. 2D Euler: line again

Now we want to find the shortest point, but encompassing all possible paths (whereas beforey had to be single valued for each x). The distance to be minimised is now:

ds =p

dx

2 + dy

2 =p

x

0(u)2 + y

0(u)2 du (61)

Where u is some variable that parametrises our path (e.g. the distance along some spiralpath); x = x(u), y = y(u). So that now:

J =

ˆu2

u1

px

0(u)2 + y

0(u)2 du (62)

must be stationary, where x

0 = dx/du now. So:

F (x, x0, y, y

0, u) =

px

0(u)2 + y

0(u)2 (63)

This is simply our equation to be minimised. As there is no x or y dependence we will dropthe first terms in the Euler equations. As before, this implies the partial derivatives wrt x0

and y

0 must equal a constant:

@F

@x

0 =x

0p

x

02 + y

02= C

1

,

@F

@y

0 =y

0px

02 + y

02= C

2

(64)

dividing one by the other, we get:

y

0

x

0 =dy/du

dx/du

=dy

dx

=C

2

C

2

= m (65)

i.e. a straight line again! This is better than before, as contains all possible paths and wasquicker!

7


Part 2: The Lagrangian

Problem 9. 1D Lagrange equation

Say we have a much simpler problem of a particle moving in 1D with mass m;

F = mx (66)

say the force is conservative, i.e. conserves mechanical energy, so that we can define:

~

F = �~rU (67)

so that:

mx = �dU

dx

(68)

getting there, but this still doesn’t look quite like the Euler equation. What about EK?

T =1

2mx

2 (69)

this could be useful to compare to the potential term above if we take a few derivatives...

@T

@x

= mx (70)

and again, but now in time, to make it look like the potential term:

d

dt

@T

@x

= mx (71)

which means that:@U

@x

+d

dt

@T

@x

= 0 (72)

almost what we want! But one part acts on T , the other on U . However, for this examplethere is no velocity dependence in the potential (conservative like gravity), and no position inkinetic energy, so we easily add in extra terms of each in both derivatives, thereby assumingsome function called the Lagrangian:

L = T � U (73)

so that our particle satisfies the Euler equation for L:

@L@x

� d

dt

@L@x

= 0 (74)

so that there must therefore be some associated functional that must be minimised tosatisfy the above:

S =

ˆtf

t1

L(x, x, t) dt (75)

8


Problem 10. Free-fall

Say we are free-falling in y-direction (+ve is up), with mass m, making our potential:U = mgy, and so:

L = T � U =1

2my

2 �mgy (76)

so, taking our Lagrangian derivatives we get:

@L@y

=@T

@y

= my (77)

@L@y

=@(�U)

@y

= �mg (78)

then taking the final time derivative:

�mg = my (79)

so we accelerate with y = �g, nice! If we tried (for some reason) to use T +U then we wouldwind up accelerating upwards in y. What about x, that we ignored?

L = T � U =1

2m(x2 + y

2)�mgy (80)

and putting into the x form of the Lagrange equations, we get:

0 = mx (81)

neat! We call redundant equations such as these as being in terms of ignorable coordinates.

Problem 11. Polar Lagragian

242 Chapter 7 Lagrange's Equations

takes the form

(generalized force) = (rate of change of generalized momentum) (7.17)

I shall illustrate these ideas in the next example.

EXAMPLE 7.2 One Particle in Two Dimensions; Polar Coordinates

Find Lagrange's equations for the same system, a particle moving in two dimen-sions, using polar coordinates.

As in all problems in Lagrangian mechanics, our first task is to write down the Lagrangian T — U in terms of the chosen coordinates. In this case we have been told to use polar coordinates, as sketched in Figure 7.1. This means the components of the velocity are v r = r and v o = 4, and the kinetic energy is T = lmv 2 r2 2). Therefore, the Lagrangian is

(r, (p) = T — U = zm (I- 2 r24; 2) — U (r, 0). (7.18)

Given the Lagrangian, we now have only to write down the two Lagrange equations, one involving derivatives with respect to r and the other derivatives with respect to 44.

The r Equation The equation involving derivatives with respect to r (the r equation) is

az d az ar dt ar

;2 au d z tnrop Or dt

— — —ttnr ) =- mr. (7.19)

Since —a U/ar is just Fr , the radial component of F, we can rewrite the r equation as

Fr = M( ; - 42), (7.20)

Figure 7.1 The velocity of a particle expressed in two-dimensional polar coordinates. As always, we need a Lagrangian for the energy, recall velocity in polars is ~v = rr+ r��,

so that the Lagrangian is:

L = T � U =1

2m(r2 + r

2

�

2)� U(r,�) (82)

9


as we have two independent variables we therefore have two Lagrange equations of motion.First, in r:

@L@r

=d

dt

@L@r

(83)

this gives:

mr�

2 � @U

@r

=d

dt

mr = mr (84)

where the dU/dr term is clearly a force in r, so that:

F

r

= m(r � r�

2) (85)

which is exactly what we had in CMI, and much faster than last time. What about inazimuth?

@L@�

=d

dt

@L@�

(86)

� @U

@�

=d

dt

(mr

2

�) (87)

To make some sense of this, we need our grad operator (the left is not a force, i.e. F

�

, dueto the derivative wrt an angle):

~r =@

@r

r +1

r

@

@�

� (88)

which we can use to read o↵ the terms in ~

F = �~rU , telling us the � component of the forceis:

F

�

= �1

r

@U

@�

=1

r

d

dt

(mr

2

�) (89)

You can probably recognise the bracketed term as angular momentum; L = mr

2

�, and rF

�

as a torque! i.e. this equation states that:

⌧ =dL

dt

(90)

i.e. that torque is the rate of change of angular momentum!

Problem 12. Field motion

A particle moves in a 2D plane under the influence of a force, f = �Ar

↵�1, directed to theorigin; A and ↵ are positive constants. Choosing some appropriate coordinates, definingU(0) = 0, we want to find the Lagrangian, and look at conserved quantities.

As our force is radial, let’s use polars. As our particle moves around, we have:

T =1

2m(x2 + y

2) =1

2m(r2 + r

2

�

2) (91)

We can derive a force from our potential (just radial, so is conservative):

F = �@U

@r

, U =A

↵

r

↵ (92)

10


so:

L =1

2m(r2 + r

2

�

2)� A

↵

r

↵ (93)

evaluating E-L for r, we get:mr = mr�

2 � Ar

↵�1 (94)

and for �:d

dt

(mr

2

�) = 0 (95)

this equation clearly is implying that angular momentum is conserved, as the term in bracketsis mr

2

! = l. We can thus re-write the r equation as:

mr � l

2

mr

3

+ Ar

↵�1 = 0 (96)

if we times the whole thing by r:

mrr � rl

2

mr

3

+ Arr

↵�1 = 0 (97)

we can see we can re-form this whole thing by the chain rule:

d

dt

1

2mr

2

�+

d

dt

l

2

2mr

2

�+

d

dt

A

↵

r

↵

�= 0 (98)

where the second term is 1

2

m

2

r

2

!

2, as l = mvr and v = r!. i.e. the rotational KE, the firstis radial KE, and the last the potential. Total energy is therefore conserved:

d

dt

(T + U) = 0 (99)

Problem 13. An moving incline

A particle moves down a smooth incline, of angle ✓. This angle increases at a rate !. If✓(t = 0) = 0 the particle is a rest, we want to find how the particle moves as the plane risesup.

Say x the distance along the moving face, measured from the ground. The angle isincreasing by ✓(t) = !t, so that the full KE is:

T =1

2m(x2 + (x!)2) (100)

The potential on the rising slope is:

U = mgh = mgx sin ✓ = mgx sin!t (101)

So that:

L = T � U =1

2m(x2 + (x!)2)�mgx sin!t (102)

11


Evaluating the E-L:@L@x

= mx,

d

dt

@L@x

= mx (103)

@L@x

= mx!

2 �mg sin!t (104)

so that:mx = mx!

2 �mg sin!t (105)

x� !

2

x = �g sin!t (106)

We’ve seen these before! We take these apart as x

p

+ x

h

, where the homogeneous is thesolution:

x� !

2

x = 0, x = Ae

!t +Be

�!t (107)

And we can then assume a x

p

= C sin!t. Plugging x

p

into our EoM we get:

� !

2

C sin!t� !

2

C sin!t = �g sin!t (108)

i.e. C = g/(2!2), and our full solution reads:

x = Ae

!t +Be

�!t +g

2!2

sin!t (109)

Evaluating BCs is trivial (x(0) = 0, x(0) = x

0

), and gives values for A and B. A bit ofre-arrangement gives:

x = x

0

cosh!t� g

!

2

sinh!t+g

!

2

sin!t (110)

and this looks like the graph below:

Problem 14. Lagrangian for pendulum

Say we have a bob on a string, of length l. We can write out our constraint as l = x

2 + y

2.We could write out a Lagrangian using the y variable, as y =

pl

2 � x

2. Easier toidentify the variable for the only DoF we have left, � (displacement from the vertical),thus parameterising both x(�) and y(�) by the new independent variable. The energies ofthe system are:

T =1

2mv

2 =1

2ml

2

�

2 (111)

12


U = mgh = mgl(1� cos�) (112)

where U is defined from U(� = 0) = 0. So:

L(�, �) = T � U =1

2ml

2

�

2 �mgl(1� cos�) (113)

so taking the di↵erent derivatives involved in the E-L equation for our generalised coordinateq = �:

@L@�

= �mgl sin� (114)

d

dt

@L@�

=d

dt

(ml

2

�) = ml

2

� (115)

�mgl sin� = ml

2

� (116)

The picture should show you that the left is the Torque (sine term included!) and the rightis I↵, so this merely encompasses ⌧ = I↵.

Thus the EoM is:�+

g

l

sin� = 0 (117)

which is exactly what we got from NL2.We could equally define x = l sin�, and y = �l cos� with T = 0.5m(x2 + y

2), noting weonly need the single parameter � (two dimensions, one constraint, so only one generalisedcoordinate), and then build the Lagrangian in this way, which gives exactly the same result.i.e.

T =1

2m(l2�2 sin2

�+ l

2

�

2 cos2 �) =1

2ml

2

�

2 (118)

Problem 15. Constrained motion on a hemisphere

Let’s find some generalised coordinates for a particle moving on a hemisphere. We musthave the following satisfied:

(x2 + y

2 + z

2)�R

2 = 0, z � 0 (119)

we can use generalised coordinates of cosines of the angles between the di↵erent axes andthe line pointing to the particle. i.e.

q

1

=x

R

, q

2

=y

R

, q

3

=z

R

, (120)

such that:q

2

1

+ q

2

2

+ q

2

3

= 1 (121)

this obviously shows that we do not have a generalised set of coordinates, because:

q

3

=q1� q

2

1

� q

2

2

(122)

13


due to our constraint removing a DoF. However, we are perfectly allowed to use q

1

= x/R

and q

2

= y/R as our generalised coordinates, so long as we have our equation of constraint:

g(q1

, q

2

) = 0 = q

2

1

+ q

2

2

+

✓z

2

R

2

� 1

◆. (123)

This should be obvious really, thinking about the Earth we only need a latitude and alongitude to define any location on the surface.

Problem 16. D’Alembert quick examples

• One quick example of d’Almebert’s formula. Trivial example is that of a falling mass.If z is up, then we have the force applied to the mass as �mg, and the virtual workdone by this force over some virtual displacement is �W

G

= �mg�z, and the inertialforce gives a work of �W

I

= �m(d2z/dt2)�z. Writing out the equation of dynamicequilibrium:

X(F

app

� p)�r = �mg�z �m

d

2

z

dt

2

�z =

✓�mg �m

d

2

z

dt

2

◆�z = 0 (124)

so that:d

2

z

dt

2

= g (125)

Not very obviously better than NL2...

• What about a bead rolling on a hoop? Component of gravity normal to the hoop doesno work on the bead, i.e. gravity component pointing out of hoop is not in directionof motion. If we have some virtual displacement along the hoop: �s = R��, then thework done by gravity along the wire is:

�W

G

= �mg sin�⇥R �� (126)

The inertial forces (p term) has two components, a

r

= �v

2

/R = �(ds/dt)2/R =�(R2

�

2)/R and a

t

= d

2

s/dt

2 = R�. The radial part is doing no work due to ourconstraint to move in �s = R��. The tangential part, however, will:

�W

I

= mR� �s (127)

14


so that, combining the virtual work from applied forces and inertial forces:

(�Wapp

� �W

I

)�s = (�mgR sin��mR

2

�) �� = 0 (128)

this gives the equation of motion for some �� 6= 0:

g sin�

R

+ � = 0 (129)

We could equally have, radially:

(�mgR cos�+R�

2) �r = 0 (130)

but this is satisfied by the �r = constraint to our motion.

Problem 17. Lagrangian Example: Atwood x1

Consider the simple Atwood machine, with two masses on a frictionless bearing (radiusR) by an inextensible string. Our main constraint is that: x + y + ⇡R = l, where x is theheight from the pivot of m

1

, and y the height of m2

. Because this length is fixed, we canmodel the system with a single variable. We can therefore express our holonomic constraintequation of the form:

g(x, y, t) = y + x+ const. = 0 (131)

so we can use x as a single generalised coordinate. Noting that y = �x:

T =1

2m

1

x

2 +1

2m

2

y

2 =1

2(m

1

+m

2

)x2 (132)

while the potential is:

U = �m

1

gx�m

2

gy = �(m1

�m

2

)gx+ const. (133)

15


The constant is uninteresting for physics (see what happens with the E-L eq) so that:

L = T � U =1

2(m

1

+m

2

)x2 + (m1

�m

2

)gx (134)

when subbed into the E-L equation:

@L@x

=d

dt

@L@x

(135)

we get:(m

1

�m

2

)g = x(m1

+m

2

) (136)

x = g

m

1

�m

2

m

1

+m

2

(137)

This was an very useful tool to measure g. Say you choose the setup so that you minimisethe di↵erence but maximise the total. This means you can accurately measure the very smallacceleration, and if you know both the masses you can therefore find g!

Unfortunately, in this example NL2 does just as good a job. For example, if the tensionis T , then the force on 1 is m

1

g � T , so:

m

1

x = m

1

g � T (138)

as block 2 has the same acceleration, mx directed UP, we then have:

m

2

x = T �m

2

g (139)

two equations an two unknowns, so equate to T gives us:

m

1

g �m

1

x = m

2

x+m

2

g (140)

x = g

m

1

�m

2

m

1

+m

2

(141)

Same thing, but we did need to deal with the force of constraint that is uninteresting tous.

Problem 18. Lagrangian Example: Atwood x2

Now lets get really interesting, what if we have a double Atwood machine? Now we haveidentified that we need two coordinates, and what they need to be, we can draw with ourtwo constraints in the form of lengths l

1

and l

2

. Say we have y, pointing up, with origin atcentre of top pulley:

y

1

= �q

1

(142)

y

2

= �l

1

+ q

1

� q

2

(143)

y

3

= �l

1

� l

2

+ q

1

+ q

2

(144)

with velocities:y

1

= �q

1

(145)

16


y

2

= q

1

� q

2

(146)

y

3

= q

1

+ q

2

(147)

So kinetic energy is:

T =1

2m

1

y

2

1

+1

2m

2

y

2

2

+1

2m

3

y

2

3

(148)

=1

2(m

1

+m

2

+m

3

)q21

+1

2(m

2

+m

3

)q22

+ (m3

�m

2

)q1

q

2

(149)

For potential, we can also just use our Cartesian coordinates (a bit simpler this time):

U = m

1

gy

1

+m

2

gy

2

+m

3

gy

3

(150)

= �(m1

�m

2

�m

3

)gq1

� (m2

�m

3

)gq2

� (m2

gl

1

+m

3

g(l1

+ l

2

)) (151)

where we can ignore the constant terms at the end. So:

L =1

2(m

1

+m

2

+m

3

)q21

+1

2(m

2

+m

3

)q22

+(m3

�m

2

)q1

q

2

+(m1

�m

2

�m

3

)gq1

+(m2

�m

3

)gq2

(152)We can then solve the two E-L equations:

@L@q

i

� d

dt

@L@q

i

= 0 (153)

So that:(m

1

�m

2

�m

3

)g = (m1

+m

2

+m

3

)q1

+ (m3

�m

2

)q2

(154)

(m2

�m

3

)g = (m2

+m

3

)q2

+ (m3

�m

2

)q1

(155)

i.e.A = Bq

1

+ Cq

2

(156)

D = Eq

2

+ Cq

1

(157)

so we can re-arrange by:(A� Cq

2

)/B = (D � Eq

2

)/C (158)

q

2

= (BD � CA)/(BE � CC) = const. (159)

so we get constant accelerations (same for the other) so integrating up is simply q

2

=at

2 + bt+ c, where the direction is determined by mass ratios.Note however from above that the algebra is sensible at equilibrium points. If m

2

= m

3

we get: q2

= 0 from the second E-L equation, which makes sense as that pulley is balanced.If we have m

1

= m

2

+m

3

and m

2

= m

3

then the whole system is balanced, and we get noaccelerations: q

1

= 0.

17


Problem 19. Rotating pendulum

Holonomic can have time dependence, here is an example. Say we have a pendulum on acircular support that rotates at some angular velocity, !, with radius a, where the pendulumhas a length l. We have conservative forces (i.e. gravity, as constraints don’t enter theLagrangian). The constraint is holonomic though, as it does not depend on the mass’svelocity, so we can still write the constraint as a function of position and time. This is aforced/rheonomous constraint, as opposed to a natural/normal/scleronomous constraint.

We want to setup a system with polars as our generalised system (though a constraint isr = l). From our diagram, we will measure !t form the x-axis upwards. As such the positionof the bob in the inertial/fixed frame is:

x = a cos!t+ l sin ✓ (160)

y = a sin!t� l cos ✓ (161)

see, only 1 generalised coordinate, as there is only 1DoF (as we expect from a pendulum).This gives:

x = �a! sin!t+ l✓ cos ✓ (162)

y = a! cos!t+ l✓ sin ✓ (163)

So, the Lagrangian in terms of Cartesian KE is easy now:

L = T � U =1

2mx

2 +1

2my

2 �mgy (164)

in terms of our angle, we can write as:

L =1

2m

ha

2

!

2(sin2

!t+ cos2 !t) + l

2

✓

2(cos2 ✓ + sin2

✓) + 2l✓a!(sin ✓ cos!t� cos ✓ sin!t)i

�mg(a sin!t� l cos ✓)

which greatly simplifies thanks to trig identifies!

L =1

2m

ha

2

!

2 + l

2

✓

2 + 2l✓a! sin(✓ � !t)i�mg(a sin!t� l cos ✓) (165)

We still need some derivatives, so:

@L@✓

= ml✓a! cos(✓ � !t)�mgl sin ✓ (166)

@L@✓

= ml

2

✓ +mla! sin(✓ � !t),d

dt

@L@✓

= ml

2

✓ +mla!(✓ � !) cos(✓ � !t) (167)

so that in full:

ml✓a! cos(✓ � !t)�mgl sin ✓ �ml

2

✓ �mla!(✓ � !) cos(✓ � !t) = 0 (168)

✓ =a!

2

l

cos(✓ � !t)� g

l

sin ✓ (169)

Phew! And that’s not even solving for ✓(t). If rotation vanished, at least you can see we getback the pendulum equation.

18


Part 3: More on the Lagrangian

Problem 20. Euler’s equation for K.E.

Say we describe the KE for some extended system:

T =1

2

nX

↵=1

3X

i=1

m

↵

x

2

↵,i

(170)

if this is transformed into a dependence on generalised coordinates: x

i

= x

i

(qj

, t) wherej = 1, 2, ..., s then by the multi-variate chain rule (noting that d(t)/dt = 1):

x

↵,i

(qj

, t) =sX

j=1

@x

↵,i

@q

j

q

j

+@x

↵,i

@t

(171)

squaring this mess gives us:

x

↵,i

(qj

, t)2 =sX

j,k

@x

↵,i

@q

j

@x

↵,i

@q

k

q

j

q

k

+ 2X

j

@x

↵,i

@q

j

@x

↵,i

@t

q

j

+

✓@x

↵,i

@t

◆2

(172)

so the kinetic energy is thus, now summed over all particles, ↵:

T =X

↵

sX

i,j,k

m

↵

2

@x

↵,i

@q

j

@x

↵,i

@q

k

q

j

q

k

+ 2X

↵

X

i,j

m

↵

@x

↵,i

@q

j

@x

↵,i

@t

q

j

+X

↵

X

i

m

↵

2

✓@x

↵,i

@t

◆2

(173)

that is an insane amount of sums... But say we only want a T (q) rather than T (x), then theabove should have a general form in terms of generalised velocities:

T =X

j,k

a

j,k

q

j

q

k

+X

j

b

j

q

j

+ c (174)

If we look at scleronomic systems (no time dependence in transformations; x = x(q), notx = x(q, t)), so that b

j

= 0, c = 0 giving:

T =X

j,k

a

j,k

q

j

q

k

(175)

Now, if we di↵erentiate wrt q

l

, then we can e↵ectively split the sum up into two parts forwhen the j, k = l:

@T

@q

l

=X

k

a

l,k

q

k

+X

j

a

j,l

q

j

(176)

then, if we times by q

l

and sum over l, we get:X

l

q

l

@T

@q

l

=X

k,l

a

l,k

q

k

q

l

+X

j,l

a

j,l

q

j

q

l

(177)

now we just have a bunch of dummy indices on the RHS, so they are e↵ectively the same,thus: X

l

q

l

@T

@q

l

= 2X

j,k

a

j,k

q

j

q

k

= 2T (178)

19


Problem 21. What is the Hamiltonian?

We want to find out what H is, where we currently have:

H =nX

i

p

i

q

i

� L (179)

First we can note that the generalised momenta can be written as:

p

i

=@L@q

i

(180)

If we are dealing again with natural systems (no time dependance on transforms) then thepotential can be written as: U = U(q

i

) and @U/@q

i

= 0, so that:

@L@q

i

=@(T � U)

@q

i

=@T

@q

i

(181)

giving a new expression for our Hamiltonian constant as:

H =nX

i

q

i

@T

@q

i

� L (182)

Recall our neat expression for the first part (hopefully you do!)?

X

l

q

l

@T

@q

l

= 2T (183)

as it simplifies things very nicely:

H = 2T � (T � U) = T + U (184)

Problem 22. Conservation example

Say we have a Lagrangian of the form:

L =1

2m(x2 + 2y2) + C(x� y) (185)

a bit weird... but note that the Lagrangian is unchanged if we apply the transform of (checkit!):

x ! x+ ✏ (186)

y ! y + ✏ (187)

i.e. there is some conserved quantity due to the symmetry of that specific transform, in thatthere is a direction in xy where the Lagrangian is translationally invariant. Let’s changevariables:

u ! x+ y (188)

v ! x� y (189)

20


so:

u ! x+ y (190)

v ! x� y (191)

we can flip this around by:

x =1

2(u+ v) (192)

y =1

2(u� v) (193)

subbing this into our expression for the our Lagrangian:

L =3

8(u2 + v

2)� 1

4muv + Cv (194)

back to the slides for what this means...

Problem 23. Lagrangian for E&M

A particle is moving, with a charge q and mass m, in an electromagnetic field. The force is:

~

F = m~r = q( ~E + ~r ⇥ ~

B) (195)

We can use the definitions of the scalar and vector potentials:

~

E = �~rV � @

~

A

@t

,

~

B = ~r⇥ ~

A (196)

Now, our normal definition of potential is that it can only be a function of position, and theassociated force is given by �~rU = ~

F . If we instead postulate something akin to a velocity-

dependent potential energy or “generalised potential” then we can formulate a Lagrangian.I postulate this will return the NL2 on the first line. E.g.

U = qV � q

~

A · ~r (197)

so that, with L = T � U we have:

L(r, r, t) = 1

2m~r

2 � q(V � ~

A · ~r) (198)

=1

2m(x2 + y

2 + z

2)� q(V � xA

x

� yA

y

� zA

z

) (199)

BUT note this generalised potential is quite di↵erent to our standard potential energyformulation. Now, reading o↵ from the Euler-Lagrange equations: Let’s check this makessense by looking at the Euler Lagrange equation in x:

@L@x

=d

dt

✓@L@x

◆(200)

21


where:@L@x

= �q

✓@V

@x

� x

@A

x

@x

� y

@A

y

@x

� z

@A

z

@x

◆(201)

and:@L@x

= mx+ qA

x

(202)

as the magnetic potential can be a function of time and space, A

x

(x, y, z, t). The timederivative of this is a bit uglier (using the chain rule):

d

dt

@L@x

= mx+ q

✓x

@A

x

@x

+ y

@A

x

@y

+ z

@A

x

@z

+@A

x

@t

◆(203)

we can then equate these terms and re-arrange (the x terms cancel) giving:

mx = �q

✓@V

@x

+@A

x

@t

◆+ qy

✓@A

y

@x

� @A

x

@y

◆� qz

✓@A

x

@z

� @A

z

@x

◆(204)

which using our knowledge of potentials, E and B: ~

E = �~rV � @

~

A

@t

, ~

B = ~r ⇥ ~

A this isclearly:

mx = q(Ex

+ yB

z

� zB

y

) (205)

which is the desired form of Newton’s law for our particle!

Problem 24. Beads and hoops

We have a bead of mass m on some hoop, which can slide frictionless along (of radius R).Gravity is the only relevant force, but the hoop can also rotate at frequency ! (say anti-clockwise from above). Firstly, in constructing a Lagrangian we need the energies of thesystem.

The kinetic energy is a bit messy. As an aside, we will derive T in cylindrical polars forgeneral systems: x = ⇢ cos�, y = ⇢ sin�, z = z (where � is measured in x/y from x). Takingthe velocities of each we get:

x = ⇢ cos�� ⇢� sin� (206)

y = ⇢ sin�+ ⇢� cos� (207)

z = z (208)

if we then sub these into T = 1

2

m(x2 + y

2 + z

2) we find that:

T =m

2(⇢2 + ⇢

2

�

2 + z

2) (209)

which is quite nice, good idea to remember this for the future1. Note the second term is justa tangential velocity (v

�

= r�).

1ASIDE: The spherical polars, r, ✓,�, one is much worse, but su�ced to say it gives (we will stick tocylindrical): T = m

2 (r2 + r2✓2 + r2 sin2 ✓ �2)

22


In terms of our system in particular, we have, noting that z and ⇢ are related by theparametric coordinate ✓, so are constraining each other:

⇢ = R sin ✓ (210)

� = !t (211)

z = R(1� cos ✓) (212)

where we placed z = 0 at the hoop base. Taking the time derivatives, we then get:

⇢ = R✓ cos ✓ (213)

� = ! (214)

z = R✓ sin ✓ (215)

We can then put this into the T equation we just derived;

T =m

2(⇢2 + ⇢

2

�

2 + z

2) =m

2(R2

✓

2 cos2 ✓ + (R sin ✓)2!2 +R

2

✓

2 sin2

✓) (216)

=1

2mR

2(✓2 + !

2 sin2

✓) (217)

Potential is easy, as is just a function of z, if we again measure from the hoop-base:

U = mgz = mgR(1� cos ✓) (218)

So our Lagrangian is then:

L =1

2mR

2(✓2 + !

2 sin2

✓)�mgR(1� cos ✓) (219)

We only have one coordinate, ✓, so we only have one E-L equation to worry about:

@L@✓

=d

dt

✓@L@✓

◆(220)

Taking the first derivative we get:

@L@✓

= mR

2

!

2 sin ✓ cos ✓ �mgR sin ✓ (221)

d

dt

@L@✓

=d

dt

(mR

2

✓) = mR

2

✓ (222)

Putting this all together we get:

✓ = (!2 cos ✓ � g/R) sin ✓ (223)

23


Problem 25. Equilibrium of the bead

Now we have the equation of motion for the bead on the hoop:

✓ = (!2 cos ✓ � g/R) sin ✓ (224)

we want to find the nature of the equilibrium points, which are found at:

✓

0

= 0, ± cos�1

⇣g

R!

2

⌘, ⇡ (225)

Let’s look at these solutions in di↵erent regimes.Slow rotation: If we rotate slowly we have !

2

< g/R and two equilibrium points.Starting at the top: we can Taylor expand about this minima as ✓ = ⇡ + ✏, where ✏ is asmall number and expanding only to first order. First simply plug in:

✏ = (!2 cos(⇡ + ✏)� g/R) sin(⇡ + ✏) (226)

we now actually need to the Taylor of our two trig functions:

f(x0

+ ✏) ⇡ f(x0

) + ✏f

0(x0

) +O(✏2) (227)

cos(⇡ + ✏) ⇡ cos ⇡ � ✏ sin ⇡ = �1 (228)

sin(⇡ + ✏) ⇡ sin ⇡ + ✏ cos ⇡ = �✏ (229)

putting back in gives us:

✏ ⇡ (!2(�1)� g/R)(�✏) = (!2 + g/R)✏ (230)

The term in brackets is always positive (for any rotation speed), so for any displacementregardless of sign, the bead will accelerate in that direction. E.g. if we move to positive ✓,the acceleration will increase in that direction, and vice-versa. So the point at the top isunstable. The gravitational forces wants to pull the bead down, and centrifugal wants topull the bead away from the axis, so the bead will never want to restore back to the top.

If we do the same at the bottom, we can expand around ✓ = 0 + ✏:

✏ = (!2 cos(0 + ✏)� g/R) sin(0 + ✏) (231)

cos(0 + ✏) ⇡ cos 0� ✏ sin 0 = 1 (232)

sin(0 + ✏) ⇡ sin 0 + ✏ cos 0 = ✏ (233)

which gives2:✏ ⇡ (!2(+1)� g/R)(✏) = (!2 � g/R)✏ (234)

The term in brackets here is always going to be negative when the speed is low, so accelerationwill always be back towards the equilibrium point (✏ / �✏). Thus this point is stable. Notice

2We could also have simply approximated the functions about a small angle, as we have done many timesbefore, so that cos ✓ ⇡ 1, and sin ✓ ⇡ ✓.

24


this is an SHM-style EoM! So we can assign some frequency; ⌦ =pg/R� !

2 that is positive,so that:

✏ ⇡ �(g/R� !

2)✏ = �⌦2

✏ (235)

with the general solution (going back to normal notation about the bottom point):

✓(t) = A cos(⌦t� �) (236)

so long as !2

< g/R.If we start to increase the speed so that !2

> g/R, then ⌦ is imaginary, and we get anexponential style solution (cos i↵ = cosh↵), indicating the bottom point becomes unstableas the centrifugal force overwhelms gravity and pulls the bead up.

Fast rotation: Now we have !

2

> g/R and gain two new equilibrium points. Ourequation is still:

✓ = (!2 cos ✓ � g/R) sin ✓ ⌘ !

2(cos ✓ � cos ✓0

) sin ✓ (237)

Looking at the point on the right ( 0 < ✓ < ⇡/2) first3. At exactly the equilibrium point,the bracketed term is zero (✓ = ✓

0

), and sin ✓ > 0. If we increase ✓ away from equilibriumthen the bracketed will be become negative (cos only decreases in this range until it getsto ⇡), and sin ✓ remains positive. As such, ✓ becomes negative, and the bead acceleratesback to equilibrium. The same goes for decreasing ✓ slightly (draw some cosines to convinceyourself!). These equilibria are therefore stable.

Looking at the motion here in a bit more detail, we can again take the Taylor expansionabout the equilibrium point ✓

0

= ± cos�1

�g

R!

2

�, and using the double angle formula

cos(A+B) = cosA cosB � sinA sinB, sin(A+B) = sinA cosB + cosA sinB,

✏ = (!2 cos(✓0

+ ✏)� g/R) sin(✓0

+ ✏) (238)

= (!2(cos ✓0

cos ✏� sin ✓0

sin ✏)� g/R)(sin ✓0

cos ✏+ cos ✓0

sin ✏) (239)

⇡ (!2(cos ✓0

� ✏ sin ✓0

)� g/R)(sin ✓0

+ ✏ cos ✓0

) (240)

the first cosine and the g/R term cancel by definition of the stable point. Leaving:

✏ ⇡ �!

2

✏ sin2

✓

0

� !

2

✏

2 sin ✓0

cos ✓0

(241)

Taking the solution to first order alone in ✏, we have:

✏ = �✏!

2 sin2

✓

0

⌘ �⌦0 2✏ (242)

where we have defined another frequency, ⌦0 = ! sin ✓0

, where:

⌦0 2 = !

2 sin2

✓

0

= !

2(1� cos2 ✓0

) = !

2(1� (gR/!

2)2) (243)

so:

⌦0 =

r!

2 �⇣

g

!R

⌘2

(244)

Which once again gives us SHM about the minimum, ✓0

, oscillating at a di↵erent frequencyto before.

3This is the only range allowed, we cannot be higher than this! Look at cos�1� gR!2

�in the limit of huge

velocities and you will see it tends to cos�1(0) = ⇡/2, 3⇡/2.

25


Problem 26. A Lagrange multiplier

Say we have a wire that is bent to follow y = 1 � x

2. A string is stretched from the originto some point on the curve (x, y). We want to find the minimum length of the string. Thereare 3 key ways to do this, and all revolve around minimising the distance F = d

2 = x

2 + y

2,but x and y are not independent and are instead constrained by the curve.

(a) Elimination: Easiest to just plug the constraint into the minimisation equation toremove y, and then minimise:

F = x

2 + y

2 = x

2 + (1� x

2)2 = x

2 + 1� 2x2 + x

4 = x

4 � x

2 + 1 (245)

then minimise:dF

dx

= 4x3 � 2x = 0 (246)

giving x = 0, x = ±p

1/2. Need to check what is a min/max, so:

d

2

F

dx

2

= 12x2 � 2 =

(�2, when x = 0 maximum

4, when x = ±p1/2 minimum

(247)

so we get minima at x = ±p

1/2, y = 1/2.

(b) Di↵erentiation: If we couldn’t perform the substitution we can still solve the problem:

dF =@F

@x

dx+@F

@y

dy = 2x dx+ 2y dy (248)

Taking our constraint we can relate our elements: y = 1� x

2, dy = �2x dx so that:

dF = (2x� 4xy) dx (249)

to minimise we then set dF/dx = 0, so that we want to solve 2x�4xy = 0. Plugging inthe constraint we get 2x� 4x(1� x

2) = 0, giving us x = 0,±p

1/2. We can again testfor max/min by finding d

2

F/dx

2 from above. This approach generalises very nicely tomore complex problems, but still requires carrying lots of algebra around...

(c) Lagrange multiplier: If we are finding the minimum of some function, F (x, y),subject to a constraint of the form f(x, y) = const., then we are really only dealing

26


with a single free variable. We can perform a similar di↵erentiation as above for bothof those functions:

dF =@F

@x

dx+@F

@y

dy = 0 (250)

df =@f

@x

dx+@f

@y

dy = 0 (251)

where the former is a result of minimisation (dFdx

= 0, so dF = 0) and the latter due tothe constraint being a constant (df = 0). In b) we plugged in df and dy into dF andsolved, which was gross. Instead we will times df by an undetermined multiplier andadd it to dF :

dF + �df = 0 (252)✓@F

@x

+ �

@f

@x

◆dx+

✓@F

@y

+ �

@f

@y

◆dy = 0 (253)

If we can then cleverly choose � to cancel the second term:

@F

@y

+ �

@f

@y

= 0 (254)

this leaves us with:@F

@x

+ �

@f

@x

= 0 (255)

the two above equations can now be used with f(x, y) = const. to solve for �, x, y. Wehave e↵ectively found a constraint, �, that removes one of the coordinates, leaving usonly with x. This is e↵ectively the same as if we had written:

F

0(x, y) = F (x, y) + �f(x, y) (256)

and take the partial derivatives to equal zero:

@F

0

@x

= 0,@F

0

@y

= 0 (257)

So for our example, F = x

2 + y

2 and f = x

2 + y = 1:

F

0(x, y) = F + �f = x

2 + y

2 + �(y + x

2) (258)

so taking the derivatives gives:

@F

0

@x

= 2x+ 2�x = 0,@F

0

@y

= 2y + � = 0 (259)

which are easy to solve simultaneous equations if we use the constraint. The firstimplies two solutions: A) x = 0, and then from f = x

2 + y = 1 we get y = 1, andform 2y + � = 0 we get � = �2, B) � = �1, so 2y + � = 0 means y = 1/2, andf = x

2 + y = 1 gives x = ±p

1/2. This is the same as before, but much faster, thoughwe still need to find which is a max/min.

27


Problem 27. Constraints on a hemisphere

A particle of mass m sits on a hemisphere of radius a, we want to find the force of constraintand the angle where the particle falls o↵. When the constraint force in r drops to zero iswhen the particle slides o↵ the slope. We start by noting our constraint using the positionof the particle as r, ✓ (measured from vertical):

f(r, ✓) = r � a = 0 (260)

i.e. the particle is constrained to the surface. We can then construct the Lagrangian as:

T =m

2(r2 + r

2

✓

2) (261)

U = mgr cos ✓ (262)

L =m

2(r2 + r

2

✓

2)�mgr cos ✓ (263)

where U(✓ = ⇡/2) = 0. We now need our augmented Lagrange equations:

@L@r

+ �

@f

@r

� d

dt

@L@r

= 0,@L@✓

+ �

@f

@✓

� d

dt

@L@✓

= 0 (264)

The constraint derivatives give:@f

@r

= 1,@f

@✓

= 0 (265)

so that our E-L equations become:

r : mr✓

2 �mg cos ✓ �mr + � = 0 (266)

✓ : mgr sin ✓ �mr

2

✓ � 2mrr✓ = 0 (267)

our constraints then imply we can remove all r derivatives and set r = a:

ma✓

2 �mg cos ✓ + � = 0 (268)

mga sin ✓ �ma

2

✓ = 0 (269)

Want to combine into 1 equation. The second is ✓ = g

a

sin ✓. We want to use this to solve

the second first equation, i.e. to get (✓)2, so integrate it up:

✓ =d

dt

d✓

dt

=d✓

dt

=d✓

d✓

d✓

dt

= ✓

d✓

d✓

(270)

So that: ˆ✓d✓ =

ˆ✓d✓ =

ˆg

a

sin ✓d✓ (271)

integrating both sides:✓

2

2= �g

a

cos ✓ +g

a

(272)

28


where the constant comes in from our constraint that we are initially stationary, and so ✓ = 0means we must be in equilibrium and ✓ = 0. Using this in our r E-L equation gives us:

� = mg cos ✓ �ma

⇣�2

g

a

cos ✓ + 2g

a

⌘= mg(3 cos ✓ � 2) (273)

Recall the constraint force (in r) expression, and sub in � from above:

F

cstr

r

= �

@f

@r

= mg(3 cos ✓ � 2) (274)

As you can see, the constraint force drops as we fall down the sphere (what about at✓ = 0?), such that when it vanishes we fall o↵:

F

cstr

r

= 0 = mg(3 cos ✓0

� 2) (275)

such that:✓

0

= cos�1(2/3) ⇡ 48� (276)

You could generalise this to a non-point mass, i.e. a rolling hoop or sphere. You wouldthen need an additional rotational kinetic energy term and to consider a new constraint; thatof rolling (now needing a �

1

and �

2

, the latter would be akin to a frictional/contact forcecontrolling the rolling). This would lead you to use r, ✓ again but also a � that determinedthe orientation of the object as it rolled (needed for the � term).

29


Problem 28. A physics multiplier: falling disc

Say we have a disc that is wrapped around a string and allowed to fall down. The string ismeasured form the top by y (increasing down), like a yo-yo. Let’s find the forces of constraintof this system. The disc rotates about its CoM by an angle of �, and has a radius a, massm. KE is:

T =1

2my

2 +1

2I

cm

�

2 =1

2my

2 +1

2(0.5ma

2)�2 (277)

where for a solid disc: I

cm

= 0.5ma

2. The potential is simply V = �mgy, where we definey = 0 at the top. So we have:

L = 0.5my

2 + 0.25ma

2

�

2 +mgy (278)

Inspection tells us that our constraint equation is:

f(y,�) = y � a� = 0 (279)

We could have used this to transform our Lagrangian into a function of a single coordinate.But we can’t asses the forces of constraint if we do that. So instead:

@L@y

+ �

@f

@y

� d

dt

@L@y

= 0,@L@�

+ �

@f

@�

� d

dt

@L@�

= 0 (280)

This gives us:y : mg �my + � = 0 (281)

� : �0.5ma

2

�� a = 0 (282)

the constraint is kinda useless as it stands, as our EoM needs double derivatives, sodi↵erentiating it twice gives:

� = y/a (283)

Plugging this into the second, and then the third gives us (3 equations, 3 unknowns):

� = �1

3mg (284)

using this value gives us our EoMs as:

y =2

3g, � =

2

3

g

a

(285)

These are the EoMs for our two generalised, dependent, coordinates. Looking at the y

equation again: my = mg + �, it is clear that � here is the tension in the string, and one ofthe generalised forces of constraint. We can check via our generalised forces equation:

Ny

= �

@f

@y

= �1

3mg (286)

N�

= �

@f

@�

= �1

3mga (287)

First is clearly tension in the rope (-ve as points against the +ve y, which is down), whilethe second is a restraining torque on the disc as it spins around.

30


Problem 29. Multiplier for Atwood machine

We already did Atwood’s machine by noting the degrees of freedom. Now we will use aLagrangian multiplier to obtain the same result. We once again have the setup in thediagram. Writing out the Lagrangian:

L = T � U =1

2m

1

x

2 +1

2m

2

y

2 +m

1

gx+m

2

gy (288)

Now, we have a constraint equation of the form:

f(x, y) = x+ y = const. (289)

Our constrained E-L equations read:

@L@x

+ �

@f

@x

=d

dt

@L@x

,

@L@y

+ �

@f

@y

=d

dt

@L@y

(290)

The derivatives of our constraint are easy, giving 1 in both cases. This gives us:

m

1

g + � = m

1

x (291)

m

2

g + � = m

2

y (292)

Which are a pair of simultaneous equations. They are easily solved if we use our constraintequation, and taking the derivatives so we can plug it into the above:

y = �x (293)

Substitute this in the second and minus the two equations from each other, gives:

x =g(m

1

�m

2

)

m

1

+m

2

(294)

Which is the same as we got before, but that required us to cleverly spot our constraint andwhat the correct generalised coordinate should be resulting from the constraint. Note thatwe can better understand what � is from the NL2 version:

m

1

g � F

t

= m

1

x, m

2

g � F

t

= m

2

y (295)

where clearly our Lagrange multiplier is the tension in the rope:

� = �F

t

= � 2g

m

�1

1

+m

�1

2

⌘ �

@f

@x

= N (296)

31


Part 4: Classical problems with the Lagrangian

Problem 30. Infinitesimal translation and central forces

Consider the central-force Lagrangian with no external forces. Say we add a tiny constanto↵set to our Lagrangian (both bodies) of +✏~a. As this is a tiny o↵set, we can Taylor expandour Lagrangian:

L(~r1

+ ✏~a,~r

2

+ ✏~a, ~r

1

, ~r

2

) = L(~r1

,~r

2

, ~r

1

, ~r

2

) + ✏~a · (~r1

+ ~r2

)L+O(✏2) (297)

In order to impose translational invariance, with other things being non-zero, we musthave that the Lagrangian be invariant under spatial translations, so cannot have a spatialdependance, i.e.

~r1

L+ ~r2

L = 0 (298)

(We saw for Lagrangians with no positional dependence, e.g. L = 0.5mx

2 + 0.5my

2 �mgz,that we get @L/@x = F

x

= 0). In components we have:

@L@x

1

+@L@x

2

= 0,@L@y

1

+@L@y

2

= 0,@L@z

1

+@L@z

2

= 0 (299)

we can write our the E-L equations for both bodies as (just in x):

d

dt

✓@L@x

1

◆=

@L@x

1

,

d

dt

✓@L@x

2

◆=

@L@x

2

(300)

adding together:d

dt

✓@L@x

1

◆+

d

dt

✓@L@x

2

◆=

@L@x

2

+@L@x

1

= 0 (301)

which must be zero given the above. If the canonically conjugate/generalised momentum isgiven by p

x

= @L/@x then:dp

x1

dt

+dp

x2

dt

= 0 (302)

so in general:

~p

x

+ ~p

y

⌘ ~

P

tot

= 0 (303)

32


Problem 31. Lagrangian in non-inertial frame

We initially have a Lagrangian of the form: L0

= m~v

0

2

/2� U . We now want to sub in thenew definition of coordinates with respect to our accelerating frame:

~v

0

= ~v

0 + ~

V (t) (304)

Plugging in we get the Lagrangian in the accelerating frame:

L0 =1

2m~v

0 2 +m~v

0 · ~V +1

2m

~

V

2 � U (305)

There is a trick here is to note that the V (t) is solely a function of time. As such it is possibleto write as the total derivative wrt time. Such terms act purely as boundary terms whenminimising the action, and play no part in the minimisation (by S =

´L dt, they will just

give a constant, and so play no role in dynamics) so can be dropped.

Here is a quick proof of that (LLp4), say we have a Lagrangian:

L0(q, q, t) = L(q, q, t) + d

dtf(q, t) (306)

Looking at the action:

S0 =

ˆt2

t1

L0(q, q, t) dt =

ˆt2

t1

L(q, q, t) dt+ˆ

t2

t1

d

dtf(q, t) dt = S + f(q(t

2

), t2

)� f(q(t1

), t1

) (307)

i.e. the di↵erence in S and S0 is a term with zero variation at the boundaries, so �S0 = �S.

Back to our Lagrangian. The above allows us to drop the V

2 term, and we can re-writethe cross term as:

mV (t) · ~v 0 = m

~

V · d~r0

dt

=d(m~

V · ~r 0)

dt

�m~r

0 · d~

V

dt

(308)

which is just the reverse product rule. We can drop the total derivative term, d(mV · r0)/dt,for the same reasoning as above (it is expressed as d(F )/dt, so integrates away). This givesus:

L0 =1

2m~v

0 2 �m

~

A(t) · ~r 0 � U (309)

where ~

A = d

~

V /dt is the acceleration of the frame. So that the EoM gives:

m

d~v

0

dt

= �@U(r0)

@r

0 �m

~

A(t) (310)

so it merely acts as if an extra force is added, an inertial/centrifugal force, that we are veryfamiliar with!

33


Problem 32. Pendulum stability

There are two points where we negate the torque on a pendulum swinging due to gravity.The combined potential is, with U(✓ = 0) = 0:

U(✓) = mgl(1� cos ✓) (311)

so the generalised force (i.e. torque) is simply:

F✓

=@L@q

= �@U

@✓

= �mgl sin ✓ (312)

implying stability at ✓ = 0, ⇡. Yet motion around each of these points is very di↵erent...Let’s expand this about the two points. For the zero point, we can construct a Lagrangianof the form:

L(✓, ✓) = ml

2

2✓

2 �mgl(1� cos ✓) (313)

we can then Taylor expand the cosine about 0, giving 1� ✓

2

/2 + ✓

4

/24 + ..., so that:

L(✓, ✓) = ml

2

2✓

2 �mlg

✓

2

2+O(✓4) (314)

⇡ mgl

✓

2

l/g

2� ✓

2

2

!(315)

we can smoosh the constant into the Lagrangian, and redefine a new time unit ⌧ = t

pl/g,

so that:

L0(✓ ⇡ 0) =2Lmgl

⇡ ✓

2 � ✓

2 (316)

where we have re-defined ✓ to be in terms of ⌧ instead of t. We can do the same for theupper point. To do so we need to Taylor expand around some small value again, so chooseour point to be around ✓ = ⇡+�✓. Giving cos(⇡+�✓) = �1+�✓

2

/2+O(�✓

4). A similarre-arrangement gives us (recall ✓ = ⇡ + �✓):

L0(✓ ⇡ 0) =2Lmgl

⇡ �✓

2

+�✓

2 (317)

where we dropped the constant as well. Here you can easily see the di↵erence in restoringforce of the e↵ective potential in the Lagrangian, in that it points in the positive directionof displacement for the upper point, but in the opposite direction in the lower point.

34


Problem 33. Green’s function solution to an impulsive force

Our EoM reads:G+G = �(t� t

0) (318)

now, we can integrate it slightly around the point t0 to take advantage of continuity:ˆt

0+✏

t

0�✏

(G+G) dt =

ˆt

0+✏

t

0�✏

�(t� t

0) dt = 1 (319)

Note the RHS must be unity, as we are integrating the delta function over the entire domainit resides in, so by definition must be unity. The left-most term is easy!ˆ

t

0+✏

t

0�✏

G dt = G(t0 + ✏)� G(t0 � ✏) (320)

This is in e↵ect the amount of the discontinuous “jump” in G at t0, i.e. is Gleft

� G

right

. Theright term in the integral is a bit trickier:ˆ

t

0+✏

t

0�✏

G(t) dt ⇡ˆ

t

0+✏

t

0�✏

(G(t0) + (t� t

0)G(t0) + ...) dt = 2✏G(t0) (321)

where we have Taylor expanded around the su�ciently small ✏, and then we can neglect allbut the first terms, leaving us with [tG(t0)]. We have already stated that this term mustvanish, due to continuity (see slide) when ✏ ! 0. This leaves us only with the first timederivative (i.e. velocity) with a non-zero value.

i.e. for t = t

0, we have the jump in the “velocity” term: G

left

� G

right

= 1. And for allother times, t 6= t

0, we must obey the homogeneous solution to G+G = 0.

Problem 34. Green’s general solutions

Ok, so we have this thing:

q(t) =

ˆt

�1F (t0)G(t� t

0) dt0 (322)

how do we use it? The Green’s function is by definition simply the solutions we were justseeking, i.e. the impulse/delta function driving force. So, for ANY arbitrary forcing termwe can work out what q should be! E.g. for critical damping we had:

G

crit

(t� t

0) = (t� t

0)e�(t�t

0) (323)

so, supremely simply, we have:

q

crit

(t) =

ˆt

�1F (t0)(t� t

0)e�(t�t

0)

dt

0 (324)

and that’s it! The same for our other two solutions:

q

over

(t) =

ˆt

�1F (t0)

⌧

+

⌧�

⌧

+

� ⌧�

he

�(t�t

0)/⌧+ � e

�(t�t

0)/⌧�idt

0 (325)

q

under

(t) =

ˆt

�1F (t0)

1

!

0 e�(t�t

0)/2Q sin[!0(t� t

0)] dt0 (326)

35


Problem 35. Superposition of waves

a) Say we have a damped SHO solution and we want q(t) if is driven by a square/boxwave of a unit magnitude for a period T for [�T/2, T/2], where q(t > T/2). We can startwith the general solution for the STEP FUNCTION driving force, when initialising FROMREST (we derived this from matching BCs in lectures).

q

0

(t) = 1� e

�t/2Q

✓cos!0

t+1

2Q!

0 sin!0t

◆(327)

where velocity is initially zero, and !

0 =p

1� 1/4Q2. Our pulse force e↵ectively has a formthat is a combination of two heavyside steps:

F (t) = ✓(t+ T/2)� ✓(t� T/2) (328)

So, logically we can maybe take the sum of our two solutions after being translated inaccordance with the step function i.e.:

q(t) =

8><

>:

0, t < �T/2

q

0

(t+ T/2), �T/2 < t < T/2

q

0

(t+ T/2)� q

0

(t� T/2), t > T/2

(329)

We can check the solution is continuous in position by looking at the jump just beforet = T/2 (i.e. just after forcing turns o↵).

limt!T/2

q�(t) = q

0

(t+ T/2)|t=T/2

= q

0

(T ) (330)

and looking at just after t = T/2:

limt!T/2

q

+

(t) = q

0

(t+ T/2)� q

0

(t� T/2)|t=T/2

= q

0

(T )� q

0

(0) = q

0

(T ) (331)

i.e. the function is continuous in position. We can do again to show q(t) is as well, whichsimilarly relies on q(0) = 0. Thus, evaluating our q

0

(t), we get4:

q(t) = � e

�(t+T/2)/2Q

✓cos!0(t+ T/2) +

1

2Q!

0 sin!0(t+ T/2)

◆(332)

+ e

�(t�T/2)/2Q

✓cos!0(t� T/2) +

1

2Q!

0 sin!0(t� T/2)

◆(333)

4Though H&F has a weird 1/(!0 2 +1/4Q2) ⌘ 1 attached to the whole thing probably there to help withc), and is missing a minus sign.

36


b) Now, we checked the solution at +T/2, what about at �T/2, and does it give anydiscontinuity? We have shown that q and q are continuous, so in the EoM:

q + 2q + q = F (t) = ✓(t+ T/2)� ✓(t� T/2) (334)

we have a jump in the acceleration, which can be seen for the above evaluated at t = �T/2,we get ✓(0)� ✓(�T ) = 1� 0, so has a height of 1 (though depending on your definition, canbe ✓(0) = 1/2).

c) Finally, let’s consider a very small pulse; T ! 0. If so:

q(t± T/2) ⇡ q(t)± T

2q(t) + ... (335)

so, for our step response:

q(t) = q

0

(t+ T/2)� q

0

(t� T/2) ⇡ (q0

(t) +T

2q

0

(t))� (q0

(t)� T

2q

0

(t)) = T q

0

(t) (336)

q(t) =T

2Qe

�t/2Q

✓cos!0

t+1

2Q!

0 sin!0t

◆� Te

�t/2Q

✓�!

0 sin!0t+

!

0

2Q!

0 cos!0t

◆(337)

= Te

�t/2Q

✓1

4Q2

!

0 sin!0t+ !

0 sin!0t

◆=

T

!

0 e�t/2Q sin!0

t (338)

but wait! recall our Green’s function for an impulse:

G(t� t

0) =

(0, t� t

0< 0

1

!

0 e�(t�t

0)/2Q sin(!0(t� t

0)), t� t

0 � 0(339)

i.e. we get, as T ! 0:q(t) ! T q(t) = TG(t� t

0) (340)

37

part 1: calculus of variationsastro3.sci.hokudai.ac.jp/~alex/teaching_materials/examplescm2.pdf ·...

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