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Chapter 10 – Lecture I
Lagrangian Mechanics
For a system of N particles, the Lagrangian is
defined as:
( i i )
N
i
L = (^) ∑ T − V
= 1
where Ti (the Kinetic Energy) and Vi (the Potential
Energy) are to be expressed in terms of convenient
coordinates.
T 1 for example might be:
( )
2 1
2 1
2 1 1 2
(^1) m x& + y& +z& or possibly
( )
2 1 1
2 2 1
2 1
2 1
2
2 1 1 sin
(^1) m r& + r θ& +r θ ϕ&
They do not have to be orthogonal coordinates.
Often there are constraints present in a problem so
that there are less than 3N independent
coordinates.
The general procedure to work a problem is to first
pick a suitable set of generalized coordinates that
uniquely specifies the systems configuration.
Second, calculate the Kinetic energy in terms of
these generalized coordinates. Third, calculate the
potential energy of each particle in terms of the
generalized coordinates.
Once you have T & V , you combine them to have
L , then compute
K
i q
L
dt
d
q
L
i i
This gives a set of D.E. that are called the equations
of motion.
Example: Two dimensional pendulum of length l
with mass m.
( )
2 2 2
2
T = 1 mr& +r θ& but r = constant so only
one coordinate is needed
to completely specify the
locates of m.
2 2
2
T =^1 ml θ &
V = −mglcos θ
= θ + cos θ 2
L 1 ml^2 &^2 mgl
This gives the velocity in terms of the generalized
velocities:
t
x q q
x x
i j j
i
m
j
i ∂
=
1
To get the KE one will need to sum up terms that
have
2 x& i in them which will give terms with
q& (^) j , q&j ,
2 and a term without q& (^) j's in them.
Let’s look at two more examples.
Textbook 10.3. Find the acceleration of a solid
uniform sphere rolling down a perfectly rough fixed
incline plane.
2 2
2
T = m x& + Iw
x R
ϕ=( x 0 −x) R, ϕ& =−^1 &
2
5
w = ϕ&, I = mR
θ
x
Y
This is point of contact
V = mgY
L = mx + Iϕ − mgY
2 2
2
−^ θ
= + sin 5
2 2 2 mgx R
x mx mR
−^ θ
= + sin 5
mx& mgx
= − sin θ 10
L mx& mgx
Now the equation of motion in
mg m x dx
L
dt
d
x
L
⇒− sin^ θ= ⋅
sin θ 7
∴ x&& =− g
y& = R&sin θ+Rcosθ θ&
( )
2 2 2
2
KE = mR&^ +R θ&
= ( + θ ) − sinθ 2
L mR& R & mgR
The equation of motion is
R
L
dt
d
R
L
(m R) dt
d mR θ&^ −mgsinθ= &
2
Now use the constant θ =wt and θ&^ =w.
mRw − mgsinwt =m R&&
2
R Rw gsin wt
2 ∴ &&− =−
To solve this D.E. we first solve
2 R&&^ − Rw = which has a solution of
wt wt R Ae Be
− = + or R =acosh wt+bsinhwt
then for a particular solution:
Rp =αsin wt
R Rw w sin wt w sinwt gsin wt
2 2 2 &&− =−α −α =−
or
2 2 2 αw = g, α =g 2 w
This gives us a solution of
wt w
g R a wt b wt sin 2
cosh sinh 2
Now to meet the boundary conditions:
at t = 0 , let R = x 0 so we have
x 0 = a+ 0 + 0 ⇒a=x 0 , also at t = 0 , letting R= 0
gives
cos 0 2
sinh 0 cosh 0 (^0 ) w w
g R& =aw +bw +
w
g bw 2
0 = + or 2 2 w
g b =−
wt w
g wt w
g R x wt sin 2
sinh 2
cosh (^0 )
Here we have solved the equation of motion to get
R ( t).
