Generalized Momentum
Starting with ΔKE = Work = F·dr = (dp/dt)·dr = dp·(dr/dt) = v·dp = v·mdv =
mv·dv = Δ(½mv²) = p·dv = ΔT , we get: px = T/x’ , where vx = dx/dt = x’ .
In rectangular coordinates:
T = Σi[½mi(xi’² + yi’² + zi’²)],
pxi = T/xi’ = mixi’ ;
(similar for y&z).
In spherical coordinates:
T = Σi[½mi(ri’² + ri²θi’² + r²sin²(θ)φi’²)],
pθi = T/θi’ = miri²θi’ = Ii = Liθ
pri = T/ri’ = miri’,
(where  = θi’)
[so pθi = Liθ];
and a similar result happens for pφi. Thus, using our expression for the kinetic energy, T, in
generalized coordinates: T = ΣkΣl{½Akqk’q’} + ΣkBkqk’ + To we can define a
generalized momentum as:
pj = T/qj’ = /qj’[ΣkΣ{½Akqk’q’} + Σk{Bkqk’} + To]
= Σ{½Ajq’} + Σk{½Akjqk’} + Bj
(but in first sum,  is dummy, replace by k; in second sum k is dummy, but keep k; also Akj=Ajk)
= ΣkAjkqk’ + Bj = pj
(which is our expression for the generalized momentum!)
Generalized Force
We will now use ΔW = F·ds to define a generalized FORCE. But first, let's define a virtual
displacement: δxi = Σk{(xi/qk)δqk } where we have NO EXPLICIT CHANGE WITH TIME
(no xi/t).
δW = Σi[Fixδxi + Fiyδyi + Fizδzi]
= Σi[Fix{Σk(xi/qk)δqk} + Fiy{Σk(yi/qk)δqk} + Fiz{Σk(zi/qk)δqk}]
= Σi[Σk{Fix(xi/qk)δqk} + Σk{Fiy(yi/qk)δqk} + Σk{Fiz(zi/qk)δqk}] .
We now DEFINE Qk = Σi[Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)]
where Qk is the generalized force component corresponding to the qk coordinate so that
δW = ΣWk = Σk[Qkδqk] where δWk is the virtual work when you move only in the qk
direction (still no time). Qk, Fix, ... are NOT considered to change while making the change of
δxi or δqk. Note: to find Qk, we need to know Fix and we need to know xi as a function of the qk.
--------Example:
For polar coordinates for a single particle (so no need to sum over i), show Qr = Fr, and see what
Qθ is and see if you can identify it.
y
F
Fy
Fx
x
= θF
direction of force
y
Fxr
Fyθ
Fxθ
Fy
Fyr
Fx
x
= θ direction for location
Fx = F cos(θF) and Fy = F sin(θF) where F is the magnitude and θF is the direction of the force.
Fxr = Fx cos(θ)
and
Fxθ = -Fx sin(θ) where  is the direction for the location of the force;
Fyr = Fy cos(90o-θ) = Fy sin(θ) , and
x = r cos(θ)
and
Fyθ = Fy sin(90o-θ) = Fy cos(θ) ; also
y = r sin(θ).
Qk = Σi[Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)]
In our case, with only one particle and 2-D, we have:
Qk = Fx(x/qk) + Fy(y/qk) , so
Qr = Fx(x/r) + Fy(y/r) = Fx cos(θ) + Fy sin(θ) =
Fxr + Fyr = Fr
Qθ = Fx(x/θ) + Fy(y/θ) = Fx (-r sin(θ)) + Fy (r cos(θ)) = r[Fxθ + Fyθ] = r Fθ = torque.
---------
IF we have conservative forces so the F = -V, where V is the potential energy, then (note a
change in virtual work implies NO change in kinetic energy since there is no change in time and
hence no change in distance for the particle; to review the del operator, see
http://www.cbu.edu/~jholmes/P380/del.html ): δW = -δV (where V = V(q1, q2, ..., qn) , and
δW = Σk[Qkδqk] from above , and
-δV = -Σi{(V/xi)δxi + (V/yi)δyi + (V/zi)δzi} = -Σk(V/qk)δqk
so we can write: Qk = -V/qk ; or alternatively:
V/qk = Σi[(V/xi)(xi/qk) + (V/yi)(yi/qk) + (V/zi)(zi/qk)]
= Σi[(-Fix(xi/qk) + (-Fiy(yi/qk) + (-Fiz(zi/qk)] = -Qk .
An Equation of Motion
Let's start with Newton's Second Law:
ΣF = dp/dt,
Eq.1
where we already have pk = T/qk’ by definition. In rectangular coordinates for T we have:
pk = T/qk’ = /qk’[Σi{½mi(xi’² + yi’² + zi’²)}
= Σi{½mi[2xi’(xi’/qk’) + 2yi’(yi’/qk’) + 2zi’(zi’/qk’)]}
or
pk = T/qk’ = Σi{mi[xi’(xi’/qk’) + yi’(yi’/qk’) + zi’(zi’/qk’)]} .
Eq. 2
Now let's consider xi’ = dxi/dt where
[since xi = xi(q1, q2, ..., qn,t), we have for dxi = Σ (xi/q)dq + (xi/t)dt]
xi’ = dxi/dt = (1/dt){Σ (xi/q)dq + (xi/t)dt} = Σ (xi/q)q’ + xi/t
and (using the above for xi’):
xi’/qk’ = /qk’{Σ (xi/q)q’ + xi/t} = xi/qk
(since xi depends only on the qk and not the qk’; and since only when =k in the sum do we have
a match!). Note then that
xi’/qk’ = xi/qk .
Eq. 3
-------Example: For polar coordinates: x = r cos(), and
x’ = dx/dt = d(r cos()/dt = (dr/dt)*cos() + r*(d cos()/d)*(d/dt) = r’ cos() – r’sin()
where r’ cos() is the x component of the radial speed and – r’sin() is the x component of the
tangential speed.
Therefore x/r = cos() and x’/r’ = cos() . .
-------Now we can use Eq. 3, xi’/qk’ = xi/qk , in the Eq. 2 expression for T/qk’
pk = T/qk’ = Σi{mi[xi’(xi’/qk’) + yi’(yi’/qk’) + zi’(zi’/qk’)]} .
Eq. 2
which was used in Eq. 1 for pk , and so we can get something for dpk/dt as an equation of motion:
dpk/dt = d/dt[Σimi{xi’(xi/qk) + yi’(yi/qk) + zi’(zi/qk)}]
= Σimi{xi’’(xi/qk) + yi’’(yi/qk) + zi’’(zi/qk)}
+ Σimi{xi’d/dt(xi/qk) + yi’d/dt(yi/qk) + zi’d/dt(zi/qk)} .
Eq. 4
In the first half of the expression, mixi’’ can be replaced by Fxi (and same for the y and z parts).
But we also need to consider what d/dt(xi/qk) is:
d/dt(xi/qk) = 1/dt [Σ{/q (xi/qk)}dq + /t(xi/qk)dt]
= Σ{/q (xi/qk)}q’ + /t(xi/qk)
= /qk{Σ [(xi/q)q’ + xi/t} = xi’/qk .
(In other words, you can simply bring the d/dt inside the /qk.)
Thus we can now write dpk/dt as:
dpk/dt = Σi{Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)}
+ Σi{mi[xi’(xi’/qk) + yi’(yi’/qk) + zi’(zi’/qk)}
Eq. 5
The first sum in the expression is what we defined as Qk.
The lower half can be identified as T/qk:
T/qk = [Σi½mi(xi’²+yi’²+zi’²)]/qk = Σi½mi{2xi’(xi’/qk) + 2yi’(yi’/qk) + 2zi’(zi’/qk)}
which (when you cancel the ½ out front with each of the 2's) is the same as the lower half of Eq. 5.
Thus we can write for Eq. 5: dpk/dt = Qk + T/qk .
And since we can write pk in terms of T as well: pk = T/qk’, we get one form of:
Lagrange's Equations:
d/dt[T/qk’] - T/qk = Qk
which is a generalized equation of motion! [Note that in rectangular coordinates, T/qk = 0
since T does not depend on the coordinates, only on the velocities.]
Another form of Lagrange's equations can be written IF all the forces are conservative so that
Qk = -V/qk, AND IF V is not a function of velocities, only positions, so V/qk’ = 0. In this
case we define what is called the: LAGRANGIAN: L = T - V .
Then the Lagrange's equation of motion is:
d/dt[L/qk’] - L/qk = 0 .
--------EXAMPLE:
Write Lagrange's Equations of Motion for plane polar coordinates and interpret what it
says.
Qk = d/dt[T/qk’] - T/qk
Where in rotating plane polar coordinates we have already seen (last section on Generalized
Coordinates):
T = ½ mr’2 + ½ mr2(θ’+ω)2
Here we do not have a rotating frame so ω = 0, so T = ½ mr’2 + ½ mr2θ’2
for qk = r:
T/r’ = mr’ , and then d/dt[T/r’] = d/dt[mr’] = mr’’ (which is the
regular radial acceleration;
T/r = mrθ’2 (where rθ’2 is the circular acceleration in the negative radial direction).
Therefore, we have Qr = mr’’ - mrθ’2 = Fr .
for qk = :
T/θ’ = mr2θ’ , and then d/dt[T/θ’] = d/dt[mr2θ’] = 2mrr’θ’ + mr2θ’’
2
(where the mr θ’’ = Iα is the moment of inertia times the tangential acceleration, and the
2mrr’θ’ is the radius times the coriolis term);
T/θ = 0 ,
Therefore, we have Qθ = mr2θ’’ + 2mrr’θ’ – 0 = Fθ r = torque.
---------
WRITTEN HOMEWORK PROBLEM #5: (Problem 9-1 in Symon)
Consider the coordinates u and w that are defined in terms of the polar coordinates r and θ:
u = ln(r/a) – θ cot(ζ)
and
w = ln(r/a) + θ tan(ζ)
where a and ζ are constants.
Sketch the curves of constant u and w: this is already done for you - see next page
a) Find the kinetic energy for a particle of mass m in terms of u, w, du/dt, and dw/dt .
b) Find expressions for Qu and Qw in terms of the polar force components Fr and Fθ .
c) Find pu and pw .
** d) extra credit: find the forces Qu and Qw required to make the particle move with
constant speed, ds/dt, along a spiral of constant u=uo .
HINT: since the transformation equations are in terms of r and θ (polar form), you can start with
the expressions for KE, Qi and pi in polar form rather than in rectangular form.
WRITTEN HOMEWORK PROBLEM #6: (Problem 9-3 in Symon)
For plane (2-D) parabolic coordinates f and h defined such that: x = f – h and y = 2(fh)1/2
(see the excel spreadsheet accessed from the course web page to see an image),
a) find the expression for the kinetic energy in terms of f, h, f’, and h’;
b) find the expression for the momenta: pf and ph ;
c) write out the Lagrange equations in these coordinates if we assume the particle is not acted on
by any force.
REVIEW
generalized coordinates
qi(x1, y1, ..., zN, t)
x1(q1, q2, ..., qn, t) where n = 3N
polar example
r = (x²+y²)
x = r cos(θ)
x1’ = dx1/dt = Σ(x1/qi)qi’ + x1/t
x’ = r’ cosθ + -rθ’ sin(θ)
Kinetic Energy:
T = ΣkΣ½Akqk’q’ + ΣkBkqk’ + To
T = ½mr’2 + ½mr²θ’²
Ak = Σi{mi[(xi/qk)(xi/q) + (yi/qk)(yi/q) + (zi/qk)(zi/q)]}
Bk = Σi{mi[(xi/qk)(xi/t) + (yi/qk)(yi/t) + (zi/qk)(zi/t)]}
To = Σi{½mi[(xi/t)² + (yi/t)² + (zi/t)²]} .
Arr = m, Arθ = Aθr = 0, and Aθθ = mr²;
Bk = 0; To = 0.
NOTES: Ak = Ak .
If Ak = δk (where δk = 1 if k= and δk = 0 - called the Dirac delta function),
then the coordinates are said to be orthogonal.
If xi/t = 0, then Bk = 0 and To = 0.
Polar coordinates are orthogonal.
Generalized Momentum:
pj = T/qj’ = ΣkAjkqk’ + Bj
pr = mr’; pθ = mr²θ’ = Lθ.
virtual work: δW = ΣiFixδxi = ΣiQiδqi
δW = Frδr + rFθδθ (τ = rFθ)
where Qk = Σi[Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)]
Qr = Fr, Qθ = rFθ = τ .
Equation of motion:
d/dt(T/qk’) - T/qk = Qk
Qr = mr’’ – mθ’²r = Fr
Qθ = mr2θ’’ + 2mrr’θ’ = Fθ r = torque
If the forces are conservative: V = V(qi) so V/qi = -Qk, and if V/qk’ = 0 then we
DEFINE the Lagrangian: L = T - V to get:
d/dt[L/qk’) - L/qk = 0 .
---------Example: Bead on a rotating hoop
Axis of
ω
rotation
r
θ
contact
force
Fcφ 
mg
Ffθ =
friction
Fcr =
contact
force
Rotating spherical coordinates:
x = r sin(θ) cos(φ+ωt)
y = r sin(θ) sin(φ+ωt)
z = r cos(θ)
NOTE: plus z direction is DOWN and
 is measured from the down direction as shown
2
T = ½ mr’ + ½ mr2θ’2 + ½ mr2sin2()(φ’+ω)2
Qr
=
Fcr + mg cos(θ)
(contact force in the radial direction plus radial component of gravity)
[see Appendix A for a detailed derivation of this term starting from the
definition of Qk]
Qθ
=
r[Ffθ – mg sin(θ)]
(frictional force along the loop plus component of gravity along the loop times the
distance to the center)
Qφ = r sin(θ)[Fcφ]
(contract force in the rotating direction times the distance to the axis which is r sin(θ).)
Lagrange's Equations:
d/dt[T/qk’] - T/qk = Qk
Radial equation:
d/dt[T/r’] - T/r = Qr
gives
d/dt[mr’] - mrθ’2 - mrsin2()(φ’+ω)2
mr’’ - mrθ’2 - mrsin2(φ’+ω)2
=
=
Fcr + mg cos(θ)
Fcr + mg cos(θ).
, or
In steady state (where r’ = 0, θ’ = 0, and φ’ = 0), we have
Fcr
=
- mg cos(θ) - mrsin2(θ) ω2
Where the first term is the force holding up the radial component
of the weight (towards the center if θ is less than 90o), and the
second term is the term balancing the “centrifugal term” and is
always towards the center (since θ is limited to between zero and
180o). Note that [r sin(θ)] in the centripetal acceleration term
is the radial distance to the axis, and the second sin(θ) factor
in the centripetal acceleration term comes from this being the
radial component of the centripetal acceleration (which is
directed away from the axis, not the center).
Theta equation (along the loop):
d/dt[T/θ’] - T/θ = Qθ
gives
(with Qθ = rFfθ)
d/dt[mr2θ’] – mr2sin(θ)cos(θ)(φ’+ω)2
=
rFfθ - mgr sin(θ),
2mrr’θ’ + mr2θ’’ – mr2sin(θ)cos(θ)(φ’+ω)2
=
rFfθ - mgr sin(θ) .
or
In steady state (where r’ = 0, θ’ = 0, and φ’ = 0), we have
Ffθ
=
mg sin(θ) – mr sin(θ)cos(θ) ω2
.
If there is no friction, then Ffθ = 0 which requires
mg sin(θ) = mr sin(θ) cos(θ) ω2 , or
g
= r cos(θ) ω2
,
or
cos(θ) = g /{ω2r}.
If g > ω2r, there is no solution to this, so we have to go back
to the equation before we divided out the sin(θ) factor. For
this equation to be true, sin(θ) must be zero, so θ must be zero,
and the ball will remain at the base of the rotating hoop.
However, if ω2r > g , then the ball will rise up to an angle θ
such that θ = cos-1[g /{ω2r}.
There is a φ equation, but we won’t bother to do that here. We
know that in the steady state, there should be no Fcφ required.
However, if we are starting up or slowing down the speed of the
hood (ω is changing), then there will be a contact force, Fcφ ,
required.
-----------
Appendix A: Derivation of Qr from basic principles:
To find Qr, we could start with the definition of generalized force:
Qk = Σi[Fix(xi/qk) + Fiy(yi/qk) + Fiz(zi/qk)]
Here we have only one particle, so we don’t need the summation over i or the i subscripts.
x = r sin(θ)cos(φ+ωt)
y = r sin(θ)sin(φ+ωt)
z = r cos(θ)
so
(x/r) = sin(θ)cos(φ+ωt)
(y/r) = sin(θ)sin(φ+ωt)
(z/r) = cos(θ).
Fx = Fcr-x + Ff-x + Fc-x
Fy = Fcr-y + Ff-y + Fc-y
Fz = Fcr-z + Ff-z + Fc-z - mg
where
Fcr-x = Fcr [sin(θ)cos(φ+ωt)]
Fcr-y = Fcr [sin(θ)sin(φ+ωt)]
Fcr-z = Fcr [cos(θ)]
Ff-x = Ff [sin(θ+90o)cos(φ+ωt)]= Ff [cos(θ)cos(φ+ωt)]
Ff-y = Ff [sin(θ+90o)sin(φ+ωt)]= Ff [cos(θ)sin(φ+ωt)]
Ff-z = Ff [cos(θ+90o)]= -Ff [sin(θ)]
Fc-x = Fc [sin(θ)cos(φ+ωt+90o)]= Fc [-sin(θ)sin(φ+ωt)]
Fc-y = Fc [sin(θ)sin(φ+ωt+90o)]= Fc [sin(θ)cos(φ+ωt)]
Fc-z = 0 (since the φ direction is always horizontal)
so Qr =
+
+
Fcr [sin(θ)cos(φ+ωt)] * sin(θ)cos(φ+ωt)
Ff [cos(θ)cos(φ+ωt)] * sin(θ)cos(φ+ωt)
Fc [-sin(θ)sin(φ+ωt)] * sin(θ)cos(φ+ωt)
(term 1)
(term 2)
(term 3)
+
+
+
Fcr [sin(θ)sin(φ+ωt)]* sin(θ)sin(φ+ωt)
Ff [cos(θ)sin(φ+ωt)] * sin(θ)sin(φ+ωt)
Fc [sin(θ)cos(φ+ωt)] * sin(θ)sin(φ+ωt)
(term 4)
(term 5)
(term 6)
+
+
Fcr [cos(θ)]* cos(θ)
Ff [-sin(θ)]* cos(θ)
(term 7)
(term 8)
-
mg * cos(θ).
(term 9)
If we add terms 1, 4, and 7, we get Fcr (using the pathagorean
theorem twice).
If we add terms 2, 5, and 8, we get zero.
If we add terms 3, 6, we get zero.
Therefore, we get Qr = Fcr – mg cos(θ).