Seminar 5: LAGRANGE MULTIPLIERS
Problem 19.
A sphere of radius a and mass m rolls over the surface of a sphere of radius b > a.
Where the first sphere will leave the second one if initially it was slightly knocked in
the top position?
Solution
Choosing the generalized coordinates r, θ, ϕ as shown in Figure, we may write the
Lagrangian as
1
1
1
L = mṙ2 + mr2 θ̇2 + Im ω 2 − mgr cos θ.
(1)
2
2
2
Here ω is the angular frequency of the rotation of a small sphere which can be
expressed in terms of the generalized coordinates ϕ and θ according to
ω = ϕ̇ + θ̇,
(2)
since the total angle of rotation of this sphere is equal ϕ + θ.
Next we notice that to describe the configuration of the system we introduced three
generalized coordinates though the system has one degree of freedom. Therefore,
there are two conditions of constraint. They follows from the equalities
r =a+b
aϕ = bθ.
(3)
By inroducing the functions of the generalized coordinates
f1 = a + b − r
f2 = aϕ − bθ,
(4)
we can rewrite Eqs. (4) in the standard form of the equations of constraint
f1 = 0
.
f2 = 0
(5)
The generalized forces of constraint are thus

∂f2
∂f1


 Qr = λ1 ∂r + λ2 ∂r = −λ1
Q =λ
∂f1
+λ
∂f2
= −λ b
θ
1 ∂θ
2 ∂θ
2


 Q = λ ∂f1 + λ ∂f2 = λ a,
ϕ
1 ∂ϕ
2 ∂ϕ
2
where λ1 and λ2 are the Lagrange multipliers.
(6)
Substituting these forces into Lagrange’s equations we obtain the set of three
equations of motion

2

 mr̈ − mr θ̇ + mg cos θ = −λ1


mr2 θ̈ + Im (θ̈ + ϕ̈) − mgr sin θ = −λ2 b
Im (θ̈ + ϕ̈) = λ2 a
(7)
On the other hand, from the equations of constraint we have
b
r = a + b → r̈ = 0 aϕ = bθ → ϕ̈ = θ̈.
a
(8)
Then the last equation in (7) gives
λ2 = Im
1
b
1
2
1 + θ̈ = Im 2 (a + b)θ̈ = m(a + b)θ̈.
a
a
a
5
(9)
Here the use of the expression for the moment of inertia for a sphere,
2
Im = ma2 ,
5
(10)
has been made. Now the second equation can be transformed as
m(a + b)2 θ̈ + 25 ma2 (a + b) a1 θ̈ − mg(a + b) sin θ = − 52 m(a + b)θ̈b
(a + b)θ̈ + 52 aθ̈ − g sin θ = − 25 bθ̈
7
(a
5
+ b)θ̈ − g sin θ = 0.
(11)
Multiply this equation by θ̇ to obtain
7
(a + b)θ̇θ̈ − g θ̇ sin θ = 0
5
(12)
7
1d 2
d
(a + b)
(θ̇ ) = −g cos θ.
5
2 dt
dt
(13)
or
It follows
θ̇2 = −
10g
(cos θ − 1),
7(a + b)
(14)
and the first of Eqs. (7) gives
λ1 =
10
17
10
mg(1 − cos θ) − mg cos θ = mg(1 −
cos θ).
7
7
10
(15)
Therefore the generalized force Qr is equal
Qr = −λ1 =
10
17
mg( cos θ − 1).
7
10
(16)
Note that at θ = 0, i.e. at the top of the larger sphere, this force is positive. As θ
increase, the force decreases, and at some angle θ = θ0 it becomes zero. What might
happen later? The answer is clear, for by definition the normal force which is exerted
by the fixed sphere on the moving one can only be directed outward, never inward.
In means nothing but the stop of rolling of a smaller sphere and the beginning of
its flying off the fixed one. So the answer to the question in the statement of the
problem follows from the condition
λ1 = 0,
(17)
which in view of Eq. (15) yields
θ = arccos
10
≈ 54◦ .
17
(18)
Note that this angle is independent of neither the radius of moving sphere radius a,
nor the radius of the fixed one b.
Problem 20.
Let stand a homogeneous plain bar of the mass m and length 2l in an unstable
position near a vertical wall as shown in Figure from the Set of Problems. Under
some disturbance the bar starts to move in the vertical xy-plane. When the bar will
loose the contact with the vertical wall ? Assume that the system is free of friction.
Solution
Let (xB , yB ) be the Cartesian coordinates of the center of mass of the bar, B, and
θ is the angle of rotation of the bar about its center of mass. Then the potential
energy of the bar is
V = mgyB ,
(19)
while the kinetic energy consists of the contributions from the motion of the center
of mass and rotation about the center of mass, that is
1
1
T = m(ẋ2B + ẏB2 ) + I θ̇2 ,
2
2
(20)
where I is the momentum of inertia of the bar,
1
I = ml2 .
3
(21)
In principle, the coordinates (xB , yB ) can be expressed in terms of the angle θ as
xB = l sin θ
,
yB = l cos θ
(22)
and the problem could be solved by using a single generalized coordinate θ in accordance with the existence of only one degree freedom. However, we can obtain
the answer to the problem in an essentialy simpler way by choosing two generalized
coordinates, x = xB and θ. With this choice, the first of Eqs. (22) will serve as an
equation of constraint,
x = l sin θ.
(23)
f (x, θ) = x − l sin θ = 0,
(24)
In standard notation,
so that
∂f
∂f
= 1,
= −l cos θ,
∂x
∂θ
and the corresponding generalized forces are
Qx = λ
∂f
∂f
= λ Qθ = λ
= −λl cos θ.
∂x
∂θ
(25)
(26)
To obtain Lagrange’s equations with these generalized forces we rewrite the potential and kinetic energies, given by Eqs (19) and (20) , in terms of (x, θ) as
V = mgl cos θ
(27)
1
1
T = m(ẋ2 + l2 sin2 θθ̇2 ) + I θ̇2 .
2
2
(28)
1
1
L = T − V = m(ẋ2 + l2 sin2 θθ̇2 ) + I θ̇2 − mgl cos θ,
2
2
(29)
and
Hence the Lagrangian is
and Lagrange’s equations,



d ∂L
dt ∂ ẋ d ∂L
dt ∂ θ̇
−
−
∂L
∂x
∂L
∂θ
=λ
= −λl cos θ,
(30)
take the form
mẍ = λ
d
(ml2 sin2 θ θ̇ + I θ̇) − ml2 sin θ cos θ θ̇2 − mgl sin θ = −λl cos θ
dt
(31)
Note that the first of this equations immediately gives the criterium for loosing
the contact between the bar and the wall: it happens if
λ = 0.
(32)
Hence to find the answer to our problem we must only express λ as a function of θ.
A great advantage of using the Lagrange’s multiplier method is that it can be done
without direct solving the equations of motion. The procedure is as follows. Use the
equation of constraint (23) to calculate
ẋ = l cos θ θ̇
ẍ = −l sin θ θ̇2 + l cos θ θ̈
ml(cos θ θ̈ − sin θ θ̇2 ) = λ
ml2 cos θ(cos θ θ̈ − sin θ θ̇2 ) = λl cos θ.
(33)
Substituting the final result of this calculation into the second equation of motion
(31) leads after some algebra to the equation
3g
sin θ.
4l
Starting from this equation we can easily calculate also the quantity θ̇2 :
θ̈ =
2θ̇θ̈ = 2
d
(θ̇2 )
dt
3g
4l
(34)
sin θ θ̇
cos θ
= 2 dtd − 3g
4l
θ̇2 = 2 − 3g
cos θ + C.
4l
(35)
From initial condition (θ̇ = 0 at θ = 0) it follows
C=
3g
,
2l
(36)
and we thus have
3g
(1 − cos θ)
(37)
2l
Finally, substituting Eqs. (35) and (37) into the expression for λ from Eq. (33),
θ̇2 =
we obtain
λ = ml(cos θ θ̈ − sin θ θ̇2 )
sin θ − sin θ 3g
(1 − cos θ)
= ml cos θ 3g
4l
2l
= mg sin θ
3
4
cos θ − 32 + 23 cos θ
= 34 mg sin θ(3 cos θ − 2).
(38)
It follows that λ takes zero values when
θ = 0 or 3 cos θ = 2.
(39)
The first value is out of our interest since it coincides with the starting value of θ,
while the second equality determines the angle
2
θ = arccos ,
3
which we search for.
(40)