Lagrange multipliers
Problem:
A heavy particle with mass m is placed on top of a vertical hoop. Calculate
the reaction of the hoop on the particle by means of the Lagrange undetermined
multipliers and Lagrange's equations. Find the height at which the particle
falls off.
Solution:
 Concepts:
Lagrange's Equations,
Lagrange multipliers
d/dt(∂L/∂(dq_{k}/dt))  ∂L/∂q_{k} = ∑_{l}λ_{l}a_{lk}, Σ_{k} a_{lk
}dq_{k
}+ a_{lt }dt = 0.
 Reasoning:
The problem requires us to use the method of Lagrange multipliers.
Imagine the particle to be constrained to move on the hoop. For a small θ,
the force of constraint points away from the origin. For a large θ it
points toward the origin. The angle for which the force of constraint is
zero is the angle at which the particle falls off. To find the force of
constraint as a function of θ, we use the technique of Lagrange multipliers.
We have two coordinates, r and θ, and one equation of constraint, r = R,
which must be cast into the form λ_{1}Σ a_{1k }dq_{k
}= 0, λ_{1}(a_{1r }dr + a_{1θ} dθ)_{ }
= 0.
The equation of constraint is r = R, dr = 0, therefore a_{1r }=
1, a_{1θ }= 0.
 Details of the calculation:
T = ½m((dr/dt)^{2} + r^{2}(dθ/dt)^{2}),
U = mgr cosθ.
L = ½m((dr/dt)^{2} + r^{2}(dθ/dt)^{2})  mgr cosθ.
∂L/∂(dr/dt) = m dr/dt, d/dt(∂L/∂(dr/dt)) = md^{2}r/dt^{2},
∂L/∂r = mr(dθ/dt)^{2}  mgcosθ.
∂L/∂(dθ/dt) = mr^{2}(dθ/dt), d/dt(∂L/∂(dθ/dt)) = mr^{2}d^{2}θ/dt^{2}
+ 2mr(dθ/dt)(dr/dt),
∂L/∂θ = mgr sinθ.
d/dt(∂L/∂(dr/dt))  ∂L/∂r = λ_{1}a_{1r}, _{
}md^{2}r/dt^{2}  mr(dθ/dt)^{2} + mgcosθ
= λ_{1}
(1)
d/dt(∂L/∂(dθ/dt))  ∂L/∂θ = λ_{1}a_{1θ},
mr^{2}d^{2}θ/dt^{2} + 2mr(dθ/dt)(dr/dt)  mgr
sinθ = 0.
(2)
The equation of constraint is dr = 0. We have three equations and three
unknowns,
r(t), θ(t), and λ_{1}.
dr/dt = 0 > d^{2}r/dt^{2} = 0, r = R.
mR^{2}d^{2}θ/dt^{2}  mgRsinθ = 0.
(from 2)
mR(dθ/dt)^{2} + mgcosθ = λ_{1},
(from 1)
dθ/dt = [(g/R)cosθ  λ_{1}/(mR)]^{½}.
The particle falls off when λ_{1} = 0. Then
dθ/dt = [(g/R)cosθ]^{½}.
To find an expression for λ_{1} we need to solve the differential
equation
d^{2}θ/dt^{2} = (g/R)sinθ
for dθ/dt
and equate the solution with dθ/dt = [(g/R)cosθ  λ_{1}/(mR)]^{½}.
Try
dθ/dt = [A + B cosθ]^{½},
then d^{2}θ/dt^{2} = (B/2) sinθ.
Therefore B = 2g/R, dθ/dt = [A  (2g/R) cosθ]^{½}.
Initial conditions: dθ/dt = 0 for θ = 0, A = 2g/R.
dθ/dt = [(2g/R)  (2g/R) cosθ]^{½}.
The particle falls off when (g/R)cosθ = (2g/R)  (2g/R)cosθ, or cosθ =
⅔,
θ = 48.19^{o}.
To find λ_{1} set (2g/R)  (2g/R)cosθ = (g/R)cosθ  λ_{1}/(mR).
λ_{1}/(mR) = (3g/R)cosθ  (2g/R), λ_{1} = 3mg cosθ  2mg.
λ_{1}a_{1r} = F_{r}, F_{r} = is the
reaction of the hoop on the particle.
Problem:
A bead of mass m slides without friction on a circular loop of radius a. The
loop lies in a vertical plane.
(a) Use the Lagrange multiplier method and find the appropriate Lagrangian
including terms expressing the constraints.
(b) Apply the EulerLagrange equations to obtain the equations of motion and
solve for θ << 1.
(c) Find the force of constraint.
Solution:
 Concepts:
Lagrangian Mechanics, Lagrange multipliers
 Reasoning:
We are instructed to use the Lagrange multiplier method
 Details of the calculation:
T = ½(m(dr/dt)^{2}
+ r^{2}(dθ/dt)^{2}),
U = mgrcosθ, L = T – U.
(a) We have two coordinates, r, and θ, and one equation of constraint, r =
a, which we want to cast into the form λ_{1}∑ a_{1k }dq_{k
}= 0. We have dr = 0, a_{1r }= 1, a_{1θ} = 0.
(b) Equations of motion:
d/dt(∂L/∂(dq_{k}/dt))  ∂L/∂q_{k} = ∑_{l}λ_{l}a_{lk}, Σ_{k} a_{lk
}dq_{k
}+ a_{lt }dt = 0.
d^{2}r/dt^{2} – r(dθ/dt)^{2} – gcosθ = λ_{1}/m.
d^{2}θ/dt^{2} = –(g/r) sinθ.
The equation of constraint is r = a, dr/dt = d^{2}r/dt^{2} =
0.
Therefore d^{2}θ/dt^{2} = – (g/a) sinθ, a(dθ/dt)^{2}
– g cosθ = λ_{1}/m.
If θ << 1, then d^{2}θ/dt^{2} = –(g/a)θ. θ = θ_{max}sin(ωt
+ θ_{0}), ω^{2} = g/a.
a(dθ/dt)^{2}  g = λ_{1}/m, λ_{1} = mg + ma(dθ/dt)^{2}
= mg  mv^{2}/a.
(c) The force of constraint has magnitude mg + mv^{2}/a and points
radially inward towards the origin.
Problem:
Two masses m_{1} and m_{2} are connected by
a massless string that runs over a frictionless pulley. The length of the
string, l, somehow increases at a constant rate, i.e. l(t) = l_{0} + l_{1}
t. Use the method of Lagrange multipliers to determine the tension of the
string at time t.
Solution:
 Concepts:
Lagrange's Equations, Lagrange multipliers
d/dt(∂L/∂(dq_{k}/dt))  ∂L/∂q_{k} = ∑_{l}λ_{l}a_{lk}, Σ_{k} a_{lk
}dq_{k
}+ a_{lt }dt = 0.
 Reasoning:
The problem requires us to use the method of Lagrange multipliers.
Let the generalized coordinate x_{i} denote the distance along the
string from the top of the pulley to the mass m_{i}.
Then L = m_{1}(dx_{1}/dt)^{2}/2
+ m_{2}(dx_{2}/dt)^{2}/2
+ m_{1}gx_{1} + m_{2}gx_{2}.
The equation of constraint is x_{1} + x_{2} = l_{0}
+ l_{1}t, dx_{1} + dx_{2}  l_{1}dt = 0.
There is only one equation of constraint. We write λ(a_{x1 }dx_{1}
+ a_{x2} dx_{2} + a_{t}dt) _{ }= 0.
a_{x1 }= 1, a_{x2} = 1, a_{t} = l_{1}.
 Details of the calculations:
The equations of motion are
m_{1}d^{2}x_{1}/dt^{2}  m_{1}g = λ,
m_{2}d^{2}x_{2}/dt^{2}  m_{2}g = λ.
From dx_{1} + dx_{2}  l_{1}dt = 0 we obtain
d^{2}x_{1}/dt^{2} + d^{2}x_{2}/dt^{2}
= 0.
Solving these 3 equations we obtain
m_{1}d^{2}x_{1}/dt^{2}  m_{2}d^{2}x_{2}/dt^{2}
= (m_{1}  m_{2})g .
m_{1}d^{2}x_{1}/dt^{2} + m_{2}d^{2}x_{1}/dt^{2}
= (m_{1}  m_{2})g .
d^{2}x_{1}/dt^{2} = (m_{1}  m_{2})g/(m_{1}
+ m_{2}).
λ = m_{1}(m_{1}  m_{2})g/(m_{1} + m_{2})
 m_{1}g = 2m_{1}m_{2}g/(m_{1} + m_{2}).
λ is the force of constraint acting on m_{1}, its magnitude is the
tension in the string, the  sign indicates that it is acting upward on m_{1}.
Problem:
Consider a particle of mass m constrained to remain on the surface of a
cylinder of radius b. Let the axis of the cylinder be the zaxis.
The particle is subject to the force of gravity mg in the negative zdirection.
(a) Use the Lagrange multiplier method and find the appropriate
Lagrangian including a term expressing the constraint.
(b) Apply
the EulerLagrange equations to obtain the equations of motion.
(c)
Find the force of constraint and briefly discuss why this value is reasonable
and to be expected.
(d) Now, repeat parts (a) and (b) without using
the Lagrange multiplier method. Instead, build the constraint into the
general coordinates.
Solution:

Concepts:
Lagrangian
Mechanics, Lagrange multipliers

Reasoning:
We are asked to find the force of constraint using
Lagrange multipliers.

Details of the calculation:
T = ½m((dr/dt)^{2} + (dz/dt)^{2}) + ½mr^{2}(dφ/dt)^{2}, U = mgz, L = T – U.
(a) We have three coordinates, r, z, and
φ,
and one equation of constraint, r = b, which we want to cast into the form λ_{1}∑
a_{1k }dq_{k }= 0. We have dr = 0, a_{1r }= 1,
a_{1z} = a_{1ɸ} = 0.
(b) Equations of motion:
d/dt(∂L/∂(dq_{k}/dt))  ∂L/∂q_{k} = ∑_{l}λ_{l}a_{lk}, Σ_{k} a_{lk
}dq_{k
}+ a_{lt }dt = 0.
md^{2}r/dt^{2} – mr(dφ/dt)^{2} = λ_{1},
md^{2}z/dt^{2} + mg = 0, mr^{2}d^{2}φ/dt^{2}
+ 2mr(dr/dt)(dφ/dt) = 0
The equation of constraint is r = b, dr/dt = d^{2}r/dt^{2}
= 0.
Therefore d^{2}z/dt^{2} = g, z(t) = z_{0}
+ v_{z0}t – ½ gt^{2},
d^{2}φ/dt^{2} = 0,
φ(t) = φ_{0} + ω_{0}t, with ω_{0} = dφ/dt,
mbω_{0}^{2}
= λ_{1}.
(c) The force of constraint has
magnitude mbω_{0}^{2} and points radially inward towards the
zaxis. It is the centripetal force providing the centripetal
acceleration mv_{φ}^{2}/b.
(d)
T = ½m(dz/dt)^{2} + ½mb^{2}(dφ/dt)^{2}, U = mgz, L = T – U.
d/dt(∂L/∂(dq_{k}/dt))  ∂L/∂q_{k} = 0.
md^{2}z/dt^{2} + mg = 0, mr^{2}d^{2}φ/dt^{2}
= 0.
d^{2}z/dt^{2} = g, z(t) = z_{0} + v_{z0}t
– ½ gt^{2}.
d^{2}φ/dt^{2} = 0, φ(t) = φ_{0}
+ ω_{0}t.