# Lagrangian Mechanics

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I would like you to solve those two questions. Make sure you show me all the work. show me every single step and explain everything in details. Make sure it's clear and understandable. good luck and show me a good quality.

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Barbartos
School: New York University

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Lagrangian Mechanics
December 18, 2017

Problem 1
a)
Since our problem is two-dimensional, we need two generalized coordinates. We can also see this
from analyzing the physical setup. As the particle moves in a plane, we can completely describe its
movement using just polar coordinates. Therefore, we choose r, the distance of the particle from
the center of the dome and θ, the angle between the vertical and the line from the center of the
dome to the particle.

b)
We are interested in the movement of the particle just until it leaves the surface of the sphere; that
is, when the particle is always r = R away from the center of the dome. Therefore, our constraint
is r − R = 0 where R is the radius of the sphere.

c)
The kinetic energy is given by the sum of the translational ( 21 mṙ2 ) and rotational ( 12 mr2 θ̇2 ) kinetic
energies. Therefore
1
m(ṙ2 + r2 θ̇2 )
2
We set the potential energy at the ground to be 0. Therefore, the potential energy of the particle
is mgh where h is the height of the particle. Using geometry, we see that h = r cos θ so
T =

U = mgr cos θ
The free Lagrangian is
1
m(ṙ2 + r2 θ̇2 ) − mgr cos θ
2
We can then add the constraint to the Lagrangian:
L=T −U =

L0 = T − U − λf (r, θ) =

1
m(ṙ2 + r2 θ̇2 ) − mgr cos θ − λ(r − R)
2

1

d)
We solve for the equations of motion by using the Euler-Lagrange equations on each generalized
coordinate. Then our equations of motion are:
d
dL
=
dr
dt



∂L
∂ ṙ



mrθ̇2 − mg cos θ − λ = mr̈
dL
d
=

dt



∂L
∂ θ̇
2
−mgr sin θ = mr θ̈



And for our Lagrangian multiplier constraint we have
∂L
=0=r−R
∂λ

e)
First, from the constraint equation we have that r = R at all points during which the Lagrangian
describes the system. Substituting into our first equation of motion we have
mRθ̇2 − mg cos θ − λ = 0
Substituting into our second eqeuation of motion we obtain
mgR sin θ = mR2 θ̈ = mR2 θ̇

dθ̇

We then solve for θ̇ as a function of θ:

Z

θ

g sin θ dθ = Rθ̇ dθ̇
Z θ0
0
0
g sin θ dθ =
Rθ̇0 dθ̇0

0

0

g
1
(1 − cos θ) = θ̇2
R
2
r
2g(1 − cos θ)
θ̇ =
R

f)
First, we see that λ = mRθ̇2 − mg cos θ. We know that the conjugate force to a variable xi is
∂f
given by λ ∂x
where f (r, θ) = 0 is our constraint and xi is our generalized coordinate. Then there
i
is no non-zero conjugate force to θ since dd fθ = 0. For our coordinate r, the conjugate force is
2



2
Fr = λ ∂f
=
mR
θ̇

mg
cos
θ
(1) = mRθ̇2 − mg cos θ. recognize this as the normal force, as it is
∂r
in the difference between the radial gravitational force and the force needed to keep the particle in
centripetal motion.
We now substitute our result from the second equation of mo...

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Anonymous
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