Download The Lagrangian Method vs

Jozef HANC
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The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
The Lagrangian Method vs. other methods
(COMPARATIVE EXAMPLE)
This material written by Jozef HANC, [email protected]
Technical University, Kosice, Slovakia
For Edwin Taylor‘s website http://www.eftaylor.com/
17 February 2003
The aim of this material is a demonstration of power of Lagrange’s equations for solving
physics problems based on comparing the solution of an example by five different methods:
“Vectorial” methods
• Using Newton’s laws (inertial frame of reference)
• Using Newton’s laws with fictitious forces (accelerated frame of reference)
• Using D’Alembert’s principle (The extended principle of virtual work)
“Scalar” methods
• Using Conservation laws
• Using Lagrange’s equations (The Lagrangian method)
Here is our problem:
MOVING PLANE:
A block of mass m is held motionless on a frictionless plane of
mass M and angle of inclination θ. The plane rests on a frictionless horizontal surface. The block is released (see Fig.1).
1
What is the horizontal acceleration of the plane?
m
M
θ
Fig. 1
SOLUTIONS:
„VECTORIAL” METHODS:
2
I. METHOD: Newton’s laws (in an inertial frame of reference) :
N
vertical
positive
direction
mg
Fig. 2 Diagram with forces applying on the
moving block showing the notation:
N – magnitude of the normal force from
the plane
mg – magnitude of gravity
horizontal positive direction
1
2
The problem was taken from section 2.7 in Morin [Ref.1], ch2.pdf, p.18
This solution according to Morin [Ref. 1], ch2.pdf, p. 22
Jozef HANC
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vertical
positive
direction
F
N
The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
Fig. 3 Diagram of the moving plane shows notation:
N – magnitude the normal force from the block
Mg – magnitude of gravity
F –magnitude of the normal force from the base
θ
Mg
horizontal positive direction
Equations of motion of the block
horizontally
vertically
Equations of motion of the plane
horizontally
vertically
N sin θ = ma x
mg − N cos θ = ma y
(1.1)
(1.2)
− N sin θ = MAx
Mg + N cos θ − F = MAy = 0
(1.3)
(1.4)
Kinematical equation resulting from the motion of the block:
a yt 2 / 2
(a x − Ax )t 2 / 2
=
ay
a x − Ax
= tan θ
(1.5)
We have five equations with five unknowns. To find the quantity Ax, it is sufficient to resolve
equations (1.1), (1.2), (1.3) and (1.5):
ay
N sin θ = ma x
mg − N cos θ = ma y
− N sin θ = MAx
= tan θ
a x − Ax
The solution for Ax has to be done by a more time consuming procedure:
Ax = −
mg tan θ
M (1 + tan 2 θ) + m tan 2 θ
(1.6)
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The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
II. METHOD: Newton’s laws with fictitious forces:
FORCES acting on the BLOCK (moving in an accelerated frame of reference)
vertical
positive
direction
N
Fig. 4 Diagram of the moving block in accelerated
frame determined by the plane.
N – magnitude of the normal force from the plane
Ftrans – magnitude of the translation fictitious force
equal to mAx and with opposite direction as Ax
mg – magnitude of gravity
Ftrans
mg
horizontal positive direction
Equations of motion of the block:
mg sin θ + (− mAx ) cos θ = ma
horizontally
mg cos θ − N − (−mAx ) sin θ = 0
vertically
(2.1)
(2.2)
Equations of motion of the plane are identical with those written in Section 1, because we
consider the plane in the same inertial frame of reference. So:
horizontally
vertically
− N sin θ = MAx
Mg + N cos θ − F = MAy = 0
(2.3)
(2.4)
We have four equations with four unknowns. To find the quantity Ax, it is sufficient to resolve
equations (2.2) and (2.3)
mg cos θ − N + mAx sin θ = 0
− N sin θ = MAx
The solution for Ax is given by a quick way:
Ax = −
mg sin θ cos θ
M + m sin 2 θ
Using known trigonometric formulas, we get the same result as (1.6).
(2.5)
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The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
III. METHOD: D’Alembert’s principle (“The extended principle of virtual work”)
SHORT SUMMARY:
3
D’Alembert’s Principle : The total virtual work of the effective forces is
zero for all reversible variations (virtual displacements), which satisfy the
given kinematical conditions.
REMARKS:
1) This principle uses the concept of the virtual displacement, which means, not actual small displacement,
but it involves a possible, but purely mathematical experiment, so it can be applied in a certain definite
time (even if such a displacement would involve physically infinite velocities). At the instant the actual
motion of the body does not enter into account. To distinguish between virtual displacement from actual
one, we use label δ instead of ∆ or d, e.g. δx, δy, δR.
2) The special name − the effective forces − represents a common name for the given external or applied
forces on a mechanical system, augmented by the inertial forces (see fictitious force in Section 2), but
without the forces of constraint.
3) Since the forces of constraint do not need to be taken into account, D’Alembert principle immediately
gives the consequence: The virtual work of the forces of reaction is always zero for any virtual displacement which is in harmony with the given kinematic constrains. Or in other words we restrict ourselves to systems for which the net virtual work of the forces of constraint is zero4. This assertion is easily understandable from the Newtonian mechanics. If e.g. a particle is constrained to move on a surface,
the force of constraint arises from the reaction forces (Newton’s third law) and is perpendicular to the
surface, while virtual displacement must be tangent to it, and hence the virtual work of the force vanishes. However, in general case it is not so evident and becomes postulate called POSTULATE A.
horizontal positive direction
vertical
positive
direction
mg
θ
Mg
Fig. 5 Diagram of “impressed” forces (mg,
Mg) − the given external forces acting on the
mechanical system. The forces of constraint
have no influence on a value of virtual work
and can be neglected in a virtual-work calculation.
Total virtual work is represented by sum of virtual works of all impressed forces together with
the inertial forces on mechanical system: block + plane (the vectorial form):
δW = (mg − ma)δr + ( Mg − MA)δR = 0
(3.1)
where a and A are accelerations of the block and the plane, Mg+(−MA), mg+(−ma) are corresponding net effective forces acting on the block and the plane, and δr, δR are virtual displacements of the block and plane in harmony with the given kinematical conditions.
3
The formulation of D’Alembert’s Principle according to Lanczos [Ref. 2], page 90, Section IV. See also Ref. 3,
Section 1.4
4
According to Goldstein [Ref.3]
Jozef HANC
The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
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We can rewrite the equation (3.1) in more appropriate, analytical language. Using Cartesian
coordinates: a=(ax, ay), A=(Ax, 0), Mg+(−MA)=(−MAx, Mg), mg+(−ma) = ( − max, mg − may),
and δr = (δx, δy), δR = (δX, 0), we get the total virtual work as a sum of virtual works of
components of the effective forces in x and y-direction:
δW = (mg − ma y )δy + (−ma x )δx + (− MAx )δX = 0
(3.2)
All these displacements must be in harmony with the kinematical conditions (the block must
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slide down the frictionless plane). It gives the condition :
δy = (δx − δX ) tan θ
(3.3)
In other words all the virtual displacements δx, δy, δX are not independent. At the same time
the previous equation (3.3) provides the following relations between accelerations ax, ay, Ax:
a y = (a x − Ax ) tan θ
(3.4)
Since one equation (3.3) connects the virtual displacements, one of the displacement, say δx,
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can by (3.3) be expressed in terms of the other two δy, δX. Substitution of δx in terms of δy,
δX into the virtual work (3.2) with an rearrangement leads to:
δW = (mg tan θ − ma y tan θ − ma x )δy − (ma x + MAx )δX = 0
(3.5)
This remaining pair δy, δX may be considered as quite independent displacements without
further restriction (Think about it!). Then the vanishing of the resultant work requires:
mg tan θ − ma y tan θ − ma x = 0
(3.6a)
ma x + MAx = 0
(3.6b)
These equations (3.6) together with equation (3.4) provide the system of three equations with
three unknowns. Solving the equations, we obtain in short time the same result for Ax as (1.6).
ADDITIONAL NOTE (This note can be skipped the first time): The analytical treatment according to
D’Alembert’s principle takes only the external and inertial forces into account. Since the inner forces
which produce constraints need not be consider, D’Alembert’s method offers a significant simplification in comparison with Newton’s laws. But it is interested to look at the forces of constraints (see Fig.
6).
horizontal positive direction
N2
vertical
positive
direction
N1
Fig. 6 Diagram of all forces of constraints N1,
N2 and F. According to D’Alembert’s principle the forces of constraint have no influence
on a value of virtual work and their work has
zero value
F
θ
5
6
This condition can be obtained by the same way as in Section IV, Figure 7 of this material.
We can use any virtual displacement and express it in terms of all remaining ones.
Jozef HANC
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The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
1. If we assume the validity of Newton’s third law, then magnitudes of forces N1, N2 are equal: N1 = N2
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= N. Simultaneously according to POSTULATE A the resultant work of these forces with respect to
δx, δy, δX must be zero. Mathematically:
δWconstraints = − N 1 sin θ ⋅ δX + F ⋅ 0 + N 2 sin θ ⋅ δx − N 2 cos θ ⋅ δy = 0
(3.7)
or since N1 = N2 = N
cos θ ⋅ δy = sin θ ⋅ δx − sin θ ⋅ δX
(3.8)
This equation (3.8) is nothing else as the kinematical condition (3.3): δy = (δx − δX ) tan θ that was derived by geometrical arguments from a figure considering motion of the block on the plane. Use of
Newton’s third law together with Postulate A yield easily the kinematical condition without considering motion of the block on the plane.
2. Consider only Newton’s third law with the kinematical condition δy = (δx − δX ) tan θ . Virtual work
of the forces of constraints equals:
δWconstraints = − N sin θ ⋅ δX + N sin θ ⋅ δx − N cos θ ⋅ δy = N . cos θ(tan θ ⋅ δx − tan θ ⋅ δX − δy )
(3.9)
Then kinematical condition (the expression in brackets) provides the zero value of virtual work. So in
that case the POSTULATE A is deducible from Newton’s third law. However, in general8 case the
third law of motion, “action equals reaction”, is not wide enough to replace Postulate A.
3. Finally we can start from POSTULATE A and the kinematical condition δy = (δx − δX ) tan θ and
come to Newton’s third law. The resultant virtual work (3.7) of the forces of constraints implies:
δy = (δx −
N1
δX ) tan θ
N2
(3.10)
Subtracting the last equation (3.10) from the δy = (δx − δX ) tan θ , we obtain the condition:
0 = (1 − N 1 / N 2 )δX tan θ and hence the use of arbitrariness of δX yields a result: N1 = N2 = N:
7
8
See Remark 3 in this section.
According to Lanczos [Ref.2], p. 77, footnote 1.
Jozef HANC
The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
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„SCALAR” METHODS
IV. METHOD: Conservation laws
We use the center-of-mass frame of reference. For easy work with potential energy we will
assume a coordinate system connected to this frame reference, which has the same horizontal
positive direction, but opposite (obvious) vertical direction, as in previous cases9. Let the
plane have the horizontal coordinate l1 + x1(t), where the l1 represents an initial x-coordinate
of the plane in center-of mass frame reference and x1(t) translation from this position. Similarly l2 + x2(t) for the block. Denote v1, a1 the velocity and acceleration of the plane and v2, a2
the velocity and acceleration of the block.
1. Conservation of momentum in x-direction:
initial



 ≡ 0 = mv 2 + Mv1
 x - momentum 
(4.1)
2. Conservation of motion of center of mass (c.m.):
 initial position  ml 2 + Ml1
m(l 2 + x 2 ) + M (l1 + x1 ) mx 2 + Mx1

 =
≡0=
=
of c.m.
m+M
m+M
m+M


(4.2)
3. Conservation of energy:
If the block translates through x2 and the plane through x1, then it is easy to see from Fig. 7:
x2 − x1
x2
x2 − x1
x1
h
θ
7.a
θ
θ
7.b
7.c
Fig.7 If the block translates trough a horizontal displacement x2 and the plane through x1 and we
find the height fallen by the block, we do not need to consider the translations of the block and the
plane simultaneously. The same effect is obtained if we first translate horizontally the block
through x2 (7.a), then the plane through x1 (7.b) and finally the block vertically through h (7.c).
that the height h fallen by the block is h = ( x 2 − x1 ) tan θ . Hence we can write:
 change in 
1
1
 potential energy 
2
2
2
2

 ≡ mg ( x 2 − x1 ) tan θ = 2 Mv1 + 2 m v2 + (v 2 − v1 ) tan θ
 of the block 


[
9
It means y increases in the direction vertically upward.
]
(4.3)
Jozef HANC
The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
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Mx1
Mv1
and x 2 = −
from (4.1) and (4.2) to (4.3), we obtain
m
m
a relationship between position x1 and v1:
Substituting expressions v 2 = −
− mg ( x1 +
M
m
x1 ) tan θ =
1
1
1
2
Mv12 + m(Mm v1 ) + m(v1 +
2
2
2
M
m
v1 ) 2 tan 2 θ
(4.5)
Because we have a need for acceleration a1 of the plane - time derivative of velocity v1- it is
sufficient to take derivative of the equation (4.5) with respect to time:
− mgv1 (1 +
M
m
Dividing by v1 (1 +
) tan θ = Mv1a1 + m(Mm ) v1a1 + mv1a1 (1 +
2
M
m
M 2
)
m
tan 2 θ
(4.6)
) , we get acceleration a1:
− mg tan θ = Ma1 + a1 (m + M ) tan 2 θ
or
a1 = −
mg tan θ
M + (m + M ) tan 2 θ
(4.7)
ADDITIONAL NOTE: (Here we present a non-calculus way how to obtain the acceleration a1 from
(4.5)). Instead derivative we can use another general expressions from kinematics, which is valid for
uniformly accelerated motion with zero initial velocity. We can write for the plane:
l1 + x1 = l1 + 12 a1t 2
⇒ x1 = 12 a1t 2 ; v1 = a1t ⇒
2 x1 a1 = v12
(4.8)
After rearranging (4.5), we get:

mg tan θ
2 x1 ⋅ −
2
 M + (m + M ) tan

2
 = v1
θ 
(4.9)
So according to (4.8) the magnitude of the a1 has a form:
a1 = −
mg tan θ
M + (m + M ) tan 2 θ
(4.10)
Jozef HANC
The Lagrangian Method vs.other methods
COMPARATIVE EXAMPLE
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V. METHOD: Lagrange’s equations
We use the same notation as in Section IV. Moreover label h0 the initial height of the block.
We know from the previous section that the height fallen by the block is ( x 2 − x1 ) tan θ . Then:
1. Potential energy of the system (block + plane):
V = mg [h0 − ( x 2 − x1 ) tan θ]
(5.1)
2. Kinetic energy of the system:
T=
[
1
1
Mv12 + m v 22 + (v2 − v1 ) 2 tan 2 θ
2
2
]
(5.2)
3. The Lagrangian of the system:
L ≡ T −V =
[
]
1
1
Mv12 + m v 22 + (v 2 − v1 ) 2 tan 2 θ − mg [h0 − ( x2 − x1 ) tan θ]
2
2
(5.3)
4. The equations of motion:
∂L d ∂L
−
=0
∂x1 dt ∂v1
(5.4a)
d ∂L
∂L
−
=0
∂x 2 dt ∂v 2
(5.4b)
After substituting Lagrangian (5.3) to (5.4a) and (5.4b), we have:
− mg tan θ = Ma1 + m(a1 − a 2 ) tan 2 θ
(5.5a)
+ mg tan θ = ma 2 − m(a1 − a 2 ) tan 2 θ
(5.5b)
Adding equations (5.5), we get Ma1 + ma2 = 0 or a 2 = −
Ma1
and substituting this result to the
m
(5.5a), we easily obtain:
a1 = −
mg tan θ
M (1 + tan 2 θ) + m tan 2 θ
(5.6)
REFERENCES:
[1] David J. Morin, Mechanics and Special Relativity,
http://www.courses.fas.harvard.edu/~phys16/handouts/textbook/ch5.pdf
[2] Cornelius Lanczos, The Variational principles of Mechanics, (Dover Publications, New
York, 1986)
[3] Herbert Goldstein, Charles Poole, and John Safko, Classical Mechanics, third edition
(Addison-Wesley, 2002)
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See the detailed description of the explanation and use of the Lagrangian method in Morin [Ref.1], ch5.pdf