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2. Solve GPS Chap 1, Derivation 8:If L is a Lagrangian of n degrees of freedom satisfying Lagranges Equations

ddt

∂L∂ q j

− ∂L∂q j

= 0 (14)

show by direct substitution that

L = L + F (q 1 ,...,q n , t ) (15)

where F is any arbitrary differentiable function, also satises Lagrange’s equations.

Substituting directly into Lagrange eqns.,

ddt

∂L∂ q j

− ∂ L∂q j

= d

dt∂L∂ q j

− ∂L∂q j

+ ddt

∂ F ∂ q j

− ∂ F ∂q j

(16)

= d

dt ∂ ∂ q j

∂F ∂t

+ ∂F ∂q k

q k − ∂ F ∂q j

(17)

= d

dt∂F ∂q j

− ∂ ∂q j

dF dt

(18)

= 0 , (19)

since ddt

∂F ∂q j

= ∂ ∂t

∂F ∂q j

+ ∂ ∂q k

∂F ∂q j

q k = ∂ ∂q j

∂F ∂t

+ ∂ ∂q j

∂F ∂q k

q k = ∂ ∂q j

dF dt

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3. Solve GPS Chap 1, Exercise 18:A Lagrangian for a particular physical system can be written as

L = m2

(a x 2 + 2 bx y + cy2) − K 2

(ax 2 + 2 bxy + cy2) (20)

where a, b, and c are arbitrary constants but subject to the condition that b2 − ac = 0. What are the equations of motion? Examine particularly the two cases a = 0 = c and b = 0, c = − a . What is the physical system describedby this Lagrangian?

Using the Euler-Lagrange equations,

ddt

∂L∂ q j

− ∂L∂q j

= 0 , (21)

for x and y :

ddt

∂L∂ x

− ∂ L

∂x =

ddt

(m (a x + by)) − (− Kax − Kby ) (22)

0 = m (a x + by) + K (ax + by); (23)

d

dt

∂L

∂ y −

∂L

∂y =

d

dt (m (bx + cy)) − (− Kbx − Kcy ) (24)0 = m (bx + cy) + K (bx + cy); (25)

the equations of motion are

m (a x + by) = − K (ax + by) (26)m (bx + cy) = − K (bx + cy) (27)

In both the particular cases a = 0 = c and b = 0, c = − a , we get

m x = − Kx (28)m y = − Ky (29)

which are the equations of motion of a 2D harmonic oscillator of mass m

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5. GPS Chap 1, Exercise 22:Obtain the Lagrangian and equations of motion for the double pendulum shown (GPS Fig 1.4), where the lengthsof the pendula are l1 and l2 with corresponding masses m 1 and m 2 .

Using

x 1 = l1 sin θ1 ; x 1 = l1 cosθ1 θ1 ; (38)

y1 = l1 cosθ1 ; y1 = − l1 sin θ1 θ1 ; (39)

x 2 = l1 sin θ1 + l2 sin θ2 ; x 2 = l1 cosθ1 θ1 + l2 cosθ2 θ2; (40)

y2 = l1 cosθ1 + l2 cosθ2 ; y2 = − l1 sin θ1 θ1 − l2 sin θ2 θ2 ; (41)

then

T 1 = m 1 l2

1

2θ2

1 , (42)

V 1 = − m 1gl1 cosθ1 , (43)

T 2 = m 2

2l21 θ2

1 + l22 θ2

2 + 2 l1 l2 θ1 θ2 cos(θ1 − θ2) (44)

V 2 = − m 2g (l1 cosθ1 + l2 cosθ2) (45)

The Lagrangian is

L = T 1 + T 2 − V 1 − V 2

= ( m 1 + m 2)gl1 cosθ1 + m 2gl2 cosθ2 + 1

2(m 1 + m 2)l2

1(θ1)2 + m 2 l1 l2 cos(θ1 − θ2)θ1 θ2 +

1

2m 2 l2

2(θ2)2 (46)

Euler-Lagrange equation for θ1 :

ddt

∂L

∂ θ1−

∂L∂θ 1

= 0 (47)

= d

dt(m 1 + m 2)l2

1 θ1 + m 2 l1 l2 cos(θ1 − θ2)θ2 − − m 2 l1 l2 sin(θ1 − θ2)θ1 θ2 − (m 1 + m 2)gl1 sin θ1

(m 1 + m 2)l1 θ1 + m 2 l2 cos(θ1 − θ2)θ2 + m 2 l2 sin(θ1 − θ2)θ22 + ( m 1 + m 2)g sin θ1 = 0 (48)

Euler-Lagrange equation for θ2 :

ddt

∂L

∂ θ2 − ∂L∂θ 2 = 0 (49)

= d

dtm 2 l1 l2 cos(θ1 − θ2)θ1 + m 2 l2

2 θ2 − m 2 l1 l2 sin(θ1 − θ2)θ1 θ2 − m 2gl2 sin θ2

m 2 l1 cos(θ1 − θ2)θ1 + m 2 l2 θ2 − m 2 l1 sin(θ1 − θ2)θ21 + m 2g sin θ2 = 0 (50)

Notice that if we make m 2 = 0 , θ2 = const = 0 , we get from the rst equation of motion θ1 = − gl 1

sin θ1 as expected.Similarly, with m 1 = 0 , θ1 = const = 0 , we get from the second θ2 = − g

l 2sin θ2 .

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