Classical Assignment 1

8/10/2019 Classical Assignment 1

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National Institute of Technology, Rourkela

Department of Physics and Astronomy

Instructor : Dr A C Pradhan

Classical MechanicsAssignment I

Auro Prasad Mohanty411PH5026

September 6, 2014

Question 1

A particle is constrained to move on the arc of a parabola z = 4ay2, (z-axis vertical) under theaction of gravity. The parabola is rotating around the z-axis with constant angular velocity .

Find the Lagrangian equation of motion.

Answer

The parabolaz = 4ay2 is rotating about the z-axis. Thus, takingx2 + y2 =r2 where r is thedistance of the particle from the z-axis. Letx= r cos and y = r sin . Hence we can writethe paraboloid equation as z = 4ar2. Calculating the kinetic energy

T =1

2m(x2 + y2 + z2)

=1

2m(r2 + r22 + z2)

(1)

Now, the potential energy V is given by mgz = 4mgar2. z = 8arr, = . Hence theLagrangian is given by,

L= T V

=1

2m(r2 + r22 + 64a2r2r2) 4mgar2

(2)

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Figure 1: Particle moves in a parabola which in turn rotates about the z-axis with gravity actingoppossite the z-xis.

The Lagrangian equation of motion is given by

d

dt

L

r

L

r = 0 (3)

d

dt(mr+ 64a2r2r)mr2 64a2rr2 + 8mgar= 0 (4)

mr+ 128a2

rr2

+ 64a2

r2

r mr2

64a2

rr2

+ 8magr= 0 (5)

mr+ 64a2rr2 + 64a2r2r mr2 + 8magr= 0 (6)

Find the Hamiltonian equation of motionAnswer

pr = L

r =mr+ 64a2r2r (7)

and,

p = L

=mr2 (8)

=r=

pr

m + 64a2r2 (9)

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and,

= p

mr2 (10)

Thus the Hamiltonian is given by

H=

piqi

L (11)

So,

H= rpr+ p L

= p2r

m + 64a2r2 mp

2r

2

m + 64a2r22 mr

22

2

32ma2r2p2r

m + 64a2r22 + 4mgar2 + mr22

=

1

2 m

1 + 64a2r2m + 64a2r2

2p

2

r+

p22mr2

(12)

Canonical transformation equations are:

r = pr

m + 64a2r2 =

H

pr

= p

mr2 =

H

p

pr = H

r

= 8mgar

mp2r

2

r

1 + 64a2r2

m + 64a2

r2

= 8mgar mp2

r

2

128a2r

(m + 64a2r2)2 21 + 64a

2r2

m + 64a2r2 128a2r

p = H

= 0

= p = constant (13)

Find at what condition the particle will be at rest.Answer

Particle is at rest when its velocity is zero. i.e. r= 0

mr2 8mgar = 0

a= 2

8g(14)

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Question 2

A particle is constrained to move under gravity on the surface of a smooth sphere which is rotatingwith a constant angular velocity around the z-axis.

Find the Lagrangian equation of motion. Find the condition so that the particle may not fly off the surface.Answer

Consider the equation of a sphere

x2 + y2 + z2 =a2 (15)

a is the radius. P(r, )is a point on the sphere.

Figure 2: Particle constrained to move under gravity on the surface of a smooth sphere rotatingwith a constant angular velocity around the z-axis.

Sphere is rotated about the z-axis with angular velocity . Also the particle rotates in radius r.Therefore,

x2 + y2 =r2 (16)

and,

z2 + r2 =a2

z =

(a2 r2)

z = rra2 r2

(17)

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K.E.=1

2m(r2 + r22 + z2)

=1

2 m(r2

+ r2

2

+ r2r2

a2 r2 )(18)

P.E.= mgz

=mg

a2 r2(19)

The Lagrangian is given by,

L= T V

=1

2m(r2 + r22 +

r2r2

a2 r2 )mg

a2 r2(20)

The Lagrangian equations of motion are,

d

dt

L

r

L

r = 0 (21)

On substituting the corresponding values in the above equation we get

m

1 +

r2

a2 r2

r+

2mr

a2 r2 + 2mr3

(a2 r2)2

r2 =mr2 + mrr2

a2 r2 + mr3r2

(a2 r2)2 + mgr

a2 r2 (22)

and,

d

dt

L

L

= 0 (23)

we have,

mr2+ 2mrr= 0 (24)

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Question 3

A particle of massmis falling under gravity. The potential is V = -mgz(z=height) and describingthe path.

z= 12

gt2

z = ctCalculate th eaction for the paths and show that minimum occurs for the path z = 1

2gt2.

Answer

Att = 0, both the paths a and b have the same end points as origin.Att = t1,

gt21

2 =ct1

=

c= gt1

2

(25)

Now, path(a) has

z= gt2

2 (26)

and, path (b)has

z = gt1t

2 (27)

Lagrangian for(a) is

La= mz2

2 +

mg2t2

2

= mg2t2

2

+mg2t2

2=mg2t2

(28)

The action integral for (a) is

Sa =

t10

Ladt= mg2t3

1

3 (29)

Now, for(b),

Lb =1

2m

gt1

2

2+

mg2tt1

2 (30)

and

Sb =

t10

Lbdt = mg2t3

1

4 +

mg2t31

8 (31)

= 3mg2t31

8 (32)

Thus, Sa < Sb.

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Question 4

A paticle of mass m is connected by a spring of unstretched length lo and of spring constant k .The other end of the spring is connected to a point in the roof and the particle is swinging like a

plane simple pendulum. Find the Langrangian equation.

Answer

: angle made by spring with vertical

F= kx= k(l l0)

(33)

Figure 3: Spring with a particle hanging from it and moving as a simple pendulum.

where, x is the stretched length

V1 = k(l l0)dl

=1

2k(l l0)2

(34)

(Potential energy due to restoring force) V2 = mgl cos Therefore,

V =12

k(l l0)2 mgl cos (35)

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Kinetic Energy:

T =1

2m(x2 + y2)

=1

2 m(l2

+ l2

2

)

(36)

Hence, the Lagrangian,

L= T V =12

m(l2 + l22) 12

k(l l0)2 + mgl cos (37)

The Lagrangian equations of motion is given by:

ml= ml2 k(l l0) + mgcosml+ 2ml= mgsin

(38)

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Question 5

The Lagrangian for a system with two degrees of freedom is L = q21

+ q22

+ q1q2 q21q22 . Obtain theLagrangian equation of motion.

Answer

The Lagrangian equations of motion are given by,

d

dt

L

q1

=

L

q1

= ddt

2q1+ q2

= 2q1

= 2q1+ q2 = 2q1

(39)

Figure 4: The horizontal table has a spring and a particle attached to it.

d

dt

L

q2

=

L

q2(40)

=d

dt

2q2+ q1

= 2q2 (41)

=

2q2+ q1 = 2q2 (42)

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Thus,

3(q1+ q2) = 2(q1+ q2)= 3X= 2X

= X+ 23

X= 0

(43)

X=q1+ q2Similarly,

q1 q2 = 2(q1 q2)= Y = 2Y

= Y + 2Y = 0(44)

Y =q1 q2

Simple Harmonic Oscillator.

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Question 6

Find the Lagrangian and Hamiltonian for a double pendulum.

Answer

x1 = l1sin 1

x2 = l1sin 1+ l2sin 2

y1 = l1cos 1

y2 = l1cos 1+ l2cos 2

(45)

Figure 5: The double pendulum.

For mass m1,

T1 =1

2m1(x

2

1+ y2

1) =

1

2m1l

2

121

V1 = m1gl1cos 1(46)

For mass m2,

T2 =1

2

m2(x2

2+ y2

2)

=1

2m2(l

2

121

+ l22

22

+ 2l1l2cos(1 2)12)V2 = m2g(l1cos 1+ l2cos 2)

(47)

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The Lagrangian is given by

L= T1+ T2 V1 V2=

1

2

(m1+ m2)l2

121+

1

2

m2l2

222

+ m2l1l212cos(1 2)+ (m1+ m2)gl1cos1+ m2gl2cos2

(48)

The Hamiltonian is given by,

H= T+ V

= (T1+ T2) + (V1+ V2)

=1

2(m1+ m2)l

2

121

+1

2m2l

2

222

+ m2l1l212cos(1 2)m1gl1cos 1 m2gl2cos 2

(49)

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and,

d

dt

L

L

= 0

= mr2

= 0= p = 0

(54)

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Question 8

Find the Lagrangian and Hamiltonian for a charged particle in an electromagnetic field.Answer

E+ 1c

B

t = 0 (55)

H = 4c

J+1

c

D

t (56)

. D = 4 (57). B = 0 (58)

The Lorentz force is given by,F =qE+

q

c(v B) (59)

. B= 0 = B = A (60)So,

E+ 1c

t( A) =

E+

1

c

A

t

= 0 (61)

E+1

c

A

t = (62)

=E= 1

c

A

t (63)

So,

F=q

1

c

A

t

+1

c

(V

(

A)) (64)

=Fx = q

x 1

c

Ax

t +

1

c(V ( A))x

(65)

d

dtAx =

tAx+

Ax

x

dx

dt +

Ax

y

dy

dt +

Ax

z

dz

dt (66)

d

dtAx =

tAx+

Ax

x vx+

Ax

y vy+

Ax

z vz (67)

v

A

x

=vy

A

z

vz

A

y

=vy

Ay

x Ax

y

vz

Ax

z Az

x

(68)

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Fx = q

x 1

c

tAx+

1

c

x

v. A

1c

d

dtAx+

1

c

tAx

=q

x

1c v. A

1c ddt vx

v. A

= x

q

1c

v. A) + d

dt

vx

v.

A

c

= Ux

+ d

dt

U

vx

(69)

U=q qc

v. A

(70)

L = T q + qc

v. A

(71)

= mq2

2 q + q

c

v. A

(72)

Taking Cartesian coordinates as generalized coordinates,

L = mx2i

2 +

q

cAixi q (73)

pi = mxi+q

cAi (74)

=xi =

pi qAi

c

m (75)

H =

pi qAi

c

2

2m + q (76)

H =

p qAc

2

2m + q (77)

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Question 9

A mass 2m is suspended from a fixed support by a spring of spring constant 2k. From this mass,another mass m is suspendedby another spring of spring constantk. Find the equation of motion

for the coupled system.

Answer

T1 = mx2

1 (78)

T2 = mx2

2

2 (79)

P1 = k

x1 x012 2mgx1 (80)

P2 = 1

2k

x2 x1 x012 mgx2 (81)

So, the Lagrangian L is given by

L = mx21+1

2mx2

2 kx1 x012 k

2

x2 x1 x01

2+ 2mgx1+ mgx2 (82)

d

dtm + 2k

x1 x01

k

x2 x1 x02

2mg= 0 (83)

=2mx1+ 2k

x1 x01

kx2 x1 x02 2mg= 0 (84)=

x1+

2

x1 x2 2x01+ x02

= 0 (85)

And similarly for x2,d

dt L

x2

L

x2= 0 (86)

=d

dt

mx2

+ k

x2 x1 x02

+ mg = 0 (87)

mx2+ k

x2 x1 x02+ mg = 0 (88)

=x2+

k

m

x2 x1 x02

= 0 (89)

where

22 =

k

m

(90)

These equations give the equations of motion.

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Question 10

Find the equation of motion for the Harmonic oscillator by

Lagrangian methodAnswer

T =1

2mq2,

V =1

2kq2,

L= T V

(91)

Figure 7: Spring with a mass attached.

Applying the Lagrangian equation of motion we get

q+ 2q= 0, (92)

where,

2

=

k

m (93)

Hamiltonian formulation

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Answer

pq = L

q =mq (94)

which is

q= pq

m (95)

Therefore,

H=pqq L

=mq2 12

mq2 +1

2kq2

= 1

2mp2q+

1

2kq2

(96)

Hence,

q= H

pq=

pq

m (97)

and,

pq = H

q = kq (98)

Method of canonical transformation (generator of the transformation is F1 = 12mq2 cot Q)Answer

H= p2

2m+

1

2kq2

=

k

m

(99)

Therefore,

H= p2

2m+

1

2m2q2 (100)

F1(q ,Q ,t) = 12mq2 cot Q (101)

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Figure 9: Diagram showing Q as function of t.

Therefore, since Q is cyclic,

P=constant= E

Q=

= Q= t +

q=

2E

m2sin(t + )

p=

2Em cos(t + )

(106)

If,

t + =

q=

2E

m2sin ,

p=

2Em cos

(107)

Hamiltonian-Jacobi method

Answer

p= sq

(108)

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Figure 10: Diagram showing p as function of q.

Therefore,

1

2m

s

q

2+ m22q2

+

s

t = 0 (109)

s(q , ,t) = (q, ) t (110)s

q =

q (111)

s

t = (112)

1

2m

q

2

+1

2m2q2 = (113)

q =

2mm22q2 (114)

=

2mm22q2dq+ C (115)

s =

2mm22q2dq t + C (116)

Now,

=

S

= 1

2

2m2mm22q2 dq t (117)

+ t = 1

arcsin q

m2

2 (118)

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Classical Assignment 1