PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 3, Solutions
1
Graded Problems
Problem 1
(1.a)
We use cylindrical coordinates and notice that z = r cot . We use cfw_r, as generalized coordinates
and write the kinetic energy of the bead as
1
1
1
r2
1
PHY5246 Midterm 1: Solution
Problem 1. Part (a) 1 m(r2 + r2 2 + 2 r2 ) mgr 2
L= Part (b)
(1)
L = m(1 + 2 )r r d dt L r
(2)
= m(1 + 2 ) r
(3)
L = mr2 mg r Euler-Lagrange equation for r m(1 + 2 ) = mr2 mg r is cyclic so L = mr2 = l
(4)
(5)
(6)
is conser
Theoretical Dynamics PHY 5246
Final Exam December 9, 2004
1. Consider a particle with charge e moving in the plane in the presence of a perpendicular magnetic eld B. If the vector potential is taken to be A = coordiantes, the Lagrangian for this particle
Theoretical Dynamics - PHY 5246
Final Exam December 15, 2005
1. (40 pts) Consider a uniform cylindrical shell of radius R and mass 3m rolling on a flat surface without slipping. A small object of mass m slides without friction inside the cylinder (see fig
Theoretical Dynamics PHY 5246
Midterm I
October 20, 2006
1. (40 pts) (40 pts) A pendulum consists of a mass m attached to a massless rod of length
l which hangs from a pivot point that is attached to a motor. The motor moves the pivot
point rp around a ci
Theoretical Dynamics PHY 5246
Midterm II
November 22, 2006
1. (60 pts) Consider a uniform at rigid body of mass M which has the shape of an
isosceles triangle with height h equal to its base (see gure).
y
h/2
h/2
h
O
Z
(out of page)
x
(a) What are the pri
1
PHY5246
Midterm 2: Solution
1.
(a) The xy and yz planes passing through the point O are symmetry planes so the z and x axes are principal axes.
The third principal axis must be orthogonal to both z and x, and thus is the y axis.
(b) The area of the tria
1
PHY5246 Final Exam: Solution
Problem 1. Part (a) The canonical momenta are pr = p Part (b) The Lagrangian can be written L= 1 (r, ) 2 m 0 0 mr2 r 1 + (0, eBr2 ) 2 r (3) L = mr r L eB r2 = = = mr2 + 2 (1) (2)
and so, using the matrix method, we find the
Theoretical Dynamics - PHY 5246
Midterm I October 22, 2003
1. A long straight wire makes an angle with the z axis. The wire rotates about the z axis with a constant angular frequency as shown in the figure. Both and are fixed. A bead of mass m slides on t
Theoretical Dynamics PHY 5246
Midterm I October 15, 2004
1. (60 pts) A small body of mass m moves under the inuence of gravity on the interior of the surface of a cone. Friction can be ignored. Taking the z direction to be up, the surface of the cone is d
Theoretical Dynamics - PHY 5246
Final Exam December 14, 2006
1. (50 pts) Three objects of equal mass m are confined to move in one dimension. The masses are connected by springs with equal spring constants k and equilibrium lengths b, as shown in the figu
1
PHY5246 Midterm 1: Solution
Problem 1. Part (a) 1 m(r2 + r2 2 + 2 r2 ) - mgr 2
L= Part (b)
(1)
L = m(1 + 2 )r r d dt L r
(2)
= m(1 + 2 ) r
(3)
L = mr2 - mg r Euler-Lagrange equation for r m(1 + 2 ) = mr2 - mg r is cyclic so L = mr2 = l
(4)
(5)
(6)
is
1
PHY5246 Midterm 2: Solution
Problem 1. Part (a) The rotational kinetic energy of the two bars about their centers of mass is TRot. = 1 1 2 2 ( M L2 )(1 + 2 ) 2 12 (1)
and the translational kinetic energy of the centers of mass is TT rans. = so the total
Theoretical Dynamics - PHY 5246
Midterm II November 30, 2005
1. (40 pts) Consider a thin uniform square tile of mass M , and side L. The tile is rotating with constant angular velocity about a fixed axis passing through the center of the tile making an an
PHY 5246: Theoretical Dynamics, Fall 2010
Assignment # 4, Solutions
1
Graded Problems
Problem 1
(1.a)
The problem has spherical symmetry and is therefore naturally solved using spherical coordinates
(see gure). The xed length of the pendulum gives the con
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 5, Solutions
1
Graded Problems
Problem 1
(1.a)
Using the equation of the orbit or force law
d2
d2
1
mr 2
1
+ = 2 F (r ) ,
r
r
l
(1)
with r () = ke one nds
2 1
mr 2
+ = 2 F (r ) ,
r
r
l
(2)
(1 + 2 )l2
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 6, Solutions
1
Graded Problems
Problem 1
The equation of the orbit for the force given in this problem is
2
2
that can be written as
1
mr 2
mk m
1
+ = 2 F (r ) = 2 + 2 ,
r
r
l
l
lr
2
2
m
1
+ 1 2
r
l
m
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 7, Solutions
1
Graded Problems
Problem 1
(1.a)
In order to nd the equation of motion of the triangle,
we need to write the Lagrangian, with generalized coordinate cfw_. The potential energy is going t
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 8, Solutions
1
Graded Problems
Problem 1
First we calculate the moments of inertia:
I1 = I2 = m
x3
I3 =
a
x1
ma2
.
2
(1.a)
The torque is zero! This can be seen in several ways: for
instance, from the
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 9, Solutions
1
Graded problems
1. In coordinates (1 , 2 ) (see gure), we have
1
1
m(b1 )2 + m(b2 )2
2
2
T=
V
1
= mgb(1 cos 1 ) + mgb(1 cos 2 ) + k (b sin 1 b sin 2 )2 .
2
Note the equilibrium length o
PHY 5246: Theoretical Dynamics, Fall 2012
November 21st, 2012
Assignment # 10, Solutions
1
Graded problems
Problem 1
(1.a)
The Lagrangian is
1
L = m(r 2 + r 2 2 + r 2 sin2 2 ) V (r ),
2
and the conjugate momenta are
L
= mr,
r
L
= mr 2 ,
=
L
=
= mr 2 sin2
PHY 5246: Theoretical Dynamics, Fall 2012
December 5th, 2012
Assignment # 11, Solutions
1
Graded problems
Problem 1 (Goldstein 9.21)
(1.a)
We have a 1D system
H=
1
p2
2.
2
2q
D=
pq
Ht
2
We want to show that
is a constant of the motion.
D
D
pq
dD
= [D, H
PHY 5246: Theoretical Dynamics, Fall 2012
Assignment # 1, Solutions
Problem 1
y
cos
r
sin
r
r=
sin
cos
r
x
Starting from the position vector for a planar motion in polar coordinates
r = r ,
r
we can derive the velocity vector as
r = r + r ,
r
where w