Physics 507
Midterm Exam
Fall 2012
Monday, November 12, 12 - 1:20 pm, ARC 203
All problems carry the same weight.
1. (from homework) Two mass points of mass m1 and m2 are connected by a string passing through
a hole in a smooth table so that m1 rests on t
Physics 507
Solutions to Homework Set 1
Fall 2012
1. Goldstein Ch. 1, problem 12.
v=
mv 2
Mm
G
= 0 escape condition
2
R
M
= 2gR 9.8 6.4 106 m/s = 11.2 km/s,
2G
R
where we used mg = GmM/R2 , which implies GM/R = gR.
2. Goldstein Ch. 1, problem 14.
Let the
Physics 507
Solutions to Homework Set 2
Fall 2012
1. Goldstein Ch. 2, problem 12.
t2
S=
t1
t2
S =
t1
L(qi , qi , qi , t)dt
L
L
L
qi +
qi +
qi dt = 0
qi
qi
qi
Integrate the rst term by parts once and the second term twice. The boundary terms vanish
bec
Physics 507
Solutions to Homework Set 3
Fall 2012
1. Goldstein Ch. 3, problem 13.
a) Let f (r) be the force and V (r) the corresponding potential. The component of the force
towards the center provides the centripetal acceleration
f (r )
mv 2
2[V (r) E ]
Physics 507
Solutions to Homework Set 4
Fall 2012
1. Goldstein Ch. 3, problem 18.
Impulse S means an instantaneous change in particles linear momentum
p = S r
The corresponding changes in the energy, angular momentum and Runge-Lentz vector are
E =
S2
,
2m
Physics 507
Solutions to Homework Set 5
Fall 2012
1. Goldstein Ch. 4, problem 6.
Let A and B be rotations in parts a) and b), respectively and let Rz be a regular rotation about
the current z -axis. The rotation about the old z axis is
C = (BA)Rz (BA)1
be
Physics 507
Solutions to Homework Set 6
Fall 2012
1. Goldstein Ch. 5, problem 13.
Let be the angle the left rod makes with the x-axis. The coordinate system is as shown in the
gure in Goldstein. Initially = /6, when the hinge hits the oor = 0. The coordin
Physics 507
Solutions to Homework Set 7
Fall 2012
1. Goldstein Ch. 6, problem 4.
The stable equilibrium position is 1 = 2 = 0. Expand the potential and kinetic energies
around this
V = m1 gl cos 1 m2 gl(cos 1 + cos 2 ) const +
2
2
mgl1 m2 gl2
+
2
2
2
2
2
Physics 507
Solutions to Homework Set 8
Fall 2012
1. Goldstein Ch. 8, problem 14.
The Lagrangian is in the form of Eq. (8.23) of Goldstein, which diers from what we considered
in class by a linear in the velocities term. So, you can use eq. (8.27) to dete
Physics 507
Solutions to Homework Set 10
Fall 2012
1. Goldstein Ch. 10, problem 6.
The vector potential is
A=
Br
2
and the Lagrangian reads
L=
mr2 mr2 2
Br
kr2
mr2 mr2 2 qBr2 kr2
+
+ q v
=
+
+
2
2
2
2
2
2
2
2
We have
qBr2
2
p = mr2 +
pr = mr,
Thus the H
Physics 507
Final Exam
Fall 2012
Monday, December 17, 12 - 2 pm, ARC 108
All problems carry the same weight.
1. Two pendulums of equal length l and masses m1 and m2 are connected by a spring with
spring constant k . Determine the frequencies of small osci
Physics 507
Solutions to Final Exam
Fall 2012
Monday, December 17, 12 - 2 pm, ARC 108
1. Let 1,2 be the angles (in the same direction) pendulums make with the vertical. Note that
there is a normal mode 1 = 2 where the spring is not stretched and pendulums
Physics 507
Solution to the Midterm
Fall 2012
1. Convenient coordinates are e.g.: x the length of the segment of the string on the table and
the angle that segment makes with any xed line in the plane of the table. The hanging part of
the string is then