Solutions Assignment 10 11.17*, 11.19, 11.27*, 11.32*
*For 11.17 add a part (c) in which you nd the normal coordinates. Show
that by using these coordinates the Lagrangian separates into two independent
Lagrangians. *For problem 11.27 assume that m1 6= m2
Solutions Assignment 9
Verify the relations in (8.52) The semimajor axis is found from
ro
2a = rmax + rmin =
a=
+
1
ro
ro
1+
=
2ro
2
1
2
1
The semiminor axis is found from
b = ymax :
Since y = r sin
We nd
dy
d
dy
d
the maximum value for y is determined by
Solutions Assignment 8
7.38 (a) Using spherical polar coordinates for an inverted cone with a half
angle the relation between z and r is z = r cos :
The Cartesian coordinates are
x = r sin cos ; y = r sin sin ; z = r cos :
The Cartesian components of the
Solutions Assignment 7
7.16 The kinetic energy of a cylinder of mass m is given by
T=
2
1
1
mx + I! 2 ;
2
2
where x is the velocity of the center of mass, I is the moment of inertia of the
disk about its center of mass, and ! is its angular velocity. If x
Solutions Assignment 6
6.16 Using the hint given in the problem, the distance between two points
on a sphere of radiuis R is
L=
Z
2
1
q
R2 d
2
+
R2
2
sin
d
2
=R
Z
2
1
q
1 + sin2
02
d:
Since in this form the integrand is independent of the EulerLagrange e
Solutions Assignment 5
5.11 From the conservation of energy we have
E
2E
m
1
1
1
1
2
mv 2 + kx2 = mv2 + kx2 ;
2121
2
22
=
2
2
= v1 + ! 2 x2 = v2 + ! 2 x2
1
2
Solving for the angular frequency yields
!2 =
2
v2
x2
1
2
v1
:
x2
2
Since E = kA2 =2 we have for
Solutions Assignment 4
4.35 (a) Assuming that the length of the string is `; then ignoring the size
of the pulley the gravitational potential energy is
U (x) =
m1 gx
m2 g (`
x) :
The kinetic energy including the rotational energy of the pulley is
T
T
2
2
Solutions Assignment 3
!
3.27 (a) The angular momentum of an object is ` = ! !. Choosing
r
p
the orbit of the planet to lie in the x y plane then ! and ! both lie in this
r
p
plane as well. In polar coordinates we nd that
!=
r
dr
dd
d
(rr) =
b
r+r
b
r;
b
PHYSICS 110A : CLASSICAL MECHANICS
HW 8 SOLUTIONS
(1) Taylor 11.14
For our generalized coordinates we will take the angles 1 and 2 .
1
2
Figure 1: Figure for 11.14.
This leads to a kinetic energy of:
T=
1
mL2 [2 + 2 ].
1
2
2
And the potential term will be
PHYSICS 110A : CLASSICAL MECHANICS
HW 7 SOLUTIONS
(1) Taylor 8.13
U vs. r
100
Centrifugal Potential Energy
Potential Energy
Effective Potential Energy
90
80
70
U
60
50
40
30
20
10
0
0
0.2
0.4
0.6
0.8
1
r
1
Figure 1: Plot of Uef f vs. r for U = 2 kr2 where
PHYSICS 110A : CLASSICAL MECHANICS
HW 5 SOLUTIONS
(1) Taylor 7.38
a
r
Figure 1: Figure for 7.38.
The kinetic energy will be:
1
1
T = mr2 + mr2 sin2 2 .
2
2
And the potential energy will be:
U = mgr cos .
So our Lagrangian is:
1
1
L = mr2 + mr2 sin2 2 mgr
PHYSICS 110A : CLASSICAL MECHANICS
HW 4 SOLUTIONS
(2) Taylor 7.14
For the yoyo the kinetic energy will have a rotational and translational motion:
1
1
T = mv 2 + I 2 .
2
2
Now in our coordinate system v = x and = . We also know the moment of inertia for
PHYSICS 110A : CLASSICAL MECHANICS
HW 3 SOLUTIONS
(1) Taylor 6.6
(a)
Here we are working with ds =
to have:
ds =
dx2 + dy 2 . For a function y = y (x) we will pull out a dx
dx2 + dy 2 = dx
dy
dx
2
1+
2
1+
dx
dy
= dx
1 + (y )2 .
= dy
1 + (x )2 .
(b)
Simila
PHYSICS 110A : CLASSICAL MECHANICS
HW 2 SOLUTIONS
(1) Taylor 5.2
Here is a sketch of the potential with A = 1, R = 1, and S = 1. From the plot we can see
2
1.5
U(r)
1
0.5
0
0.5
1
0
2
4
6
8
10
r
Figure 1: Plot for problem 1.
the minimum of the potential wi
PHYSICS 110A : CLASSICAL MECHANICS
HW 1 SOLUTIONS
(2) Taylor 1.46
(a) The equations of motion for the puck are:
r = R vt
=0
Assuming the puck is launched from the position = 0. Technically with the polar coordinates this should only be correct until the p
INSTRUCTIONS:
Physics 110A: Final Exam
Do four problems.
of problems 3, 4, and 5.
You must do problems 1 and. 2. Then choose any two
Central Forces
Two particles attract each other via a central force 170) = Cr3/2, where C. is a constant.
(8) Write
Physics 110A: Midterm Exam #1
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PHY4324Electromagnetism II
Fall 2011
Final exam 2 hours
Dec. 13, 2011
No other materials except calculators allowed. If you cant do one part of a problem, solve subsequent parts in terms of unknown answerdene clearly. Do 4 of 6
problems, CLEARLY indicatin
PHY4324Electromagnetism II
Fall 2011
Test 2 55 minutes
Nov. 7, 2011
No other materials except calculators and one review sheet allowed. If you cant do
one part of a problem, solve subsequent parts in terms of unknown answerdene
clearly. Do 2 of 4 problems