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Physics 3309 Homework 8
Chapter 7
7-9.
Using the generalized coordinates given in the figure, the Cartesian coordinates for the disk are (x cos a, –x sin a), and for the bob they are (l sin f + x cos a, –l cos f – x sin a). The kinetic energy is given by
 (1)
Substituting the coordinates for the bob, we obtain
 (2)
The potential energy is given by
 (3)
Now let us use the relation x = Rq to reduce the degrees of freedom to two, and in addition substitute  for the disk. The Lagrangian becomes
 (4)
The resulting equations of motion for our two generalized coordinates are
 (5)
 (6)
7-15.
b = unextended length of spring
l = variable length of spring
Taking Lagrange’s equations for l and q gives
This reduces to
7-17.
Using q and q (= w t since q (0) = 0), the x,y coordinates of the particle are expressed as
 (1)
from which
 (2)
Therefore, the kinetic energy of the particle is
 (3)
The potential energy is
 (4)
Then, the Lagrangian for the particle is
 (5)
Lagrange’s equation for the coordinate is
 (6)
The complementary solution and the particular solution for (6) are written as
 (7)
so that the general solution is
 (8)
Using the initial conditions, we have
 (9)
Therefore,
d = 0,  (10)
and
 (11)
or,
 (12)
In order to compute the Hamiltonian, we first find the canonical momentum of q. This is obtained by
 (13)
Therefore, the Hamiltonian becomes
so that
 (14)
Solving (13) for  and substituting gives
 (15)
The Hamiltonian is therefore different from the total energy, T + U. The energy is not conserved in this problem since the Hamiltonian contains time explicitly. (The particle gains energy from the gravitational field.)
7-23. The Hamiltonian function can be written as [see Eq. (7.153)]
 (1)
For a particle which moves freely in a conservative field with potential U, the Lagrangian in rectangular coordinates is
and the linear momentum components in rectangular coordinates are
 (2)
 (3)
which is just the total energy of the particle. The canonical equations are [from Eqs. (7.160) and (7.161)]
 (4)
These are simply Newton’s equations.
7-27.
a)
Using l, q as polar coordinates
 (1)
 (2)
If we substitute (1) and (2) into
the result is
The equations of motion are
So  = constant
So  = constant
which reduces to
which reduces to
b) As was shown in (a)
 (total linear momentum)
c) Using L from part (a)
Inverting these equations gives (after much algebra)
Since the coordinate transformations are time independent, and U is velocity independent,
Substituting using the above equations for  in terms of  gives
The equations of motion are
Note: This solution chooses as its generalized coordinates what the student would most likely choose at this point in the text. If one looks ahead to Section 8.2 and 8.3, however, it would show another choice of generalized coordinates that lead to three cyclic coordinates ( ,  , and q ), as shown in those sections.
7-29.
b = unextended length of spring
l = variable length of spring
a) 
Substituting into 
gives
Lagrange’s equations give:
Upon simplifying, the equations of motion reduce to:
b)  (1)
or  (2)
Since the transformation equations relating the generalized coordinates to rectangular coordinates are not time-independent, the Hamiltonian is not the total energy.
Substituting (1) and (2) for  and  and simplifying gives
The equations for  and  are
The equations for  and  are
or
or
c) 
Substitute into Lagrange’s equations of motion
For small oscillations,  . Dropping all second-order terms gives
For q,
The solution to the equation for l is
So for l,
7-33.
Neglect the masses of the pulleys
We redefine the zero in U such that the constant in L = 0.
 (1)
 (2)
Solving (1) and (2) for  and  gives
where 
Substituting for  and  and simplifying gives
The equations of motion are
Chapter 8
8-3. When  , the potential energy will decrease to half its former value; but the kinetic energy will remain the same. Since the original orbit is circular, the instantaneous values of T and U are equal to the average values,  and . For a  force, the virial theorem states
 (1)
Hence,
 (2)
Now, consider the energy diagram
where
 original total energy
 original potential energy
 original centrifugal energy
The point B is obtained from  . According to the virial theorem,  or  Therefore,
Hence, if U suddenly is halved, the total energy is raised from B by an amount equal to  or by  . Thus, the total energy point is raised from B to C; i.e., E(final) = 0 and the orbit is parabolic.
At this point, you can go to the 3309 page,
the UH Space Physics Group
Web Site, or my personal Home Page.
Edgar A. Bering, III ,
Edgar A. Bering, III , <eabering@uh.edu>
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Homework 8
