Solutions for Homework 4
1. (problem 7.8)(a) L = T  U = ½m(x^{·}_{1}^{2} + x^{·}_{1}^{2})  ½kx²
(b) x_{1} = X + ½x + ½l x_{2} = X  ½x  ½l
Differentiate, substitute into L: L = mX^{·}^{2} + ¼mx^{·}^{2}  ½kx²
Hence: X equation: 2mX^{¨} = 0
x equation: ½mx^{¨} =  kx
(c) X(t) = X_{0} + V_{0}t  uniform motion (of Center of Mass)
x(t) = A cos (2k/m)^{½}(tt_{0})  simple harmonic motion relative to each other
2. (Problem 7.16)
Since ω = v/R the K.E is T = ½mv² + ½Iω² = ½(m + I/R²)x^{·}^{2}, and the PE is U = mgx sinα. From L = T  U we then get
(m+I/R²)x^{¨} = mg sinα , or x^{¨} = (mg sinα)/(m+I/R²).
3. (Problem 7.18)
We'll consider the case when the "pendulum part" of the cylinder hangs down from a fixed point (ring). Let x be the length of that part, and θ its angle from the vertical. The angular velocity of the cylinder is ω = x^{·}/R, and the square of the mass is the usual polar coordinate expression (from 1.44). The PE depends only on the height of m, U = mgx cosθ. Therefore
L = ^{1}/_{2}I( 
×
x

/R)^{2} + ^{1}/_{2}m( 
×
x
 2

+ x^{2}  ×
q
 2

) + mgx cosq 

So we get the equations of motion
xequation: (m + I/R^{2}) 
××
x

 mx 
×
q
 2

 mg cosq = 0 

qequation: m(x^{2} 
×
q

)  ×
 + mgx sinq = 0 = m(2x 
×
x
 
×
q
 + x^{2} 
××
q
 ) + mgx sinq. 

4. (Problem 7.21)
See back of book for L. EOM works out to be r^{¨} – ω² r = 0. The given initial conditions imply A = B = r_{0}/2. so for large r, r ® ½r_{0} e^{wt}.
The conserved quantity (Hamiltonian) is ½mr^{·2}  ½mω²r²
5. (Problem 7.38)
(a) In spherical polar coordinates, the angle q is fixed at q = a, so the KE is just T = ^{1}/_{2}m(r^{·2} +r^{2} sin^{2} a f^{·2}). Since the PE is U = mgz = mgrcosa,
L= ^{1}/_{2}m( 
×
r

2

+ r^{2} sin^{2} a 
×
f

2

) mgrcosa 

(b) Since L does not depend on f, the f equation
says simply that ¶L/¶f^{·} = mr^{2} sin^{2}a f^{·} is constant. That is, l_{z} is conserved.
The r equation is
m 
××
r

= mrsin^{2}a 
×
f

2

 mg cosa or 
××
r

= 
l_{z}^{ 2}
m^{2}r^{3}sin^{2}a

 g cosa 
 (1) 
where, in the second version, I have canceled an m and
replaced f^{·} by l_{z}/(m r^{2} sin^{2}a). If
l_{z} = 0, the particle simply slides in the radial direction
with the wellknown acceleration g cosa. (Remember
a is the angle of the incline with the vertical.) The
condition that the particle can remain in a horizontal circle is
that r^{·} = 0 and hence that r = r_{o}
= [l_{z}^{2}/(m gsin^{2}acosa)]^{1/3}.
(c) If we write r = r_{o}+ e, Eq.(1)
becomes

××
e

= 
l_{z}^{2}
m^{2}r_{o}^{3}sin^{2}a


æ è

1 + 
e
r_{o}

ö ø

3

 g cosa »

l_{z}^{2}
m^{2}r_{o}^{3} sin^{2}a


æ è

1  3 
e
r_{o}

ö ø

 g cosa = 
3l_{z}^{2}
m^{2}r_{o}^{4} sin^{2}a

e 

where the first and last terms in the penultimate expression
cancel because r_{o} is the equilibrium radius. This is the
equation of simple harmonic oscillations, so the circular path is
stable and the particle oscillates about this path with frequency
w
= Ö3l_{z}/(mr_{o}^{2}sinα).