Mecanica Clasica

Goldstein Chapter 1 Derivations
Michael Good
June 27, 2004

1

Derivations

1. Show that for a single particle with constant mass the equation of motion
implies the follwing differential equation for the kinetic energy:
dT
=F·v
dt
while if the mass varies with time the corresponding equation is
d(mT )
= F · p.
dt

Answer:
d( 1 mv 2 )
dT
˙
=2
= mv · v = ma · v = F · v
dt
dtwith time variable mass,
d(mT )
d p2
˙
= ( ) = p · p = F · p.
dt
dt 2
2. Prove that the magnitude R of the position vector for the center of mass from
an arbitrary origin is given by the equation:
M 2 R2 = M

2
mi ri −
i

1
2

2
mi mj rij .
i,j

Answer:

MR =

mi ri

1

M 2 R2 =

mi mj ri · rj
i,j

Solving for ri · rj realize that rij = ri − rj . Square ri −rj and you get
2
2
2
rij = ri − 2ri · rj + rj

Plug in for ri · rj
12
2
2
(r + rj − rij )
2i
1
1
2
2
mi mj ri +
mi mj rj −
2 i,j
2
ri · rj =

M 2 R2 =
M 2 R2 =

1
2

i,j

1
M
2

i

1
2
mi r i + M
2

M 2 R2 = M

2
mj rj −
j

2
mi ri −
i

1
2

1
2

2
mi mj rij
i,j
2
mi mj rij
i,j

2
mi mj rij
i,j

3. Suppose a system of two particlesis known to obey the equations of motions,
d2 R
( e)
Fi ≡ F(e)
M 2=
dt
i
dL
= N(e)
dt
From the equations of the motion of the individual particles show that the internal forces between particles satisfy both the weak and the strong laws of action and reaction. The argument may be generalized to a system with arbitrary
number of particles, thus proving the converse of the argumentsleading to the
equations above.
Answer:
First, if the particles satisfy the strong law of action and reaction then they
will automatically satisfy the weak law. The weak law demands that only the
forces be equal and opposite. The strong law demands they be equal and opposite and lie along the line joining the particles. The first equation of motion tells
us that internal forces have no effect. Theequations governing the individual
particles are
( e)

˙
p1 = F1 + F21
( e)

˙
p2 = F2 + F12

2

Assuming the equation of motion to be true, then
( e)

( e)

˙
˙
p1 + p2 = F1 + F21 + F2 + F12
must give
F12 + F21 = 0
Thus F12 = −F21 and they are equal and opposite and satisfy the weak law
of action and reaction. If the particles obey
dL
= N(e)
dt
then the time rate ofchange of the total angular momentum is only equal to
the total external torque; that is, the internal torque contribution is null. For
two particles, the internal torque contribution is
r1 × F21 + r2 × F12 = r1 × F21 + r2 × (−F21 ) = (r1 − r2 ) × F21 = r12 × F21 = 0
Now the only way for r12 × F21 to equal zero is for both r12 and F21 to lie
on the line joining the two particles, so that theangle between them is zero, ie
the magnitude of their cross product is zero.
A × B = ABsinθ

4. The equations of constraint for the rolling disk,
dx − a sin θdψ = 0
dy + a cos θdψ = 0
are special cases of general linear differential equations of constraint of the form
n

gi (x1 , . . . , xn )dxi = 0.
i=1

A constraint condition of this type is holonomic only if an integrating functionf (x1 , . . . , xn ) can be found that turns it into an exact differential. Clearly the
function must be such that
∂ (f gi )
∂ (f gj )
=
∂xj
∂xi
for all i = j . Show that no such integrating factor can be found for either of the
equations of constraint for the rolling disk.
Answer:

3

First attempt to find the integrating factor for the first equation. Note it is
in the form:
P dx +Qdφ + W dθ = 0
where P is 1, Q is −a sin θ and W is 0. The equations that are equivalent to
∂ (f gi )
∂ (f gj )
=
∂xj
∂xi
are
∂ (f P )
∂ (f Q)
=
∂φ
∂x
∂ (f P )
∂ (f W )
=
∂θ
∂x
∂ (f Q)
∂ (f W )
=
∂θ
∂φ
These are explicitly:
∂ (f )
∂ (−f a sin θ)
=
∂φ
∂x
∂ (f )
=0
∂θ
∂ (−f a sin θ)
=0
∂θ
Simplfying the last two equations yields:
f cos θ = 0
Since y is not...