Derivacion

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 dt with timevariable 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
i 2 mi ri −

1 2

2 mi mj rij . i,j

Answer:

MR =

mi ri

1

M 2 R2 =
i,j

mi mj ri · rj

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 1 2 2 2 (r + rj − rij ) 2 i 1 1 2 2 mi mj ri + mi mj rj − 2 i,j 2 ri · rj = 1 2 mi r i + M 2
2 mj rj − j

M 2 R2 = M 2 R2 =

1 2

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

i,j

1 M 2

i

1 2

M 2 R2 = M
i

2 mi ri −

1 2

2 mi mj rij i,j

3. Suppose a system of two particles is known to obey the equations of motions, d2 R (e) Fi ≡ F(e) M 2 = dti 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 arguments leading to the equations above. Answer: First, if the particles satisfy the strong lawof 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. The equations governing the individual particles are ˙ p1 = F1 + F21 ˙ p2 = F2 + F12
(e)(e)

2

Assuming the equation of motion to be true, then ˙ ˙ 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 of change of the total angular momentum is only equal to the total external torque; that is, the internal torque contribution isnull. 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 the angle between them is zero, ie the magnitude of their cross product is zero. A × B = ABsinθ
(e) (e)

4. The equations ofconstraint 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 function f (x1 , . . . , xn ) can be found that turns it into an exact differential. Clearly the function must be such that ∂(f gi ) ∂(fgj ) = ∂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 even in this first equation, the integrating factor does not depend on y and because of ∂f = 0 it does not depend on θ either. Thus ∂θ f = f (x, φ) The only way for f to satisfy this equation is...