By E. Poisson

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**Extra resources for Advanced Mechanics [Phys 3400 Lecture Notes]**

**Sample text**

1 Introduction: From Newton to Lagrange The methods of Newtonian mechanics, based on the vectorial equation F = ma, are very powerful and they can be applied to all mechanical systems. But they lack in efficiency when Cartesian coordinates (x, y, z) do not give the simplest description of a mechanical system. An example is the problem of the pendulum (Sec. 7), which is best analyzed in terms of the swing angle θ; we have seen that to derive the equation of motion for θ(t) requires somewhat laborious calculations, and the reason is precisely that θ is not a Cartesian coordinate.

A) Derive an equation of motion for R, the position of the centre of mass. (b) Derive an equation of motion for r, the position of body 1 relative to body 2. (c) Prove that h = r × r˙ is a constant vector; conclude that the motion takes place in a fixed plane. ˙ (d) Introduce the polar coordinates (r, φ) and prove that |h| = h = r2 φ; conclude that Kepler’s law — the law of areas — is valid for all central forces, and not just for gravity. (e) Show that the equation of motion for r reduces to r¨ + f h2 − 3 = 0, µ r where µ = m1 m2 /(m1 + m2 ) is known as the reduced mass of the twobody system.

In Eq. 13) we have the reduction of our original two-body problem to a simpler effective one-body problem. The effective body is fictitious; it moves with a position vector r(t) in the gravitational field of another fictitious mass M = m1 + m2 situated at the centre of mass of the original system. 3 Conservation of angular momentum From Eq. 13) we can immediately derive the statement r × r¨ = −GM r×r = 0. r3 But d(r × r)/dt ˙ = r˙ × r˙ + r × r¨ = r × r¨, and it follows that d(r × r)/dt ˙ = 0. 15) is therefore constant during the motion.