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# phys documents (dragged) 12 - T = 1 2 M tot ˙ R 2 1 2 μ...

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Chapter 1: Mechanics 5 1.4.2 Tensor notation Transformation of the Newtonian equations of motion to x α = x α ( x ) gives: dx α dt = x α ¯ x β d ¯ x β dt ; The chain rule gives: d dt dx α dt = d 2 x α dt 2 = d dt ± x α ¯ x β d ¯ x β dt ² = x α ¯ x β d 2 ¯ x β dt 2 + d ¯ x β dt d dt ± x α ¯ x β ² so: d dt x α ¯ x β = ¯ x γ x α ¯ x β d ¯ x γ dt = 2 x α ¯ x β ¯ x γ d ¯ x γ dt This leads to: d 2 x α dt 2 = x α ¯ x β d 2 ¯ x β dt 2 + 2 x α ¯ x β ¯ x γ d ¯ x γ dt ± d ¯ x β dt ² Hence the Newtonian equation of motion m d 2 x α dt 2 = F α will be transformed into: m ³ d 2 x α dt 2 + Γ α βγ dx β dt dx γ dt ´ = F α The apparent forces are taken from he origin to the effect side in the way Γ α βγ dx β dt dx γ dt . 1.5 Dynamics of masspoint collections 1.5.1 The centre of mass The velocity w.r.t. the centre of mass ± R is given by ± v - ˙ ± R . The coordinates of the centre of mass are given by: ± r m = m i ± r i m i In a 2-particle system, the coordinates of the centre of mass are given by: ± R = m 1 ± r 1 + m 2 ± r 2 m 1 + m 2 With ± r = ± r 1 - ± r 2 , the kinetic energy becomes:
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Unformatted text preview: T = 1 2 M tot ˙ R 2 + 1 2 μ ˙ r 2 , with the reduced mass μ given by: 1 μ = 1 m 1 + 1 m 2 The motion within and outside the centre of mass can be separated: ˙ ± L outside = ± τ outside ; ˙ ± L inside = ± τ inside ± p = m ± v m ; ± F ext = m ± a m ; ± F 12 = μ ± u 1.5.2 Collisions With collisions, where B are the coordinates of the collision and C an arbitrary other position, holds: ± p = m ± v m is constant, and T = 1 2 m ± v 2 m is constant. The changes in the relative velocities can be derived from: ± S = Δ ± p = μ ( ± v aft-± v before ) . Further holds Δ ± L C = ± CB × ± S , ± p ± ± S = constant and ± L w.r.t. B is constant....
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