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By A. Frank D'Souza

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Mv) dt I. 53) F dt rPdt d(mv) =I mv 2 - mv 1 where t 2 is the time at which the collar comes to rest again. I! mg d(mv) mv 2 p mv 1 I 0 mv is the linear momentum. This equation can be written as P dt = .. 54) I p Po where the subscripts l and 2 on the right-hand side designate the velocities at times It and 12 , respectively. The left-hand side of this equation is called the linear impulse and the right-hand side is the change in linear momentum. This is called the principle of impulse and momentum.

3. , knowing (H}c. We have (H}. 3 by direct Now, (J; dm) X follows that Vc = ; pc X Bp = (~c+m f (;PC +· ;) ; X cos r~, + (ih)2mco. cos r~, metho~. and the scalar components of the angular momentum vector H"' + f; + mX ;) dm 0 since the integral vanishes. 25) The foregoing result is called the translation theorem for angular momentum. H, =/,co, B. 28) where, as shown in Chapter 2, the rotation transformation matrix [C) is made up of direction cosines between the x 1 y 1 z 1 and xyz axes. 24) is the angular momentum of the body about the center of mass.

58) it is seen that h is the angular momentum per unit mass. 69) and (3. 70) become It is concluded that·~. 69) m1 . _ dr dO r - d(J dt .. 72). 75) u where the right-hand side forcing function is a constant. Its solution is obtained by adding the particular solution G(m 1 + m 2 )/h 2 to the complementary solution C cos (0- 0 to obtain 0) ... + = u = G(m 1ht mz) + C cos (0 00 ) (3. 76) 68 Dynamics of Particles: Newton's Law, Energy, and Momentum Methods Sec. 8 Chap. 3 G(ml ,j- mz) [1 = + e cos ((J Case 1: e > I, E > 0.

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