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Mech1Sols - Physics 105B Problem Set 1 April 8 2008 Jeff...

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Physics 105B Problem Set 1 April 8, 2008 Je ff Schonert: schonert (at) physics.ucsb.edu Taylor 8.7 a) Mass m 1 is orbiting mass m 2 , and the radially directed force between the two masses is given by Newton’s law of gravitation. If the distance of separation is r , and mass m 1 rotates with angular velocity ω , then by Newton’s second law we have Gm 1 m 2 r 2 = m 1 ω 2 r. (1) But ω = 2 π /T , where T is the period of one revolution, so plugging this in and rearranging gives T 2 = 4 π 2 r 3 Gm 2 . (2) b) Now we have the same masses m 1 and m 2 , but the mass m 2 is allowed to move freely. The only di ff erence in this problem is that in the expression for the centripetal force, m 1 μ , where μ is the reduced mass. Then we have Gm 1 m 2 r 2 = μ ω 2 r, (3) which implies that T 2 = 4 π 2 r 3 GM , (4) where M = m 1 + m 2 . If m 2 is much greater than m 1 , M = m 1 + m 2 m 2 , so that the result in part b) reduces to the result in part a). c) Since m 1 = m 2 , M = 2 m 2 = 2 M sun . Therefore, T = 2 π r 3 / 2 2 GM sun = 1 2 (1 year) = . 71 years . (5) Taylor 8.9 a) For two particles of equal mass m , the kinetic energy is m 2 r 2 1 + ˙ r 2 2 ) . Using equation (8.10), this is equal to M 2 ( ˙ R ) 2 + μ 2 r ) 2 , 1
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where M = 2 m and μ = m/ 2. Now we are told to express ˙ r 2 in terms of two-dimensional polar coordinates ( r, φ ). This means that the kinetic energy is T = M 2 ( ˙ R ) 2 + μ 2 ˙ r 2 + r 2 ˙ φ 2 . (6) The displacement from the equilibrium length L is r - L , so the potential energy is V = k/ 2( r - L ) 2 . The Lagrangian is L = T - V . b) We are asked to express the Lagrangian in terms of the cartesian center-of-mass coordinates ( X, Y ). Since R 2 = X 2 + Y 2 , L = M 2 ˙ X 2 + ˙ Y 2 + μ 2 ˙ r 2 + r 2 ˙ φ 2 - k 2 ( r - L ) 2 . (7) The Lagrangian is independent of X and Y
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