Analytical Mech Homework Solutions 170

Analytical Mech Homework Solutions 170 - 7 V ′′ = mg (...

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Unformatted text preview: 7 V ′′ = mg ( b − a ) cos θ mg ( b − a ) V0′′ 5g = = 7 2 M 7 (b − a ) m (b − a ) 5 ω= T0 = ∴V0′′ = mg ( b − a ) 2π ω 7 (b − a ) 5g = 2π 11.12 The potential energy of the satellite shaped like a “thin rod” is (See Example 11.2.2, Figure 11.2.1) … M e dm m but dm = dx where 2a is the length of the rod. 2a r a GM e m GM e m a dx V =∫ − dx = − −a 2a ∫− a r r 2a V = ∫ −G 1 r = ( r0 2 + x 2 + 2r0 x cos φ ) 2 1 1 r = ( r0 2 + x 2 ) 2 (1 + ε cos φ ) 2 where ε = 2 xr0 r0 2 + x 2 1 GM e m a (1 + ε cos φ ) 2 dx V =− 1 2a ∫− a r0 2 + x 2 ) 2 ( − For r0 >> x, r0 2 + x 2 ≈ r0 2 , ε≈ 1 2x and r ≈ r0 (1 + ε cos φ ) 2 r0 Thus, for small ε , the expression for the potential energy, V, can be approximated … 2 GM e m a 1 2 x 3 2x 1 − cos φ + cos 2 φ dx V ≈− 2ar0 ∫− a 2 r0 8 r0 2 3 GM e 3cos φ 2a V ≈− 2a + 0 + 2ar0 2r0 2 3 GM e a 2 cos 2 φ V ≈− 1 + r0 2r0 2 GM e a 2 GM e a 2 V′≈ 2 cos φ sin φ = sin 2φ r0 2r0 2 2r03 ...
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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