HW11_prob3_GPS11_G24

HW11_prob3_GPS11_G24 - Problem Goldstein 5-24 (3rd ed 5.11)...

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Problem Goldstein 5-24 (3 rd ed 5.11) We will designate the body frame unit vectors as i , j ,and k , and assume they lie along the principal axes of the body. In this body frame the angular momentum L is expressed as L = I 1 ω 1 i + I 2 ω 2 j + I 3 ω 3 k , (1) and the magnitude is simply L 2 = L · L = I 2 1 ( ω 2 1 + ω 2 2 )+ I 2 3 ω 2 3 , (2) wh e r ew ehav epu t I 1 = I 2 for the symmetric top. Using the body frame expressions for the components of ω ,we f nd easily that ω 2 1 + ω 2 2 = · θ 2 + · φ 2 sin 2 θ . (3) Note that the same expression shows up in the kinetic energy for the symmetric top, Eq.(5.50). We can take advantage of this “coincidence” by rewriting the energy equation, Eq.(5.55), as I 1 ( ω 2 1 + ω 2 2 )=2 E I 3 ω 2 3 2 Mgl cos θ , and substituting this result into Eq.(2) to obtain L 2 =2 I 1 ( E Mgl cos θ )+ I 3 ( I 3 I 1 ) ω 2 3 .
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HW11_prob3_GPS11_G24 - Problem Goldstein 5-24 (3rd ed 5.11)...

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