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Goldstein_4_6_7_26

# Goldstein_4_6_7_26 - Homework 8 5.4 5.6 5.7 5.26 Michael...

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Homework 8: # 5.4, 5.6, 5.7, 5.26 Michael Good Oct 21, 2004 5.4 Derive Euler’s equations of motion, Eq. (5.39’), from the Lagrange equation of motion, in the form of Eq. (1.53), for the generalized coordinate ψ . Answer: Euler’s equations of motion for a rigid body are: I 1 ˙ ω 1 - ω 2 ω 3 ( I 2 - I 3 ) = N 1 I 2 ˙ ω 2 - ω 3 ω 1 ( I 3 - I 1 ) = N 2 I 3 ˙ ω 3 - ω 1 ω 2 ( I 1 - I 2 ) = N 3 The Lagrangian equation of motion is in the form d dt ( ∂T ˙ q j ) - ∂T ∂q j = Q j The kinetic energy for rotational motion is T = 3 i 1 2 I i ω 2 i The components of the angular velocity in terms of Euler angles for the body set of axes are ω 1 = ˙ φ sin θ sin ψ + ˙ θ cos ψ ω 2 = ˙ φ sin θ cos ψ - ˙ θ sin ψ ω 3 = ˙ φ cos θ + ˙ ψ Solving for the equation of motion using the generalized coordinate ψ : d dt ( ∂T ˙ ψ ) - ∂T ∂ψ = N ψ 3 i I i ω i ∂ω i ∂ψ - d dt 3 i I i ω i ∂ω i ˙ ψ = N ψ 1

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Now is a good time to pause and calculate the partials of the angular veloc- ities, ∂ω 1 ∂ψ = - ˙ θ sin ψ + ˙ φ sin θ cos ψ ∂ω 2 ∂ψ = - ˙ θ cos ψ - ˙ φ sin θ sin ψ ∂ω 3 ∂ψ = 0 and ∂ω 1 ˙ ψ = ∂ω 2 ˙ ψ = 0 ∂ω 3 ˙ ψ = 1 Now we have all the pieces of the puzzle, explicitly 3 i I i ω i ∂ω i ∂ψ - d dt 3 i I i ω i ∂ω i ˙ ψ = N ψ I 1 ω 1 ( - ˙ θ sin ψ + ˙ φ sin θ cos ψ ) + I 2 ω 2 ( - ˙ θ cos ψ - ˙ φ sin θ sin ψ ) - d dt I 3 ω 3 = N ψ This is, pulling out the negative sign on the second term, I 1 ω 1 ( ω 2 ) - I 2 ω 2 ( ω 1 ) - I 3 ˙ ω 3 = N ψ I 3 ˙ ω 3 - ω 1 ω 2 ( I 1 - I 2 ) = N ψ And through cyclic permutations I 2 ˙ ω 2 - ω 3 ω 1 ( I 3 - I 1 ) = N 2 I 1 ˙ ω 1 - ω 2 ω 3 ( I 2 - I 3 ) = N 1 we have the rest of Euler’s equations of motion for a rigid body. 2
5.6 Show that the angular momentum of the torque-free symmetrical top ro- tates in the body coordinates about the symmetry axis with an angular frequency ω . Show also that the symmetry axis rotates in space about the fixed direction of the angular momentum with angular frequency ˙ φ = I 3 ω 3 I 1 cos θ where φ is the Euler angle of the line of nodes with respect to the angular momentum as the space z axis. Using the results of Exercise 15, Chapter 4, show that ω rotates in space about the angular momentum with the same frequency ˙ φ , but that the angle θ between ω and L is given by sin θ = Ω ˙ φ sin θ where θ is the inclination of ω to the symmetry axis. Using the data given in Section 5.6, show therefore that Earth’s rotation axis and axis of angular momentum are never more than 1.5 cm apart on Earth’s surface.

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