Analytical Mech Homework Solutions 54

Analytical Mech Homework Solutions 54 - A, Aε are the...

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Unformatted text preview: A, Aε are the accelerations of the asteroid and the Earth in the fixed, inertial frame of reference. 1st : examine: A − Aε − Ω × Ω × r = ω × ω × R − Ω × Ω × Rε − Ω × Ω × ( R − Rε ) = (ω × ω − Ω × Ω ) × R = − (ω 2 − Ω 2 ) R ˆ ˆ note: ω = ω k , Ω = Ω k Thus: a = ( Ω 2 − ω 2 ) R − 2Ω × v Therefore: ˆ jy ˆ ˆ ix + ˆ = ( Ω 2 − ω 2 ) iR cos ( Ω − ω ) t − ˆR sin ( Ω − ω ) t −2 ˆΩx + 2i Ωy j j ( ) Thus: x = ( Ω 2 − ω 2 ) R cos ( Ω − ω ) t + 2Ωy y = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx Let x = ( Ω − ω ) y and Then, we have y = ( Ω − ω ) R cos ( Ω − ω ) t + y = (Ω − ω ) x 2Ω y (Ω − ω ) which reduces to y = − ( Ω − ω ) R cos ( Ω − ω ) t Integrating … y = − R sin ( Ω − ω ) t → 0 at t = 0 Also, or − x ( Ω − ω ) = − ( Ω 2 − ω 2 ) R sin ( Ω − ω ) t − 2Ωx x = ( Ω + ω ) R sin ( Ω − ω ) t + 2Ωx x = − ( Ω − ω ) R sin ( Ω − ω ) t Integrating … x = R cos ( Ω − ω ) t + const x = R cos ( Ω − ω ) t − Rε → R − Rε at t = 0 5.8 Relative to a reference frame fixed to the turntable the cockroach travels at a constant speed v’ in a circle. Thus y′ v ′2 ˆ a′ = − er ′ . ω b Since the center of the turntable is fixed. x′ b A =0 The angular velocity, ω, of the turntable is constant, so 4 ...
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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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