homework3sol.20100214.4b78b7b9b73619.77664923

homework3sol.20100214.4b78b7b9b73619.77664923 - σ = σ ∗...

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TAM 412, Homework 3, Selected answers Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 1. c. Let f ( σ , τ )= μ στ 1 2 ¡ τ 2 σ 2 ¢ The conditions of the inverse function theorem are satis f ed at σ = τ = 1 and, in fact, everywhere except at σ = τ = 0 corresponding to the point O .Le t C consists of any set that contains C 0 but not the point O . ¡ C , ¯ ξ ¢ is then a coordinate chart. In fact, given x and y , σ = x q y + p x 2 + y 2 , τ = q y + p x 2 + y 2 is a unique inverse to f on C . e. The constraint implies that ( στ 1) 2 1 2 ¡ τ 2 σ 2 ¢ =0 The conditions of the implicit function theorem with ξ 1 =( σ )and ξ 2 =( τ )a r esa t i s f ed at σ = τ = 1 and, in fact, everywhere along the curve given by the above equation except at σ = 1 8 ³ 17 1 ´ 3 / 2 , τ = 1 2 q 17 1 < 1 Let b C consist of all points on this curve for which τ > τ . ³ b C , τ ´ is then a coordinate chart. f. The curve corresponding to
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Unformatted text preview: σ = σ ∗ , τ = τ ∗ . 1 3. The relationship between the coordinates of a tangent vector relative to the cartesian and cylindrical coordinate charts is x y z = ρ cos ϕ − ϕ ρ sin ϕ ρ sin ϕ + ϕ ρ cos ϕ h The relationship between the coordinates of the tangent vector on b C relative to the chart ³ b C , ( y,z ) T ´ and the correspondint tangent vector on C relative to the Cartesian coordinate chart is x y z = y − 2 z y z 4. The virtual power is given by F x x + F y y = R ( F x cos θ + F y sin θ ) θ = F θ θ , such that F x /F y = − tan θ ⇒ F θ = 0 which states that a force pointing towards O has no virtual power, since it is orthogonal to the motion of the particle. 5. 2...
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This note was uploaded on 11/17/2011 for the course TAM 412 taught by Professor Weaver during the Spring '08 term at University of Illinois, Urbana Champaign.

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homework3sol.20100214.4b78b7b9b73619.77664923 - σ = σ ∗...

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