219_Dynamics 11ed Manual

219_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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Unformatted text preview: Engineering Mechanics - Dynamics Chapter 13 t1 = 1 s g = 32.2 ft 2 s t = t1 Solution: θ = ct 2 θ' = 2c t θ'' = 2c Find the angel ψ using rectangular coordinates. The path is tangent to the velocity therefore. x = r cos ( θ ) = ( a)cos ( θ ) + b cos ( θ ) y = r sin ( θ ) = ( a)sin ( θ ) + ψ = θ − atan ⎛ ⎜ y' ⎞ ⎟ ⎝ x' ⎠ 1 2 x' = ⎡−( a) sin ( θ ) − 2b cos ( θ ) sin ( θ )⎤ θ' ⎣ ⎦ 2 b sin ( 2θ ) y' = ⎡( a)cos ( θ ) + b cos ( 2θ )⎤ θ' ⎣ ⎦ ψ = 80.541 deg Now do the dynamics using polar coordinates r = a + b cos ( θ ) F = 1 lb Guesses Given r' = −b sin ( θ ) θ' 2 F N = 1 lb F − FN cos ( ψ) = ⎛F⎞ ⎜ ⎟ = Find ( F , FN) ⎝ FN ⎠ r'' = −b cos ( θ ) θ' − b sin ( θ ) θ'' ⎛ W ⎞ ( rθ'' + 2r' θ' ) ⎜⎟ ⎝g⎠ F N = 0.267 lb −F N sin ( ψ) = F = 0.163 lb Problem 13-97 The smooth particle has mass M. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = b sinθ. If the cord has stiffness k and unstretched length δ determine the force of the guide on the particle when θ = θ1. The guide has a constant angular velocity θ'. Given: M = 80 gm b = 0.8 m k = 30 N m δ = 0.25 m θ 1 = 60 deg 217 ⎛ W ⎞ ( r'' − rθ' 2) ⎜⎟ ⎝g⎠ ...
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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