By Fowles G.R., Cassiday G.L.
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Extra resources for Analytical mechanics, 7ed., Solutions manual
005g g f = − µ mg is the frictional force acting on the G A0 G f box, so G G G f − mA0 = ma′ (b) (a) G ( a′ is the acceleration of the box relative to the truck. 6 (a) r = iˆ ( xD + R cos Ωt ) + ˆjR sin Ωt K r = −iˆΩR sin Ωt + ˆjΩR cos Ωt K K r ⋅ r = v 2 = Ω 2 R 2 ∴ v = ΩR circular motion of radius R K K K K K ˆ ′ + ˆjy′ (b) r′ = r − ω × r ′ where r ′ = ix ˆ ′ + ˆjy′ = −iˆΩR sin Ωt + ˆjΩR cos Ωt − ω kˆ × ix ( ) 52 = −iˆΩR sin Ωt + ˆjΩR cos Ωt − ˆjω x′ + iˆω y′ x ′ = ω y′ − ΩR sin Ωt y ′ = −ω x′ + ΩR cos Ωt (c) Let u′ = x′ + iy′ here i = −1 !
2 inches should not cause the outfielder any difficulty. 10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference. Thus … G G G da G G da = +ω × a r = dt fixed dt rot G G G G G G G G G a = r ′ + ω × r ′ + 2ω × r′ + ω × (ω × r ′ ) 57 G G G G G G G G G G da = r ′ + ω × r ′ + ω × r ′ + 2ω × r ′ + 2ω × r ′ dt rot G G G G G G G G G +ω × (ω × r ′ ) + ω × ω × r ′ + ω × ω × r′ ( ) ( ) G G G G G G G G G G G G G G ω × a = ω × r ′ + ω × (ω × r ′ ) + 2ω × (ω × r′ ) + ω × ω × (ω × r ′ ) G G is ⊥ to ω and r ′ .
24 The equation of motion is F ( x ) = x − x 3 = mx . For simplicity, let m=1. Then (a) (b) x = x − x 3 . This is equivalent to the two first order equations … x = y and y = x − x3 The equilibrium points are defined by x − x 3 = x (1 − x )(1 + x ) = 0 Thus, the points are: (-1,0), (0,0) and (+1,0). We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points. We’ll do this in part (c).