Problem 3.126 - So the slope at the free surface is slope 5 ft ⋅ 20 rev min ⋅ min 60 s ⋅ × 2 π ⋅ rad ⋅ rev × ⎛ ⎝ ⎞ ⎠ 2 × s 2

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Problem *3.126 [Difficulty: 3] H /2 D = 2.5 in. H = 5 in. r z R = 5 ft SG = 1.05 Given: Half-filled soft drink can at the outer edge of a merry-go-round ω 0.3 rev s = Find: (a) Slope of free surface (b) Spin rate necessary for spillage (c) Likelihood of spilling versus slipping Solution: We will apply the hydrostatics equations to this system. a g p G G ρ = + Governing Equations: (Hydrostatic equation) r r a g r p = + (Hydrostatic equation radial component) z z a g z p = + (Hydrostatic equation z component) Assumptions: (1) Incompressible fluid (2) Rigid body motion (3) Merry-go-round is horizontal 2 ω r r p = g z p = a r V 2 r = r ω () 2 r = r ω 2 = a z 0 = g r 0 = g z g = Thus: So p = p(r,z) g r g r z p r p dr dz 2 2 ω = = dz z p dr r p dp + = For the free surface the pressure is constant. Therefore: So the slope at the free surface is
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Unformatted text preview: So the slope at the free surface is slope 5 ft ⋅ 20 rev min ⋅ min 60 s ⋅ × 2 π ⋅ rad ⋅ rev × ⎛ ⎝ ⎞ ⎠ 2 × s 2 32.2 ft ⋅ × = slope 0.681 = H D To spill, the slope must be slope sp H D = slope sp 5 2.5 = slope sp 2.000 = Thus, ω sp g r dz dr ⋅ = ω sp 32.2 ft s 2 ⋅ 1 5 ft ⋅ × 2 × = ω sp 3.59 rad s ⋅ = This is nearly double the original speed (2.09 rad/s). Now the coefficient of static friction between the can and the surface of the merry-go-round is probably less than 0.5.Thus the can would not likely spill or tip; it would slide off!...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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