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 merrygoround is probably less than 0.5.Thus the can would not likely spill or tip; it would slide off!...
View
Full Document
 Fall '07
 Lear
 Fluid Mechanics

Click to edit the document details