p0188 - 94 CHAPTER 1. INTRODUCTION 1.88 Chapter 1, Problem...

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Unformatted text preview: 94 CHAPTER 1. INTRODUCTION 1.88 Chapter 1, Problem 88 Problem: A solid circular cylinder of radius r and length slides inside a vertical smooth pipe having an inside radius R. A frictionless guide wire (not shown) maintains the cylinder’s orientation. The small space between the cylinder and pipe is lubricated with a thin film that has viscosity µ. Derive a formula for the acceleration, dU/dt, of the cylinder in the vertical pipe. Assume that the cylinder has weight W and is concentric with the pipe as it falls. Use the formula to find the terminal velocity (the velocity when dU/dt → 0) of a 99 mm diameter cylinder that slides inside a 101 mm diameter pipe. The cylinder is 200 mm long and weighs 18 N. The lubricant is glycerin at 20o C. R . ..... .... ........... .. . ....... . .... ................................. .. . .. ..... ...... . .. . ..... ................................................. . ............................... ..... ............................................... . . .. .. . ....................................... ...................................................... .. .... . . . . ............... .......... . . . ........................................... . .. . .. ......................................... ................................................. ... .. . ..... . . ....................... ................. ................................................. . . . . . . .. .... ....... ........................ ..................................................... ............ ... .................. .................................................. .... .. . ... . . ........................................ ............................................. ... .... . . . . . . ....................... ............... .............................................. .. . . ........................................... .................................................. ..... .. . . .............. .. ........................... .. ........ ................................................... . . . ....... . ......................... ......... . ................................................ . .. ..... .. .............................. ..................................................... . ....................................... . .... ... . ............................................ . .. . . .. ........................................ ............................................... . . . .. .... ............................... ....... .................................................. . .. ..... . . . . . .............................. .. .... .................................................. . . . .... . ..... ........... ............................. .................................................... . . .. ..... .......... ............ .... .............................................. . ... .... .. . . .... ........................................... .......................................................... ... . .. ..................................... ....................................................... ................................... .. ....... .. ...... . . . . .. . . ................................................ .... . . .. ....... ............... .............. ................................................. .... ................................. .. . .. . .. .................................................... .............. .................. . .. ................................................ . ....... .. . ........ . . ........................................... ................................................. . ......... ........ . .. ........ ............................................. .. . . .. . .......................................... ................................................ . . .. ................. .... ................ .................................................. .. . . . .. . ... . ........... ......................... ................................................. .. ........................................... .................................... .................................. . .......................................... ..... ....... ..... ....................................... ........................ . ......... r W = mg U Solution: The forces acting on the cylinder are its weight, W and the friction force Since the area of the cylinder is 2π r , the friction force is 2π r τ where τ is the shear stress on the surface of the cylinder. Thus, since the mass of the cylinder is W/g , W dU = W − 2π r τ g dt Because the gap between the cylinder and the pipe is very small compared to the pipe radius, the flow looks locally planar. Thus, we can use the Couette-flow solution to compute the shear stress, i.e., µU τ= R−r Hence, the differential equation governing the motion of the cylinder is dU 2π r g µU 2π r g µ =g− =g− U dt W R−r W (R − r) 1.88. CHAPTER 1, PROBLEM 88 95 which can be rewritten as 2π r µ dU =g 1− U dt W (R − r) The terminal velocity is reached when dU/dt → 0, wherefore U→ W (R − r) 2π r µ First, note that the viscosity of glycerin is µ = ρν = 1260 kg m3 1.19 · 10−3 m2 sec = 1.4994 kg m · sec Then, substituting the given numerical values, we find U= (18 N)(0.0505 − 0.0495) m m = 0.191 2π (0.05 m)(0.2 m)[1.4994 kg/(m · sec)] sec ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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