#### Transcript Lecture 1: Rotation of Rigid Body

Chapter 15: Fluid Motion Fluids Characteristics of fluids • Microscopically molecules of a fluid do not have long-range order. But liquids do have short-range order unlike gases • Fluids can flow and conform to the boundaries of a container • Fluids cannot sustain a shearing stress Fluids (cont’d) Some useful quantities Density for mass m ,volume V m m V V for uniform solid or liquid unit kg/m3 (at STP for a gas) in SI unit water 1.0 103 kg/m3 ; air 1.2 kg/m3 styrofoam 0.1103 kg/m3 ; lead 11.35103 kg/m3 Pressure for area F dF p A A0 dA A and normal force F unit pascal Pa 1 N / 1 m2 in SI unit other useful units: 1 atm 1.01105 Pa 760 torr 14.7 lb / in 2 height of a column of Hg corresponding to this pressure (mm) Pressure Pressure of a liquid at rest (uniform density) F1 p1 A F2 p2 A fluid level y1 y2 h F1 F2 ( y2 y1 ) gA 0 A imaginary box F1 F2 mg A weight of the imaginary box p2 p1 ( y2 y1 ) g hg p2 p1 hg mg Pressure (cont’d) Pressure p0 of a fluid at rest : gauge pressure atmospheric pressure gauge pressure: fluid level y1 y2 h p2 p1 hg p1 p0 p1 A imaginary box F1 F2 mg A gauge p2 p0 p2 gauge atmospheric pressure A simple model for atmospheric pressure kp ) p2 p1, y2 y1 0, Pressure When of a gas at rest ( dp p2 p1 g ( y1 y2 ) gdy dp gdy kpgdy dp / dy kpg p2 p1e kg ( y2 y1 ) Barometer Pressure measurement using a liquid: (measures absolute pressure) p0 From as h atmospheric p0 pressure p2 p1 g ( y1 y2 ) p2 0, y2 h, p1 p0 p0 gh If the liquid is mercury, for 1 atm : p0 1 atm 101.3 103 P a, Hg 13.6 103 kg/m3 , g 9.83 m/s h p0 / g 0.758m 760 mm Manometer Pressure measurement using a liquid : (measures gauge pressure) p2 p1 g ( y1 y2 ) p0 p2 p, y2 h, level 1 pg pg p p0 gh h level 2 liquid tank manometer p1 p0 , y1 0 gauge pressure Pascal’s law incompressible Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and walls of the containing vessel p pext p0 gh weight piston (area A) pressure at P pext w liquid h P p pressure due atmospheric to weight w: pressure w/A p pext pressure due to liquid above P Pascal’s law (cont’d) Hydraulic lever p F1 / A1 F2 / A2 F2 ( A2 / A1 ) F1 A2 / A1 1 F2 F1 x1 F1 x2 x1 A1 x2 A2 x2 ( A1 / A2 )x1 A2 A1 W2 F2 x2 ( A2 / A1 ) F1 ( A1 / A2 )x1 F2 F1x1 W1 Buoyancy Origin of buoyancy • Consider a submerged massless object filled with the same fluid as the fluid that surrounds the object The object is at rest Fnet ( F2 F1 ) m f g 0 buoyancy P1 F1 mfg P2 F2 mass of fluid in the object opposite dir. buoyant force Fb m f g • Now fill the object with another material Fnet ( F2 F1 ) mobj g 0 Fb m f g Buoyancy Origin of buoyancy (cont’d) • Consider a portion of fluid at rest in a container surrounded by an imaginary boundary represented in dashed line. • Since the portion of the fluid defined by the surface in dashed line is at rest, the net force on this portion due to pressure must be equal to that of the weight of the fluid inside the surface, and opposite in direction. dF dF dF Fnet dF mfluid g Fb fluidVg 0 dF dF Fnet X mg dF dF dF Fb mfluid g fluidVg buoyant force • The same argument can be applied when the imaginary portion of the fluid is replaced by an object that occupies the same space. Buoyancy Archimedes’s principle • The buoyant force on a partly or completely submerged object is equal to the weight of the displaced fluid: Fnet Fb mobj g m f g mobj g ( f obj )Vg • If f obj , Fnet 0 apparent weight • If f obj , Fnet 0 Object sinks W ( f obj )Vg mobj g Object floats The object will rise until a part of it comes out above the fluid Surface when the average density increases to f Example What fraction of an iceberg is submerged in the sea water? Let’s assume that the total volume of the iceberg is Vi. Then the weight of the iceberg Fgi is Fgi iVi g Let’s then assume that the volume of the iceberg submerged in the sea water is Vw. The buoyant force B caused by the displaced water becomes B wVw g Since the whole system is at its static equilibrium, we obtain Therefore the fraction of the volume of the iceberg submerged under the surface of the sea water is iVi g wVw g Vw i 917kg / m3 0.890 3 Vi w 1030kg / m About 90% of the entire iceberg is submerged in the water!!! Example • A fake or pure gold crown? Is the crown made of pure gold? Tair =7.84 N T mg 0 air Twater =6.86 N Twater mg B 0 Twater Tair B 0 B Tair Twater water gVwater 0.980 N 4 Vwater Vcrown 1.0010 m m Tair / g 0.800kg 3 gold=19.3x103 kg/m3 crown m / Vcrown 8.00103 kg/m3 Example • Floating down the river What depth h is the bottom of the raft submerged? wood=6.00x102 kg/m3 B mraft g 0 B mraft g mraft g ( raftVraft ) g B mwater g ( waterVwater ) g ( water Ah) g (water Ah) g (raftVraft ) g raftVraft h 0.0632m water A A=5.70 m2 Density and Pressure Example • Oil and water P1 P0 gh1 1.01105 P a (7.00102 kg/m3 )(9.80 m/s2 ) (8.00 m) 1.5610 P a 5 Pbot P1 gh2 2.06105 Pa =0.700 g/cm3 h1=8.00 m =1025 kg/m3 h2=5.00 m Ideal fluid flow Ideal fluids in motion • Incompressible ( density is constant at any position) • No internal friction (no viscosity) • Steady (non-turbulent) flow- the velocity at a point is constant in time. flow tube A flow line : The path of an individual particle in a moving fluid steady flow: A flow whose pattern does not change with time. Every element passing through a given point follows the same flow line streamline : A curve whose tangent at any point is in the direction of the fluid velocity at that point flow tube : The flow lines passing through the edge of an imaginary area such as A flow lines Continuity equation Continuity equation I (incompressible fluid) v2 The mass of a moving fluid does not change as it flows. The volume of the fluid that passes through area A during a small time interval dt : dV Avdt A2 v2dt v1 dV / dt Av A1 v1dt dV1 A1v1dt ; dV2 A2v2dt •In an ideal fluid the density is constant. •In a time interval dt the mass that flows into Area 1 is the same as the mass that flows out of Area 2. A1v1dt A2v2dt A1v1 A2v2 Continuity equation v2 A2 Volume flow rate dV Avdt v2dt dV / dt Av volume flow rate Continuity equation II (compressible fluid) v1 1 A1v1dt 2 A2v2dt 1 A1v1 2 A2v2 A1 v1dt Bernoulli’s equation Work v2 dW p1 A1ds1 p2 A2ds2 F2 p2 A2 ( p1 p2 )dV dV A2 v1 A1 Change ds2 v2dt dV A1ds1 A2ds2 ds1 v1dt y1 in kinetic energy dK (1 / 2) dV (v22 v12 ) Change in potential energy dU dVg( y2 y1 ) dV F1 p1 A1 done by pressure y2 Bernoulli’s equation Energy conservation dW dK dU dW ( p1 p2 )dV dK (1 / 2)dV (v22 v12 ) dU dVg( y2 y1 ) ( p1 p2 )dV (1 / 2)dV (v22 v12 ) dVg( y2 y1 ) p1 p2 (1 / 2) (v22 v12 ) g( y2 y1 ) p1 gy1 (1/ 2) v12 p2 gy2 (1/ 2) v22 const. Example Venturi meter 1 2 1 2 p1 v1 p2 v2 2 2 A1 v2 v1 A2 1 2 A12 p1 p2 v1 ( 2 1) 2 A2 p1 p2 gh v1 2 gh ( A1 / A2 )2 1 h Torricelli’s theorem Example The velocity of the fluid coming out of a hole in a tank as shown in the figure can be calculated using Bernoulli’s equation. At the top surface the velocity of the fluid is zero. The pressures at the top surface and at the hole are the same, namely, the atmospheric pressure. 2 2 ptop gytop (1/ 2)vtop phole gyhole (1/ 2)vhole 2 g ( ytop yhole ) (1 / 2)vhole vhole 2 gh Siphon Example Suppose a U-shaped piece of pipe is completely submerged in water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of water, the water does not run out of the pipe. WHY? Siphon Example Suppose a U-shaped piece of pipe is completely submerged in water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of water, the water does not run out of the pipe. WHY? Air cannot enter the pipe. As the water starts running out of the pipe, a near vacuum is created in the topmost region of the inverted U. The pressure here drops to near zero. The atmospheric pressure on the surface of the water in the bucket pushes the water into the U-shaped pipe. Siphon Example If a U-shaped hose or pipe connects a liquid-filled container at a higher altitude to a container at a lower altitude over a barrier, the liquid can be siphoned into the container at the lower altitude. Atmospheric pressure helps to push the liquid over the barrier. Siphon Example When P1>P2, the fluid can be siphoned from the left to the right bucket. Example • Water garden A water hose 2.50 cm in diameter is used by a gardener to fill a 30.0-liter bucket. The gardener notices that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected? 30.0 L 1.00 10 3 cm 3 1.00 m volume flow rate 1.00 min 1.00 L 100.0 cm 3 1.00 min 3 3 5.00 10 m /s 60.0 s A1v1 A2v2 A2v0 x (v0 x : the x - componentof theinitialvelocity) A1v1 5.00104 m3 / s v0 x 10.0 m/s 4 2 A2 0.50010 m 1 2t 2(1.00 m) y v0 y t gt2 t 0.452s; x v0 xt 4.52 m 2 2 g 9.80 m/s • Example : A water tank Consider a water tank with a hole. (a) Find the speed of the water leaving through the hole. P0 h =0.500 m y1 =3.00 m 1 2 v1 gy1 P0 gy 2 2 v1 2 g ( y2 y1 ) 2 gh 3.13 m/s (b) Find where the stream hits the ground. 1 2 y 0 y1 gt v0 y t t 0.782 s 2 x v0 xt v1t 2.45 m y x • Example : Fluid flow in a pipe Find the speed at Point 1. A1v1 A2v2 v2 A2=1.00 m2 A1=0.500 m2 h =5.00 m A1 v1 A2 1 2 1 2 P0 v1 gy1 P0 v2 gy 2 2 2 2 P0 1 2 1 A v1 gy1 P0 1 v1 gy2 2 2 A2 A 2 2 gh v12 1 1 2 g ( y2 y1 ) 2 gh v1 11.4 m/s A2 1 ( A1 / A2 ) 2 Viscosity and turbulence Viscosity Viscosity is internal friction in a fluid, and viscous forces oppose the motion of one portion of a fluid relative to another. Viscosity and turbulence Drag If a fluid in laminar flow flows around an obstacle, it exerts a viscous drag on obstacle. Frictional forces accelerate the fluid backward against the direction of flow and the obstacle forward in the direction of flow. adjacent layers of fluid slide smoothly past each other and flow is steady laminar flow Viscosity and turbulence Turbulence When the speed of a flowing fluid exceeds a certain critical value the flow is no longer laminar. The flow patter becomes extremely irregular and complex, and it changes continuously in time. There is no steady flow pattern. This chaotic flow Is called turbulence. Problems Problem 1 The upper edge of a gate in a dam runs the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along the horizontal line through its center. Calculate the torque about the hinge arising from the force due to the water. 2.00 m Solution Denote the width and depth at the bottom of the gate by w and H. The force on a strip of vertical thickness dh at a depth h is: dF gh(wdh) and the torque about the hinge is d gwh(h H / 2)dh. After integrating from h=0 to h=H, you get the torque: gwH 3 / 12 2.61104 N m. Problem 2 An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density . (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object in equilibrium. (b) A downward force with magnitude F is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (c) Calculate the period of the oscillation when the force F is suddenly removed. Solution (a) From Archimedes’s principle gLA Mg, so L M /( A). (b) The buoyant force is: gA( L x) Mg F. With the result of part (a) solving for x gives: x F /( gA). (c) The force is always in the direction toward the equilibrium, namely, a restoring force Fres Fbuoy Mg gAx. Therefore the “spring constant” is k gA and the period of the oscillation is T 2 M / k 2 M /(gA). Problem 3 You cast some metal of density m in a mold, but you are worried that there might be cavities within the casting. You measure the weight of the casting to be w, and the buoyant force when it is completely surrounded by water V0 B /(water g ) w /(m g ) to be B. (a) Show that is the total volume of any enclosed cavities. (b) If your metal is copper, the casting’s weight is 156 N, and the buoyant force is 20 N, what is the total volume of any enclosed cavities in your casting? What fraction is this of the total volume of the casting? Solution (a) Denote the total volume V. If the density of air is neglected, the buoyant force in terms of the weight is: B water gV water g[(w / g ) / m V0 ] Therefore V0 B /(water g ) w /(m g ). (b) B /(water g) w /(Cu g) 2.52104 m3. The total volume of the casting is B /( water g ), The cavities are 12.4% of the total volume. Problem 4 A A U-shaped tube with a horizontal portion of length contains a liquid. What is the difference in height between the liquid columns in the vertical arms (a) if the tube has an acceleration a toward the right? (b) if the tube is mounted on a horizontal turntable rotating with an angular speed with one of the vertical arms on the axis of rotation? Solutions A (a) Consider the fluid in the horizontal part of the tube. This fluid with mass A , is subject to a net force due to the pressure difference between the ends of the tube, which is the difference between the gauge pressures at the bottoms of the ends of the tubes. Now this difference is g ( yL yR ), and the net force on the horizontal part of the fluid is g ( yL yR ) A Aa, or ( yL yR ) (a / g ). (b) Similarly to (a) consider the fluid in the horizontal part of the tube. As in (a) the fluid is accelerating. The center of mass has a radial acceleration of magnitude arad 2 / 2, so the difference in heights between the columns is (2 / 2)( / g ) 22 /(2g ).