Initials: Problem 5 - 25 points The device shown in the figure is used to measure fluid speeds. The two coaxial tubes are designed such that the inner tube has a small opening at point 1, which is Let h be the difference in liquid height facing the fluid flow. There is another between the two sides of the manometer small opening at point 2, in the outer tube, tube. The acceleration due to gravity has and the difference in height between the magnitude g. two holes is considered negligible. The fluid being studied has density p, while the manometer tube contains a fluid of density Pm. Both are considered non- viscous and incompressible fluids. The flow of fluid in motion is assumed to be laminar. Because of the static fluid in the manometer tube, there cannot be any fluid 1=0 h flow in the coaxial tubes, and you may neglect the change in pressure in each tube between the hole and the top surface of the manometer fluid. a- Based on the description provided above, explain why the velocity v1 of the fluid is zero at point 1, called stagnation point.Initials: b- Determine the pressure difference P1-P2. Express your answer in terms of p and v2. Hint : Since points 1 and 2 are part of the same external fluid flow, we can use conservation of energy in a flow tube to express a relationship between the two pressures. - Express the same pressure difference P1-P2 using the static fluid present in the manometer (U-shaped tube). Express your answer in terms of pm, g, and h. d- Determine the speed v2 of the external fluid at point 2.Initials: ing to the 139 ~4 Initials: Simple Marmonk Physics 8A Final Equation Sheet 7 (t ) = x(t) i + y(t)j AT = 12-11 =4xi+ 4yj Newton's 1s' Law AT B = constant = > F = 0 VAV = At - = VAU-x1 + VAD-y) B = lim 47 = vxi + vyj Newton's 2nd Law At - 0 4t [F = ma aAv = = QAV-xi + aAV-yl a = lim 40 = Newton's 3rd Law At to at = axi + ay ) FMe/You = -Fyou/Me B (t ) = vo + alt - tol Tension: r (t) = ro + volt - to] + salt - to]2 T Ax = X2 -X1 Ax X2 - X1 VAV-x At t2 - t1 Normal Force Ax dx Vx = lim - N At-0 At dt Avx Vx2 - Vx1 aAV-x - At Gravity t2 - t1 AVX dux Fo = mg ax = lim - At-0 At dt Ux (t) = Vox + axlt - tol Static Friction fs S HSN x (t) = xo + Vox[t - tol+ 5 ax[t - to]2 v2 = Vox + 2ax[x - xol Kinetic Friction Ay = y2 - y1 fx = HKN Ay _ yz - y1 VAV-y At tz - t1 Spring Force Ay dy Fap = - kx (x measured from relaxed position) Vy = lim At-0 4t dt AVy Vy2 - Vy1 Uniform Circular Motion aAv-y At t2- 1 a = Avy dvy r ay = lim At-0 At dt v = rw Vy (t) = Voy + dy[t - tol 2 TT 1 T = W y(t) = yo + Voy[t - to] + 5ay[t - to]2 vy = Voy vay + 2ayly - yolInitials: Fext = 0 = AP = 0 Work and Energy 2mz B mi - miz unit - Uzi W = F . di mi + mz my + mz 2m1 m2 - m1 W = FL cose (constant angle and magnitude) V26 mi + mz m + mz W. = -mgAh (height axis pointing upward) WSp = - =(X3 - XA) (x measured from relaxed position) K = =mv Rotational Kinematics Wnet = > Wi = 4K U + K =E e (t) = angular position AE = WNC 40 = 02 - 01 Ug = mgh (height axis pointing upward) 40 02 - 01 WAD Usp = = kx2 (x measured from relaxed position) At t2- t1 40 do w = lim - Universal Gravitation At-0 At dt AW W 2 - W 1 GMm aAV =- At t2 FG = 12 Aw dw GM a = lim g = At-0 At dt GMm w (t ) = wo+ alt- to] UG = r 0 (t) = 0 0 + wolt - to] += alt- to]2 GM W2 = wo+ 2a[0 - Oo] Vorb = s = ro (arc length) v = rw 2TY 3/2 atan = ra T = 12 2 VGM arad = -= rw2 Rotational Dynamics 2GM Vesc = 1 = Et=1 miri (point masses) Ip = ICM + Md2(parallel axis theorem) Linear Momentum It| = rF sin 0 p = mv Text = la P = P L = rp sino; L = Iw (@=angular velocity) Fext = dp W = | Tdo JO1 Krot = =102Initials: nitials: Cu. Cp and Y = Cp/Cy for an Ideal Gas Simple Harmonic Oscillator dzx 2+ w 2 x = 0 1. For a monoatomic gas, d=3 dt2 x(t) = A cos(wt + ) Cy = 38, Cp = 38 and Y = = 1.67 w = 2nf = ~ (@=angular frequency) 2. For a diatomic gas, d=5 Mass-spring: w = Cy = SR, Cp = 12 and Y = = = 1.4 3. For a triatomic or polyatomic gas, d=6 Simple pendulum: w = Cy = 3R, Cp = 4R and y = = = 1.33 Physical pendulum: w = mgLem Quadratic Equation Torsional pendulum: w = ax2 + bx + c = 0 has the solutions 20 (-b + vb2 - 4ac Fluids Derivatives P = V; P = perp Pgauge = Pabs - Patm d(xn) - = nxn-1 P = Po + pgh dx FB = PgVsub d(cos (ax + b)) -a sin(ax + b) Qm = PVA = pQv = const. dx 1' P + pgy + 5 puz = const. Lengths, areas and volumes Circumference of circle: 2nR Thermodynamics Area of disk: TR Surface area of sphere: 4TR PV = nRT = NKBT Lateral area of cylinder: 2TRh Eint = nCUT Volume of cylinder: TRZh AEint = Q - W R = NAKB; Co = AR Volume of sphere: 4TR3/3 2Cp Cp = CD + R ; Y = Cy Trigonometry e =- Wnet = 1 - Lout (efficiency) sin (90) = cos (09) = - cos (180) = 1 Qin Qin sin (09) = cos (909) = sin (180) = 0 TL eideal = 1 - TH cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 W Isobaric CPRAT PAV cos (180- 0) = - cose CVRAP P=const. Isochoric V= const. sin (180- 0) = sine Isothermal nRT In ( vo) nRT In T=const. cos (90 + 0) = - cos (90 - 0) = - sine Adiabatic 0 - # ( P, VJ - POVO) PV=const. sin (90 + 0) = sin (90 - 0) = cose cos2 0 + sin- 0 = 1Initials: Table of rotational inertias Axis Axis Axis Solid cylinder Hoop about Annular cylinder (or disk) about central axis (or ring) about central axis central axis 1=MR2 ( @ ) 1 = +M(R, 2 + R2? ) ( b ) 1 = 1MR (c) Axis Axis Axis Solid cylinder Thin rod about Solid spherea (or disk) about axis through center bout any central diameter perpendicular to diameter length 2R 1=!MR + $ ML2 (d ) 1= AML? (e) 1=3MR (D) Axis Axis Axis Thin spherical shella Hoop about any bout any R diameter Slab about 2 F perpendicular diameter axis through center 1 = AMR (8) 1= ;MR? (h ) 1 = 12 M(2 + 6? ) (1 )