Lecture 25Angular Momentum ExerciseSlide 3Example: Throwing ball from stoolSlide 5Ideal FluidA water fountainSlide 8Fluids BuoyancySlide 10Slide 11A new trickSlide 13ExampleSlide 15Slide 16Fluids Buoyancy & SHMSlide 18Slide 19Underdamped SHMCh. 12Hooke’s Law Springs and a Restoring ForceSimple Harmonic MotionResonance and dampingFluid FlowDensity and pressureResponse to forcesStates of Matter and Phase DiagramsIdeal gas equation of statepV diagramsThermodynamicsWork, Pressure, Volume, HeatGas ProcessesLecture 25Physics 207: Lecture 25, Pg 1Lecture 25 Today Review:Today Review:•Exam covers Chapters 14-17 plus angular momentum, rolling Exam covers Chapters 14-17 plus angular momentum, rolling motion & torquemotion & torque•AssignmentAssignment HW11, Due Tuesday, May 6th For Thursday, read through all of Chapter 18Physics 207: Lecture 25, Pg 2Angular Momentum ExerciseA mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed i = 5 rad / s in a circle of radius ri = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m. What is the final angular velocity ?Underlying concept: Conservation of MomentumriiPhysics 207: Lecture 25, Pg 3Angular Momentum ExerciseA mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed i = 5 rad / s in a circle of radius ri = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m. What is the final angular velocity ?No external torque implies L = 0 or Li = LcIi i = If fI for a point mass is mr2 where r is the distance to the axis of rotationm ri2i = m rf2 ff = ri2i / rf2 = (0.20/0.10)2 5 rad/s = 20 rad/sriiPhysics 207: Lecture 25, Pg 4Example: Throwing ball from stoolA student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector has a perpendicular distance d from the axis of rotation. What is the angular speed F of the student-stool system after they throw the ball ?Top view: before afterdMvMI I F rPhysics 207: Lecture 25, Pg 5Example: Throwing ball from stoolWhat is the angular speed F of the student-stool system after they throw the ball ?Process: (1) Define system (2) Identify Conditions (1) System: student, stool and ball (No Ext. torque, L is constant) (2) Momentum is conserved (check |L| = |r| |p| sin for sign) Linit = 0 = Lfinal = M v d + I fTop view: before afterdvMI I FPhysics 207: Lecture 25, Pg 6 Bernoulli Equation P1+ ½ v12 + g y1 = constantA 5 cm radius horizontal pipe carries water at 10 m/s into a 10 cm radius. ( water =103 kg/m3 )What is the pressure difference?P1+ ½ v12 = P2+ ½ v22 P = ½ v22 - ½ v12P = ½ v22 - v12 and A1 v1 = A2 v2P = ½ v22 – (A2/A1 2 = 0.5 x 1000 kg/m x 100 m2/s2 (1- (25/100)2 ) = 47000 Pay1y2v1v2p1p2VIdeal FluidPhysics 207: Lecture 25, Pg 7A water fountainA fountain, at sea level, consists of a 10 cm radius pipe with a 5 cm radius nozzle. The water sprays up to a height of 20 m.What is the velocity of the water as it leaves the nozzle?What volume of the water per second as it leaves the nozzle?What is the velocity of the water in the pipe?What is the pressure in the pipe?How many watts must the water pump supply?Physics 207: Lecture 25, Pg 8A water fountainA fountain, at sea level, consists of a 10 cm radius pipe with a 5 cm radius nozzle. The water sprays up to a height of 20 m.What is the velocity of the water as it leaves the nozzle? Simple Picture: ½mv2=mgh v=(2gh)½ = (2x10x20)½ = 20 m/s What volume of the water per second as it leaves the nozzle? Q = A vn = 0.0025 x 20 x 3.14 = 0.155 m3/s What is the velocity of the water in the pipe? An vn = Ap vp vp = Q /4 = 5 m/s What is the pressure in the pipe?1atm + ½ vn2 = 1 atm + P + ½ vp2 1.9 x 105 N/m2How many watts must the water pump supply? Power = Q g h = 0.0155 m3/s x 103 kg/m3 x 9.8 m/s2 x 20 m = 3x104 W (Comment on syringe injection)Physics 207: Lecture 25, Pg 9Fluids BuoyancyA metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a massles rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm3 and the water is 1.0 gm/cm3What is the average density of the cylinder?What was the tension in the rope when the cylinder was submerged in the oil?Physics 207: Lecture 25, Pg 10Fluids Buoyancyr = 0.5 m, h= 4.0 m oil = 0.9 gm/cm3 water = 1.0 gm/cm3What is the average density of the cylinder?When T = 0 Fbuoyancy = Wcylinder Fbuoyancy = oil g ½ Vcyl.+ water g ½ Vcyl. Wcylinder = cyl g Vcyl. cyl g Vcyl. = oil g ½ Vcyl.+ water g ½ Vcyl. cyl = ½ oill.+ ½ water What was the tension in the rope when the cylinder was submerged in the oil?Use a Free Body Diagram !Physics 207: Lecture 25, Pg 11Fluids Buoyancyr = 0.5 m, h= 4.0 m Vcyl. = r2 hoil = 0.9 gm/cm3 water = 1.0 gm/cm3What is the average density of the cylinder?When T = 0 Fbuoyancy = Wcylinder Fbuoyancy = oil g ½ Vcyl.+ water g ½ Vcyl. Wcylinder = cyl g Vcyl. cyl g Vcyl. = oil g ½ Vcyl.+ water g ½ Vcyl. cyl = ½ oill.+ ½ water = 0.95 gm/cm3 What was the tension in the rope when the cylinder was submerged in the oil?Use a Free Body Diagram! Fz = 0 = T - Wcylinder + Fbuoyancy T = Wcyl - Fbuoy = g ( cyl .- oil ) Vcyl T = 9.8 x 0.05 x 103 x x 0.52 x 4 .0 = 1500 NPhysics 207: Lecture 25, Pg 12A new trickTwo trapeze artists, of mass 100 kg and 50 kg respectively are testing a new trick and want to get the timing right. They both start at the same time using ropes of 10 meter in length and, at the turnaround point the smaller grabs hold of the larger artist and together they swing back to the starting platform. A model of the stunt is shown at right.How long will this stunt require if
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