Classical Mechanics Lecture 26 Today’s Concepts: A) Moving Fluids B) Bernoulli’s Equation

Clicker Question This picture is A) Real B) Fake

DP = rgDy Clicker Question Through which hole will the water come out fastest? A B C DP = rgDy

CheckPoint Water flows through a pipe that has a constriction in the middle as shown. How does the speed of the water in the constriction compare to the speed of the water in the rest of the pipe? A) It is bigger B) It is smaller C) It is the same Water is incompressible so when it goes through the constriction the water has to speed up in order to squeeze all the water through  In the last prelecture we began the study of waves. We saw that when a wave propagates through a medium, the elements of the medium only make small displacements from their equilibrium position while the distance traveled by the wave itself is unlimited. We studied the example of a wave on a string and we derived the wave equation [show it] – a differential equation that must be satisfied by whatever function describes the displacements of the medium as a function of both position and time – and we showed that the solution we guessed for a harmonic wave satisfied this equation [show Acos(kx-wt)] as long as the speed of the wave was given by w/k. We will now show that waves of any shape – not just harmonic waves – will satisfy this equation as long as the speed of the wave is constant. We start by re-writing the argument of the cosine in our harmonic solution in terms of the velocity of the wave. This solution clearly has the general form shown – it is a function which depends on the argument x-vt. Now lets see if this more general functional form is a solution of the wave equation. Using the chain rule to calculate the derivatives [not sure how much to show] we find that this is also a solution. But we now realize that something interesting has happened. The function f(x-vt) does not have to be harmonic at all – it can represent a wave of any shape moving with speed v. To see this consider the shape shown which his described by the function y=f(x) [left diagram]. Shifting this to the right a distance d along the x axis is described by the same function if the argument is simply changed from x to x-d since the new argument will have the same value when x=d as the old one did when x=0 [center diagram]. This means that we can have the pulse move to the right with a constant speed v just by saying that d = vt [right diagram]. Since we have just shown that this is a solution to the wave equation, it must be the case that a single pulse, or any other shape wave, will move through the medium with the same speed as a harmonic wave. This is certainly consistent with our experience. Whether we wiggle end of a string harmonically or we just wiggle it once, we see the shape we create propagate down the string with the same speed. Of course, we could describe a pulse moving to the left in the same way just by changing the sign of v [show f(x+vt)], and this will also be a solution to the wave equation since the mathematics cant care which end of the string we are wiggling. 4

CheckPoint Water flows through a pipe that has a constriction in the middle as shown. How does the pressure of the water in the constriction compare to the pressure of the water in the rest of the pipe? A) It is bigger B) It is smaller C) It is the same In the last prelecture we began the study of waves. We saw that when a wave propagates through a medium, the elements of the medium only make small displacements from their equilibrium position while the distance traveled by the wave itself is unlimited. We studied the example of a wave on a string and we derived the wave equation [show it] – a differential equation that must be satisfied by whatever function describes the displacements of the medium as a function of both position and time – and we showed that the solution we guessed for a harmonic wave satisfied this equation [show Acos(kx-wt)] as long as the speed of the wave was given by w/k. We will now show that waves of any shape – not just harmonic waves – will satisfy this equation as long as the speed of the wave is constant. We start by re-writing the argument of the cosine in our harmonic solution in terms of the velocity of the wave. This solution clearly has the general form shown – it is a function which depends on the argument x-vt. Now lets see if this more general functional form is a solution of the wave equation. Using the chain rule to calculate the derivatives [not sure how much to show] we find that this is also a solution. But we now realize that something interesting has happened. The function f(x-vt) does not have to be harmonic at all – it can represent a wave of any shape moving with speed v. To see this consider the shape shown which his described by the function y=f(x) [left diagram]. Shifting this to the right a distance d along the x axis is described by the same function if the argument is simply changed from x to x-d since the new argument will have the same value when x=d as the old one did when x=0 [center diagram]. This means that we can have the pulse move to the right with a constant speed v just by saying that d = vt [right diagram]. Since we have just shown that this is a solution to the wave equation, it must be the case that a single pulse, or any other shape wave, will move through the medium with the same speed as a harmonic wave. This is certainly consistent with our experience. Whether we wiggle end of a string harmonically or we just wiggle it once, we see the shape we create propagate down the string with the same speed. Of course, we could describe a pulse moving to the left in the same way just by changing the sign of v [show f(x+vt)], and this will also be a solution to the wave equation since the mathematics cant care which end of the string we are wiggling. 5

CheckPoint Water flows through a pipe that has a constriction in the middle as shown. How does the pressure of the water in the constriction compare to the pressure of the water in the rest of the pipe? A) It is bigger B) It is smaller C) It is the same A) because it's being forced through a smaller pipe it has greater pressure. In the last prelecture we began the study of waves. We saw that when a wave propagates through a medium, the elements of the medium only make small displacements from their equilibrium position while the distance traveled by the wave itself is unlimited. We studied the example of a wave on a string and we derived the wave equation [show it] – a differential equation that must be satisfied by whatever function describes the displacements of the medium as a function of both position and time – and we showed that the solution we guessed for a harmonic wave satisfied this equation [show Acos(kx-wt)] as long as the speed of the wave was given by w/k. We will now show that waves of any shape – not just harmonic waves – will satisfy this equation as long as the speed of the wave is constant. We start by re-writing the argument of the cosine in our harmonic solution in terms of the velocity of the wave. This solution clearly has the general form shown – it is a function which depends on the argument x-vt. Now lets see if this more general functional form is a solution of the wave equation. Using the chain rule to calculate the derivatives [not sure how much to show] we find that this is also a solution. But we now realize that something interesting has happened. The function f(x-vt) does not have to be harmonic at all – it can represent a wave of any shape moving with speed v. To see this consider the shape shown which his described by the function y=f(x) [left diagram]. Shifting this to the right a distance d along the x axis is described by the same function if the argument is simply changed from x to x-d since the new argument will have the same value when x=d as the old one did when x=0 [center diagram]. This means that we can have the pulse move to the right with a constant speed v just by saying that d = vt [right diagram]. Since we have just shown that this is a solution to the wave equation, it must be the case that a single pulse, or any other shape wave, will move through the medium with the same speed as a harmonic wave. This is certainly consistent with our experience. Whether we wiggle end of a string harmonically or we just wiggle it once, we see the shape we create propagate down the string with the same speed. Of course, we could describe a pulse moving to the left in the same way just by changing the sign of v [show f(x+vt)], and this will also be a solution to the wave equation since the mathematics cant care which end of the string we are wiggling. B) Pressure must be the same throughout the pipe. C) Since the fluid is all at the same height the sum (P + .5pv²) is constant throughout the pipe. The velocity is greater in the constriction, so the pressure there must be smaller. 6

Clicker Question Two empty pop cans are placed about ¼” apart on a frictionless surface. If you blow air between the cans, what happens? A) The cans move toward each other. B) The cans move apart. C) The cans don’t move at all. Blowing air

Clicker Question A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe is even with the other hole, like in the picture below: Through which drain is the water spraying out with the highest speed? A) The hole B) The pipe C) Same CORRECT

CheckPoint hL hR PL PR Bernoulli’s Equation Water flows from left to right along a pipe as shown. The right end of the pipe is twice as high as and also has four times the area of the left end.  Which of the following statements best relates the pressures at the ends of the pipe? A) PL = 2PR B) PL = PR C) PL = ½ PR D) The relative size of PL and PR depends on the speed of the flow. hL hR PL PR Bernoulli’s Equation

Clicker Question hR PR hL PL Bernoulli’s Equation Suppose the water isn't moving. The right end of the pipe is twice as high as and also has four times the area of the left end.  Which of the following statements best relates the pressures at the ends of the pipe? A) PL = 2PR B) PL = PR C) PL = ½ PR D) PL - PR = rg (hr - hL) hR PR hL PL Bernoulli’s Equation

Clicker Question hL hR PL PR Water flows from left to right along a pipe as shown. The right end of the pipe is twice as high as and also has four times the area of the left end.  Which of the following statements best relates the pressures at the ends of the pipe? A) PL = 2PR B) PL = PR C) PL = ½ PR D) PL - PR =rg (hR - hL) + 1/2r (vR2 - vL2) hL hR PL PR Bernoulli’s Equation

Clicker Question hL hR PL PR We just saw that PL - PR can be written in the following way: Is there any reason this has to mean PL = 2PR or PL = PR or PL = ½ PR A) Yes B) No hL hR PL PR Bernoulli’s Equation

CheckPoint Water flows from left to right along a pipe as shown. The right end of the pipe is twice as high as and also has four times the area of the left end.  Which of the following statements best relates the pressures at the ends of the pipe? A) PL = 2PR B) PL = PR C) PL = ½ PR D) The relative size of PL and PR depends on the speed of the flow. hR PR hL PL