Download Wednesday, October 21st

1206 - Concepts in
Physics
Wednesday, October 21st
Notes
• Reminder: additional tutorial for anybody
interested this coming Monday (October
26th) at 10:30 am in F055
• I will show a few more examples and you
can ask questions
• If you have a specific example or problem
from the textbook, please email me
beforehand - it is easier to prepare that
way
Notes
• Assignment #4 solution will be topic for
this Friday’s lecture and you will be getting
it back in class as well.
• Next Monday is the last day, you can ask
me questions in person for the midterm
• But email will still be possible for the rest
of reading week
Notes
•
Best preparation for midterm is to really
understand the problems in all assignment and the
problems discussed as examples in the lecture
notes
•
It doesn’t hurt to open the textbook - try to solve
the example problems by yourself - you can then
easily check your calculations against the book
•
Discuss amongst yourself - ask the yourself if you
can explain for example Newton’s second law or
Energy conservation in your own words.
Have a good study time and
good luck for November 02nd
Buoyant force
Have you ever tried pushing one of these under water? If so,
you know that you have to apply quite a bit of force. The
water pushes back with a strong upward force. This upward
force is called the buoyant force, and all fluids apply such a
force to objects that are immersed in them, because fluid
pressure is larger at greater depth.
P1A
A
h
P2A
Here is a cylinder filled with a liquid. Inside is a smaller
cylinder of height h. The pressure P1 on the top face
generates the downward force P1A, where A is the area of
the face (top of cylinder). And the pressure P2 on the
bottom face generates the upward force P2A. Since the
pressure is greater at greater depths, the upward force
exceeds the downward force. Consequently the liquid
applies to the cylinder a net upward force, or buoyant
force, whose magnitude FB is:
FB = P2A - P1A = (P2-P1)A = ρghA = ρgV
The quantity hA is the volume the smaller cylinder, the volume of liquid that gets moved aside
and ρ is the density of the liquid. ρV is the mass of the displaced fluid. (FB = mg)
Archimedes Principle
So the buoyant force FB is the product of the “mass of the displaced” fluid and the acceleration
due to gravity. Mass of the displaced fluid can be easily pictured. It is the amount of liquid that
would spill out of the container, if it were filled to the brim before the cylinder is inserted. The
shape of the inserted object is not important, this was first described by the Greek scientist
Archimedes (ca. 287 - 212 B.C.) and is know as Archimedes’ principle:
Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the
magnitude of the buoyant force equals the weight of the fluid that the object displaces:
FB = Wfluid
Magnitude of buoyant force = Weight of displaced fluid
There is a story how he discoved
this principle. It is said he yelled
“Eureka” and ran to the king naked
when he found out.
Floating
The effect the buoyant force has depends on its strength compared with the strengths of the
other forces that are acting. For example, if the buoyant force is strong enough to balance the
force of gravity, an object will float in a fluid. But even when an object sinks, a buoyant force is
acting, it is just not strong enough to balance the weight.
Example:
A solid, square pinewood raft measures 4.0 m on a side and is 0.30 m thick. Determine
whether the raft floats in the water (density of pine is 550 kg/m3). If you find the raft floats
determine how much of thraft is beneath the surface.
To determine whether the raft floats, we will compare the weight of the raft to the
maximum possible buoyant force and see whether there could be enough buoyant force to
balance the weight. If so, then the value of the distance h can be obtained by utilizing the fact
that the floating raft is in equilibrium, with the magnitude of the buoyant force equaling the
raft’s weight.
(a) The weight of the raft can be calculated from the density ρpine = 550 kg/m3, the volume of
the wood, and the acceleration due to gravity.
Vpine = 4.0 m x 4.0 m x 0.3 m = 4.8 m3
Wraft = ρpine Vpine g = (550 kg/m3)(4.8 m3)(9.80 m/s2) = 26000 N
The maximum possible buoyant force occurs when the entire raft is under the surface, displacing
4.8 m3 of water. According to Archimedes’ principle, the weight of this volume of water is the
maximum buoyant force FBmax. It can be calculated as follows:
FBmax = ρwater Vpine g = (1000 kg/m3)(4.8 m3)(9.80 m/s2) = 47000 N
Since the maximum possible buoyant force exceeds the 26000 N weight of the raft, the raft will
float partially submerged at a distance h beneath the water.
(b) Now we are looking for the value of h. The buoyant force balanced the raft’s weight, so
FB = 26000 N. But the buoyant force is also defined as the weight of the displaced fluid
according to the Archimedes principle. Wfluid = ρwater Vwater g where the volume of the
displaced water is the same as the earlier calculated volume of the pine raft.
26000 N = Wfluid = ρwater Vwater g = ρwater (4.0 m x 4.0 m x h)g
h = 26000 N / ρwater (4.0 m x 4.0 m)g = 0.17 m
Example:
Archimedes’ principle can also be applied to gases, we will take a look at a airship.
A airship contains about 5.40 x 103 m3 of helium
with a density of 0.179 kg/m3. Find the weight of the
load WL that the airship can carry in equilibrium at
and altitude where the density of air is 1.20 kg/m3.
The airship and its load are in equilibrium. Thus, the
buoyant force FB applied to the airship by the
surrounding air balances the weight WHe of the
helium and the weight WL of the load, including the
solid parts of the airship. The free-body diagram is
the first step and shows all the forces.
FB
WHe
WL
The forces in the free body diagram balance, so we can write:
WHe + WL = FB or WL = FB - WHe
According to Archimedes’ principle, the buoyant force is the weight of the displaced
air: FB = Wair = ρair Vship g. The weight of the helium is WHe = ρHeVship g, where we
assume that the volume of the helium and the volume of the airship are nearly the
same.
WL = ρair Vship g - ρHeVship g = (ρair - ρHe) Vship g =
(1.20 kg/m3 - 0.179 kg/m3)(5.40 x 103 m3)(9.80 m/s2) = 5.40 x 104 N
Fluids in motion
Fluids can move in many different ways. Let’s look at water - it can flow quiet and smoothly or
violently over a waterfall. We can observe similar behavior for air (wind) in nature - from a
gentle breeze to a tornado. We will now define some basic types of flow for fluids.
Steady or unsteady flow: steady flow means that the velocity of the fluid particles at a
specific point is constant over time. For example a river usually has a steady flow. It can still be
faster in the middle than at the edges. Therefore unsteady flow is when the velocity at a
point in the fluid changes as time passes. Turbulent flow is an extreme kind of unsteady
flow and occurs when there are sharp obstacles or bends in the path of a fast-moving fluid,
such as rapids in a river.
Compressible or incompressible: incompressible
means, that the density of a liquid remains almost
constant as the pressure changes. This is true for
most liquids. Gases however are highly
compressible in most cases.
A fluid can be viscous or nonviscous. An example for a fluid with high viscosity is honey - it
doesn’t flow easily. Compared to water, which flows readily. If a fluid had zero viscosity, if would
flow unhindered which translated to with o dissipation of energy. This is called an ideal fluid.
The larger the viscosity, the more dissipation of energy takes place for it to flow.
We can use streamlines to represent the trajectories of the fluid particles. A streamline is
drawn such that a tangent to the streamline at any point is parallel to the fluids velocity at that
point.
Equation of Continuity
Have you used your thumb to cover part of a water hose? If so, you know that the water flowing
from the end of the hose will increase its velocity in that case. This behavior is described by the
equation of continuity. It expresses the following simple idea: If a fluid enters one end of a pipe at
a certain rate, then the fluid must leave at the same rate, assuming that there are no places
between the two points to add or remove fluid. The mass of fluid per second (e.g. 5 kg/s) is
called the mass flow rate.
Let’s look at the two ends of this bottle like tube. The left
side has an cross-sectional area A1 and the fluid enters
with a velocity v1. On the right side (exit) the crosssectional area is A2 and the fluids velocity is v2. The crosssectional times the distance is the volume of fluid that has
flowed past this point in a time interval Δt: A2v2Δt.
The mass Δm2 of this fluid element is given by the product of the density and volume:
Δm2 = ρ2A2v2Δt. Dividing by Δt yield the mass flow rate on the right (or left):
Mass flow rate position 1: Δm1/Δt = ρ1A1v1
Mass flow rate position 2: Δm2/Δt = ρ2A2v2
Since no fluid can cross the side walls of the tube, the mass flow rates at position 1 and
position 2 must be equal. They have been selected arbitrarily, therefore we can conclude that
the mass flow rate is the same everywhere in the tube. This result is known as the equation of
continuity. It basically tells you that mass is conserved - it is neither created nor destroyed as
the fluid flows.
EQUATION OF CONTINUITY
The mass flow rate (ρAv) has the same value at every position along a tube that has a single
entry and a single exit point for fluid flow. For two positions along such a tube:
ρ1A1v1 = ρ2A2v2
where ρ = fluid density (kg/m3)
A = cross-sectional area of tube (m2)
v = fluid speed (m/s)
SI unit of Mass Flow Rate: kg/s
For an incompressible fluid we know, that the density does not change during flow, so that
ρ1 = ρ2 and the equation of continuity reduces to A1v1 = A2v2
The quantity Av represents the volume of fluid per second that passes through the tube and
is referred to as the volume flow rate Q:
Example:
A garden hose has an unobstructed opening with a cross-sectional area of 2.85 x 10-4 m2, from
which water fills a bucket in 30.0 s. The volume of the bucket is 8.00 x 10-3 m3. Find the speed
of the water that leaves the hose through
(a) and unobstructed opening and
(b) an obstructed opening with half as much area
If we can determine the volume flow rate Q, the speed of the water an be calculated as
v = Q/A, since the area is given. The volume flow rate can be found from the volume of the
bucket and the time it take to fill it.
(a) The volume flow rate Q is equal to the volume of the bucket VB divided by the filling time:
v = Q/A = (VB/t)/A = [(8.00 x 10-3 m3)/(30.0 s)]/(2.85 x 10-4 m2) = 0.936 m/s
(b) Water can be considered incompressible, so the equation of continuity can be applied in
the form A1v1 = A2v2. We know that A2 = 1/2A1, so we find:
v2 = (A1/A2)v1 = (A1/0.5 x A1)v1 = 1.87 m/s
Note! Using half the area doubles the velocity.
Example: YOUR TURN
In the condition known as atherosclerosis, as deposit or atheroma forms on the arterial wall
and reduces the opening through which blood can flow. In the carotid artery in the neck, blood
flows three times faster through a partially blocked region than it does through an
unobstructed region. Determine the ratio of the effective radii of the artery at the two places.
Example: YOUR TURN
In the condition known as atherosclerosis, as deposit or atheroma forms on the arterial wall
and reduces the opening through which blood can flow. In the carotid artery in the neck, blood
flows three times faster through a partially blocked region than it does through an
unobstructed region. Determine the ratio of the effective radii of the artery at the two places.
Blood like most liquids, is incompressible, and the equation of continuity in the form of
A1v1 = A2v2 can be applied. We also have to use the fact that the area of a circle is πr2
Example: YOUR TURN
In the condition known as atherosclerosis, as deposit or atheroma forms on the arterial wall
and reduces the opening through which blood can flow. In the carotid artery in the neck, blood
flows three times faster through a partially blocked region than it does through an
unobstructed region. Determine the ratio of the effective raii of the artery at the two places.
Blood like most liquids, is incompressible, and the equation of continuity in the form of
A1v1 = A2v2 can be applied. We also have to use the fact that the area of a circle is πr2
unobstructed volume flow rate (πrU2)vU = (πrO2)vO
The ratio of radii then is given by:
obstructed volume flow rate
rU/rO = sqrt[vO/vU] = sqrt(3) = 1.7
Bernoulli’s equation
For steady flow, the speed, pressure and elevation of an
incompressible and nonviscous fluid are related by an equation
discovered by Daniel Bernoulli. To derive Bernoulli’s equation we
will use the work-energy theorem, which states that the net work
W done on an object by external nonconservative forces is equal
to the change in the total mechanical energy of the object. We
learned last lecture, that pressure is caused by collisions inside the
fluid - they are nonconservative. Therefore when a fluid is
accelerated because of a difference in pressures, work is being
done by nonconservative forces, and this work changes the total
mechanical energy of the fluid from an initial value of E0 to a final
value of Ef. This will lead directly to Bernoulli’s equation.
Daniel Bernoulli (1700 - 1782)
dutch-swiss mathematician
Fluid flowing in a horizontal pipe encounters a region of reduced cross-section and the
pressure of the fluid drops due to Newton’s second law. When moving from a wider to a
narrower region the fluid accelerates, consistent with the conservation of mass. We need an
unbalanced force to cause the acceleration.
Another observation is, that fluid moving at a hight elevation has a lower pressure level.
Let’s look at a tube that goes upstream with a smaller opening on top (1) as on the bottom (2).
Both cross-sectional area and elevation are different at different places along the pipe. The total
mechanical energy is the sum of kinetic energy KE and gravitational potential energy PE, so we
can write: E = KE + PE = 1/2mv2 + mgh. When work W is done, the total mechanical energy
changes according to the work-energy theorem:
W = E1 - E2 = (1/2mv12 + mgh1) - (1/2mv22 + mgh2)
On the top surface of any fluid element, the surrounding fluid exerts a pressure P. This
pressure give rise to a force of magnitude F = PA, where A is the cross sectional area. On the
bottom surface, the surrounding fluid exerts a slightly greater pressure P + ΔP, where ΔP is the
pressure difference between the ends of the element. Therefore the force at the bottom has
the magnitude of F + ΔF = (P + ΔP)A. The magnitude of the net force pushing the fluid
element up the pipe is ΔF = ΔPA. Now we can also write the work done as the product of the
magnitude of the net force and the distance s: W = ΔF * s = ΔP * A * s = ΔP * V
With this expressing the work-energy theorem becomes:
W = ΔP * V = (1/2mv12 + mgh1) - (1/2mv22 + mgh2)
Dividing by the volume and recognizing that m/V is density ρ of the fluid and rearranging:
BERNOULLI’S EQUATION
In the steady flow of a nonviscous, incompressible fluid of density ρ, the pressure P, the fluid
speed v, and the elevation h at any two points (1 and 2) are related by:
P1 + 1/2ρv12 + ρgh1 = P2 + 1/2ρv22 + ρgh2
Why things fly
Many wonder, since air is invisible, how it can support an airplane. Even though it is invisible, air
is stuff. There are many examples. Feel the wind and watch the wind blow snow, leaves and trees,
and the stuff that is air becomes more apparent. The terrible destruction of hurricanes and
tornadoes is tragically clear evidence that air contains a lot of stuff, stuff that can apply a great deal
of pressure, easily enough to lift an airplane.
There are many explanations of How Airplanes Fly in books, museums, flying lesson material and
other sources. Most rely on the relation between fluid velocity and fluid pressure given many years
ago (in 1738) by Daniel Bernoulli. That relation is of fundamental importance in aeronautics and
is called Bernoulli's Theorem. Important in an explanation of flight, Bernoulli's Theorem states
that: where the velocity is higher, the pressure is smaller; and where the velocity is lower, the
pressure is larger.
Since the velocity above the wing is higher than the velocity below the wing, it follows directly
from Bernoulli's Theorem that airplane wings provide lift. To explain why the velocity is higher above the wing than below, it may help to recollect some
common observations. When rowing a boat or paddling a canoe one often notices the swirls that
the oars or paddles make in the water. These swirls occur off the relatively sharp edge of the oar or
paddle.
When fluid flow, air or water, is around a relatively sharp edge, or even when two different flow
rates meet, swirls or VORTICES form (turbulent flow). It is an important natural phenomenon.
Similarly a vortex is formed by the sharp trailing edge of an airfoil. This vortex is termed the
starting vortex.
Since every action has an equal and opposite re-action, when a vortex starts, behind an airfoil, the
equal and opposite reaction is a vortex in the opposite direction around the airfoil. Thus, there are
two velocities that affect the pressure on the airfoil: 1) the velocity of the airfoil through the air, and
2) the vortex or circulation around the airfoil. Above an airfoil, the velocity through the air and the
circulation are in the same direction and are additive. Below the airfoil the circulation is subtracted
from the velocity through the air. Consequently, above the airfoil the velocity is higher and, from
Bernoulli's Theorem, the pressure is lower. Below the airfoil there is a lower velocity and higher
pressure. The pressure differential produces lift. The lift is proportional to the velocity through the
air and the magnitude of the circulation. The circulation increases with the angle of attack of the
airfoil.
Curve of a Baseball
A non-spinning baseball or a stationary baseball in an airstream exhibits symmetric flow. A
baseball which is thrown with spin will curve because one side of the ball will experience a
reduced pressure. This is commonly interpreted as an application of the Bernoulli principle and
involves the viscosity of the air and the boundary layer of air at the surface of the ball.
There are some difficulties with this picture of the curving baseball. The Bernoulli equation cannot really be used to predict the amount of
curve of the ball; the flow of the air is compressible, and you can't track the density changes to quantify the change in effective pressure. The
experimental work of Watts and Ferrer with baseballs in a wind tunnel suggests another model which gives prominent attention to the
spinning boundary layer of air around the baseball. On the side of the ball where the boundary layer is moving in the same direction as the
free stream air speed, the boundary layer carries further around the ball before it separates into turbulent flow. On the side where the boundary
layer is opposed by the free stream flow, it tends to separate prematurely. This gives a net deflection of the airstream in one direction behind
the ball, and therefore a Newton's 3rd law reaction force on the ball in the opposite direction. This gives an effective force in the same
direction indicated above.