11/28/2007Meenakshi Narain - Physics 5 - Lecture 22 1 Lecture 22

11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

1

Lecture 22


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Announcements Homework due TODAY at the end of lecture Homework 12 based on ch 12 & 13, due on Dec 5th.

– I may post some practice problem suggestions for ch 14 and 15.

Final Exam on Thursday, Dec 13, 9am-12pm, in BARHOL 168

Close book, close lecture notes– Chapters 1-15, with emphasis on post midterm 2 material– Emphasis of the exam: problem solving (6-8 problems)– Simple calculator and one-page formula sheet allowed– Questions? Suggestions?


3

Simple Harmonic Motion

Chapter 15


4

A block of mass 1 kg is attached to a spring with k=100 N/m and free to move on a frictionless horizontal surface. At t=0, the spring is extended 5 cm beyond its equilibrium position and the block is moving to the left with a speed of 1 m/s. What is the displacement of the block as a function of time?


5

m

k

xA

x

v

Atv

Atx

and

cos

0

0

0tan

sin0

cos0

What are the values of A, , and ?A, are determined by initial conditions (x,v at t=0) –

they are not properties of the oscillating system.

)1.1 /s10cos( m 11.0

m 0.110.45

m 0.05

cos

)0(

rad 1.1

2m 0.05 10/s

m/s 1

)0(

)0(tan

s/10

)cos()(

tx(t)

xA

-

x

vm

k

tAtx

)cos()( tAtx


6

Three 10,000 kg ore cars are held at rest on a 30o incline using a cable. The cable stretches 15 cm just before a coupling breaks detaching one of the cars. Find (a) the frequency of the resulting oscillation of the remaining two cars and (b) the amplitude of the oscillation.


7

m = 10000 kgd = 0.15 mfirst compute spring constant of rope:

m

N108.9

sin3 5d

mgk

then compute frequency of oscillation:

Hz 1.12

sin3

2

1

22

1

2

d

g

m

kf

the amplitude equals the original displacement from the equilibrium position of two cars on the rope:A = d d0 = 0.05 m


8

1.0 kg

A 10 g bullet strikes a 1 kg pendulum bob that is suspended from a 10 m long string. After the collision the two objects stick together and the pendulum swings with an amplitude of 10o.

What was the speed of the bullet when it hit the bob?

1.0 kgv=?

10o

10 g


9

Energy of simple pendulum

2cos

2 small mgLLLmgmghU using ...

!4!21cos

42

xx

x

22

2

1

2

1 mgLIE with

t

t

00max

0max

sin

cos

2max

22max

2

022

max022

02max

22

2

1

cos2

1

2

1

sin2

1sin

2

1

2

12

0

mgLUKE

tmgLmgLU

tmgLtmLIKL

g

J 1.50.17m 10 s

m9.8 kg 1

2

1

2

1 2

22max MgLE

Get velocity of bob at equilibrium point from total energy

kgm/s 1.7m/s 1.72

fff MvpM

Ev

get velocity of bullet from momentum conservation

m/s 170kg 0.01

kgm/s 1.7

m

pv

fi


10

CO2 molecule Carbon dioxide is a linear molecule. The C-O bonds act

like springs. This molecule can vibrate such that the oxygen atoms move symmetrically in and out, while the carbon atom is at rest. The frequency of this vibration is observed to be 2.83x1013 Hz. – What is the spring constant of the C-O bond?– In which other way can the molecule vibrate and at what

frequency?– If the amplitude is the same, in which mode is the energy of the

molecule higher?

m=12u m=16u


11


12

CO2 molecule a)

b)

c)

the energy is the same for both modes.

N/m 844 kg/u 101.6716u Hz102.834

42

1

2

272132

22

O

O

mfkm

kf

Hz104.62kg/u101.6712u

N/m 844 2

2

1

2

2

1'

2

1

2

''2'2

13

27

CC m

k

m

kfkkkxF

22

2

21

2

'2

12

1kAAkE

kAEkAE

Parallel springsSee Lecture 10


13

Which of the following spring arrangements will oscillate with the smallest angular frequency? Assume that all springs are identical.

(1) (2) (3) (4)


14

Driven Harmonic Oscillator What happens if you have an oscillator, such as a mass

on a spring, where an external force is acting on the system?– Example: Motion of a building or bridge during an earthquake

Essentially all objects have one or more “natural frequencies” that they will oscillate at if they are initially displaced from equilibrium– Example: mass on a spring has a natural frequency given by

If an oscillating external force is applied with angular frequency close to the natural frequency 0, the results can be dramatic

m

k0


15

Driven Harmonic Oscillator To get motion, use Newton’s 2nd law:

Suppose the external force is sinusoidal:

Eventually, the object’s motion will oscillate with frequency w since that’s the frequency of the applied force

m

tFx

dt

dx

dt

xd

tFkxdt

dxb

dt

xdm

tFkxbvFma

ext

ext

extnet

)(2

)(

)(

202

2

2

2

m

b

2

m

k0

)cos(2

)cos()(

0202

2

0

tm

Fx

dt

dx

dt

xd

tFtFext


16

Driven Harmonic Oscillator Solution for driven harmonic oscillator is somewhat

more complicated than what we have done so far in Physics 5

With some effort, one can solve the equation of motion

where the frequency and amplitude are given by:

)()cos(

)cos(2

0

0202

2

txtxx

tm

Fx

dt

dx

dt

xd

Damped

20

2

222220

00

2)tan(

4)(

1

m

Fx

)'cos()( 00 textx tDamped

220'


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Driven Harmonic Oscillator Example: Mass on spring

– Notice how amplitude and phase change as the frequency of the external force crosses the natural frequency

Example: Vibrations of a solid object– Solid objects typically have one or more natural frequencies that

they oscillate at– If the damping is small, large oscillations occur when driven at

these natural frequencies– These vibrations can be a considerable source of stress on the

object!

220

22220

00

2)tan(

4)(

1

m

Fx


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Oscillations Summary We get periodic motion when force acts to push object

back towards equilibrium position Many problems exhibit simple harmonic motion

Energy exchanged between kinetic and potential energy, total mechanical energy unchanged for undamped oscillations

Correspondence between simple harmonic motion and uniform circular motion

Amplitude of oscillations decays with a damping force Driven oscillations exhibit a “resonance” at the natural

frequency

xdt

xd 22

2

)cos()( txtx m


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Fluids Chapter 14


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Fluids

What is a fluid?– A substance that flows– Examples include liquid, gas, plasma, etc– A simple fluid can withstand pressure but not shear

Density – Density– Unit: kg/m3

• Examples: density of water 1000 kg/m3 ; air 1.21 kg/m3

A material’s specific gravity is the ratio of the density of the material to the density of water at 4°C. – What is special about water at 4°C?

• Water is most dense at that temperature.– Aluminum has a specific gravity of 2.7 – it is 2.7 times more denser

than water at 4°C.

m

V


21

A table of densities

Material Density (kg/m3)

Interstellar space 10-20

Air (20°C, 1 atm.) 1.21

Water (4°C, 1 atm.) 1000

Sun (average) 1400

Earth (the planet) 5500

Iron 8700

Mercury (the metal) 13600

Black hole 1019


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Pressure Pressure

– Unit: pascal (Pa)=N/m2 • Examples: 1 atm=1.01x105 Pa=760 torr=14.7 lb/in2

• Torricelli (torr) is defined as the pressure of mm Hg. Blood pressure: 70/120 torr

At any point in a fluid at rest, the pressure is the same in all directions. – If this were not true there would be a net force on the fluid and it

could not be at rest.

The force due to fluid pressure acts perpendicular to any surface. – Else there would be a force component along the surface which

would accelerate the fluid.

Fp

A


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Atmospheric pressure

At atmospheric pressure, every square meter has a force of 100,000 N exerted on it, coming from air molecules bouncing off it!

Why don’t we, and other things, collapse because of this pressure?

We have an internal pressure of 1 atmosphere. Objects like tables do not collapse because forces on

top surfaces are balanced by forces on bottom surfaces, etc.


24

Fluid-Statics Static equilibrium

A simpler expression

– Where p0 is the pressure at the surface, and h is depth of the liquid

The pressure at any point in a fluid is determined by the density of the fluid and the depth. It does not depend on any horizontal dimension of the fluid or its container. It also does not depend on the shape of the container.

2 1

2 1 1 1 2

2 1 1 2

1 1 2 2

( )

( )

F F mg

p A p A Vg p A A y y g

p p y y g

p gy p gy const

0p p gh


25

Measuring pressure The relationship between pressure and

depth is exploited in manometers (or barometers) that measure pressure.

A standard barometer is a tube with one end sealed. – The sealed end is close to zero pressure,

while the other end is open to the atmosphere.

– The pressure difference between the two ends of the tube can maintain a column of fluid in the tube, with the height of the column being proportional to the pressure difference.

– pressure at bottom of column = atmospheric pressure

2 1P P gh


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Mercury/Water barometer Mercury: atmospheric pressure pushes Hg column up unit

mm-Hg (=torr)

– Thus Atmospheric pressure pushes the Hg column up by– 101.3 kPa/133 Pa/mm = 760 mm

Water:

– Thus atmospheric pressure pushes the water column up by– 101.3 kPa/9.8 Pa/mm = 10.3 m

another unit: 1 bar = 105 N/m2

in calculations only use N/m2 = Pa (SI unit)

Pa 133m 0.001m/s 9.8kg/m 1013.6Hgmm 1 233 gh

Pa 9.8m 0.001m/s 9.8kg/m 10OHmm 1 2332 gh


27

Gauge Pressure Gauges measure pressure relative to atmospheric

pressure absolute pressure = gauge pressure + atmospheric pressure

manometer (height of column of liquid measures gauge pressure)


28

Blood pressure A typical reading for blood pressure is 120 over 80. What do the two numbers represent? What units are they in?

120 mm Hg (millimeters of mercury) is a typical systolic pressure, the pressure when the heart contracts.

80 mm Hg is a typical diastolic pressure, the blood pressure when the heart relaxes after a contraction.

760 mm Hg is typical atmospheric pressure. The blood pressure readings represent gauge pressure, not absolute pressure – they tell us how much above atmospheric pressure the blood pressure is.


29

A delicious drink sits on the patio. From your balcony several stories up you manage to lower a straw into the glass, which is 15 m below you. Can you syphon up the drink?

(1) yes, but I will have to suck really hard(2) probably not, but a vacuum pump could(3) no, this is not possible(4) I don’t know

15 m


30

Water pressure At the surface of a body of water, the pressure you

experience is atmospheric pressure. Estimate how deep you have to dive to experience a pressure of 2 atmospheres.

h works out to 10 m. Every 10 m down in water increases the pressure by 1 atmosphere.

2 1P P gh

3200000 Pa 100000 Pa (1000 kg/m ) (10 N/kg) h


31

Rank by pressure A container, closed on the right side but open to the

atmosphere on the left, is almost completely filled with water, as shown. Three points are marked in the container. Rank these according to the pressure at the points, from highest pressure to lowest.

A = B > C B > A > C B > A = C C > B > A C > A = B some other order


32

Blaise Pascal (1623-1662)

A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished everywhere in the fluid and to the walls of the container

ext

ext

p p gh

p p


33

A container is filled with oil and fitted on both ends with pistons. The area of the left piston is 10 mm2; that of the right piston is 10,000 mm2. What force must be exerted on the left piston to keep the 10,000 N car on the right at the same height?

(1) 10 N(2) 100 N(3) 10,000 N(4) 106 N(5) 108 N

=?

=10 mm2

=10000 mm2

=10000 N


34

Pascal’s Principle

Hydraulic lever (see diagram on right)

Something has to give…– Since the liquid is incompressible,

the volume drop on the left is equal to that rises on the right, ie.

ext

ext

p p gh

p p

i o

i o

oo i

i

F Fp

A A

AF F

A

0thereforei i o o o ii

AA h A h h h

A


35

Pascal placed a long thin tube vertically into a wine barrel. When the barrel and tube were filled with water to a height of 12 m, the barrel burst.

(a) what is the mass of the water in the tube?

(b) what is the net force exerted onto the lid of the barrel?


36

Imagine holding two identical bricks under water. Brick A is just beneath the surface. Brick B is at a greater depth. The force needed to hold brick B in place is

(1) larger

(2) the same as

(3) smallerthan the force required to hold brick A in place


37

The Buoyant Force

With fluids, we bring in a new force.

The buoyant force is generally an upward force exerted by a fluid on an object that is either fully or partly immersed in that fluid.

Let’s survey your initial ideas about the buoyant force.


38

The Buoyant Force A wooden block with a weight of 100 N floats exactly

50% submerged in a particular fluid. The upward buoyant force exerted on the block by the fluid …

has a magnitude of 100 N has a magnitude of 50 N depends on the density of the fluid depends on the density of the block depends on both the density of the fluid and the

density of the block


39

Learning by Analogy

Our 100 N block is at rest on a flat table. What is the normal force exerted on the block by the table?

To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance.


40

Apply this to Buoyant force Apply the same method when the block floats in the

fluid. What is the magnitude of the buoyant force acting on

the block?

To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance.


41

Reviewing the normal force We stack a 50-newton weight on top of the 100 N block.

What is the normal force exerted on the block by the table?

To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. The block presses down farther into the table (this is hard to see).


42

Buoyant force We stack a 50-newton weight on top of the 100 N block.

What is the buoyant force exerted on the block by the fluid?

To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. The block presses down farther into the fluid (this is easy to see).


43

Apply Newton’s Second Law

Even though we are dealing with a new topic, fluids, we can still apply Newton’s second law to find the buoyant force.


44

Three Beakers The wooden block, with a weight of 100 N, floats in all

three of the following cases, but a different percentage of the block is submerged in each case. In which case does the block experience the largest buoyant force?

4. The buoyant force is equal in all three cases.


45

Three Beakers

What does the free-body diagram of the block look like?

What is the difference between these fluids?– The density


46

A block of weight mg = 45.0 N has part of its volume submerged in a beaker of water. The block is partially supported by a string of fixed length. When 80.0% of the block’s volume is submerged, the tension in the string is 5.00 N. What is the magnitude of the buoyant force acting on the block?


47

Apply Newton’s Second Law The block is in equilibrium – all the forces balance. Taking up to be positive:

0B TF F mg

y yF ma

45.0 N 5.00 N 40.0 NB TF mg F


48

Water is steadily removed from the beaker, causing the block to become less submerged. The string breaks when its tension exceeds 35.0 N. What percent of the block’s volume is submerged at the moment the string breaks?


49


The buoyant force is proportional to the volume of fluid displaced by the block. If the buoyant force is 40 N when 80% of the block is submerged, when the buoyant force is 10 N we must have 20% of the block submerged.

0B TF F mg

y yF ma

45.0 N 35.0 N 10.0 NB TF mg F


50

After the string breaks and the block comes to a new equilibrium position in the beaker, what percent of the block’s volume is submerged?

what does the free-body diagram look like now?


51


The buoyant force is proportional to the volume of fluid displaced by the block. If the buoyant force is 40 N when 80% of the block is submerged, when the buoyant force is 45 N we must have 90% of the block submerged.

0BF mg

y yF ma

45.0 NBF mg


52

Archimedes’ Principle

While it is true that the buoyant force acting on an object is proportional to the volume of fluid displaced by that object.

But, we can say more than that. The buoyant force acting on an object is equal to the weight of fluid displaced by that object. This is Archimedes’ Principle.

B disp fluid dispF m g V g

mass is our symbol for mass density:

volume

m

V


53

A Floating Object When an object floats in a fluid, the downward force of

gravity acting on the object is balanced by the upward buoyant force.

Looking at the fraction of the object submerged in the fluid tells us how the density of the object compares to that of the fluid.

fluid dispmg V g

object object fluid dispV g V g

object object fluid dispV V

object disp

fluid object

V

V


54

Beaker on a Balance A beaker of water sits on a scale. If you dip your little

finger into the water, what happens to the scale reading? Assume that no water spills from the beaker in this process.

1. The scale reading goes up 2. The scale reading goes down 3. The scale reading stays the same


55

Three Blocks We have three cubes of identical volume but different density. We

also have a container of fluid. The density of Cube A is less than the density of the fluid; the density of Cube B is exactly equal to the density of the fluid; and the density of Cube C is greater than the density of the fluid. When these objects are all completely submerged in the fluid, as shown, which cube displaces the largest volume of fluid?

1. Cube A 2. Cube B 3. Cube C 4. The cubes all displace equal volumes of fluid


56

Three Blocks

Each cube displaces a volume of fluid equal to its own volume, and the cube volumes are equal so the volumes of fluid displaced are all equal.


57

Three Blocks Which object has the largest buoyant force acting on it?

1. Cube A 2. Cube B 3. Cube C 4. The cubes have equal buoyant forces


58

Three Blocks

Each cube displaces an equal volume of the same fluid, so the buoyant force is the same on each.


59

1. The glass without the ball

2. The glass with the ball

3. The two weigh the same

Two identical glasses are filled to the brim with water. One of the two glasses has a ball floating in it. Which glass weighs more?


60

A boat carrying a large boulder is floating in a lake. The boulder is thrown overboard and sinks. What happens to the water level in the lake (relative to the shore)?

(1) it sinks

(2) it rises

(3) it remains the same


61

Cartesian diver– The diver is an object in a sealed container of water. – Air in the diver makes it buoyant enough to barely float at the

water's surface. – When the container is squeezed, the pressure compresses the

air and reduces its volume. This permits more water to enter the diver, resulting in it being less buoyant and sinking.

regular coke and diet coke


62

The origin of the buoyant force The net upward buoyant force is the vector sum of the

various forces from the fluid pressure.

Because the fluid pressure increases with depth, the upward force on the bottom surface is larger than the downward force on the upper surface of the immersed object.

2 1P P gh

net fluid fluidF P A gh A gV

This is for a fully immersed object. For a floating object, h is the height below the water level, so we get:

net fluid dispF gV


63

When the object goes deeper If we displace the object deeper into the fluid, what

happens to the buoyant force acting on it? Assume the fluid density is the same at all depths. The buoyant force:

1. increases2. decreases3. stays the same


64

When the object goes deeper

2 1P P gh

If the fluid density does not change with depth, all the forces increase by the same amount, leaving the buoyant force unchanged!


65

Archimedes’ Principle Buoyant force

Objects that float – Dry wood, ice, some plastics, oil, wax (candles)

…– Boats made of woods, ceramic, steel, or any

other materials, as long as they are hollow enough

Objects that sink– Rocks, sands, clay, metal, etc. – Any material with density larger than water

Apparent weight (example: submerged object weighs less– Apparent weight=actual weight - buoyant force– What is your apparent weight in water? (no

more than a few pounds!)

B fF m g


66

Unbalancing the forces If we remove the balance between forces, we can

produce some interesting effects. Demonstrations of this include:

1. The Magdeburg hemispheres (see below)2. Crushing a can


67

Crush a can Remember that this is just the collective effect of a

bunch of air molecules!


68

Summary Density and pressure of fluids

Air pressure, blood pressure and underwater pressure

Pascal’s Principle

Archimedes’ PrincipleB fF m g

0p p gh

ext

ext

p p gh

p p

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11/28/2007Meenakshi Narain - Physics 5 - Lecture 22 1 Lecture 22