Madonna H S Physics Tutorials

7/27/2019 Madonna H S Physics Tutorials

1/174

1

Madonna H S

Physics Tutorials

Edited by jay p mcdonald 2011

Forward

The tutorials are intended to be a nonlinear collection of topics in H S

physics. The document is to be used with an appropriate course book but it is

a stand alone device.

Not intended for commercial distribution but it is intended as a

supplementary teaching document.

2

L-001 Nature of Physics

From the dictionary ..physics [fiz-iks]noun ( used with a singular verb )

The science that deals with matter, energy, motion, and force.

Physicists work in three areas:1. Measurement,2. Analysis,3. Computation.

Consider the problem of determining distance of theEarth to the moon.

Greek astronomers, beginning with Aristarchus of Samos (310-230 B.C.,)

came up with a method of finding the moons distance by observations of a

lunar eclipse. This occurs when the moon enters the Earths shadow.

1. The measurement

A coin (quarter) of 1-inch in diameter was held at the distance where the

coin just blocks the just blocks out the light from the Sun from the

observers one eye.

(Never look at the sun directly ---youll damage your eye!)

The measured distance was about nine feet, or 108 inches

The shadow of the coin forms a cone with its point at the maximum distanceaway. Similarly, the earths shadow must also be conical. It must also

be proportional to the quarter since the sun is fixed in size. That is the

Earths shadow be 108 earth diameters long!


2/174

3

The Earths diameter was known to be about 8,000 miles at this time. Thismeans the cone is 108 earth diameters long, or a distance of 864,000 miles

from earth.

2. The analysis

Fact two: The Moon has to be much smaller than the Sun in order to be

eclipsed by the Earths shadow.

Fact two: The Moon has to be much smaller than the Earth in order to be

eclipsed by its shadow. The Sun has to be larger than the Moon and farther

away from the Earth in order to block the Sun during a solar eclipse.

Fact three: The Earths cone of darkness at the Moons distance from earth

during a lunar eclipse was observed to be 2.5 the size of the Moonsdiameter.

3. The computation (by Geometry)

4

Since the Moon and the Sun have the same apparent size in the sky then theangle ECD is the same as the angle EAF as depicted in the above figure.

This is true since AFE and EDC are similar and both are isosceles.

Notice now that the length FE is the diameter of the earths shadow at the

distance of the Moon, and the length ED is the diameter of the moon.

The Greeks found by observation of the lunar eclipse that the ratio of FE to

ED was 2.5 to 1, so looking at the similar isosceles triangles FAE and DCE,

we deduce that AE is 2.5 times as long as EC, from which AC is 3.5 times as

long as EC. AC must be 108 earth diameters in length, and taking theearths diameter to be 8,000 miles, the furthest point of the conical shadow,

A is 864,000 miles from earth.

From the above argument, this is 3.5 times further away than the moon is, so

the distance to the moon is 864,000/3.5 miles, about 240,000 miles. This iswithin a few percent of the right figure.


3/174

5

Note similar triangles have the same shape but different sizes.

Similar triangles

6

L-002 Scientific Numbers

Physics uses the SI system metric (decimal) system in scientific notation

format.

The SI system (International System of Units) is the modern metric systemof measurement and the dominant system of international commerce and

trade. The SI is maintained by the International Bureau of Weights and

Measures (BIPM, for Bureau International des Poids et Mesures) in Paris.

The core of the SI system is a short list of base units defined in an absolute

way without referring to any other units. The International System of Units

(SI) is founded on seven base units.

Quantity Name of Unit Symbol

Length meter m

Mass kilogram kg

Time second s

Electrical current ampere A

Thermodynamictemperature

Kelvin K

Luminous intensity candela cd

Amount of substance mole mol


4/174

7

Scientific notation is the way that scientists easily handle very large numbers or

very small numbers. For example, instead of writing 0.0000000056, we write

5.6 x 10-9. So, how does this work?

We can think of 5.6 x 10-9 as the product of two numbers: 5.6 (the digit term)

and 10-9 (the exponential term).

To write the number 123,000,000,000 in scientific notation:

Put the decimal after the first digit and drop the zeroes.

The coefficient will be 1.23

To find the exponent count the number of places from the decimal to the endof the number.

In 123,000,000,000 there are 11 places. Therefore we write 123,000,000,000

as:

8

Write 0.000 000 000 043 6 in scientific notation.

In scientific notation, the number part (as opposed to the ten-to-a-power part) will be "4.36". So I will count how many places thedecimal point has to move to get from where it is now to whereit needs to be:

Then the power on 10 has to be 11: "eleven", because that's how many

places the decimal point needs to be moved, and "negative", because I'mdealing with a SMALL number. So, in scientific notation, the number is

written as 4.36 1011

Here are some examples of scientific notation.

10000 = 1 x 104

24327 = 2.4327 x 104

1000 = 1 x 103 7354 = 7.354 x 103

100 = 1 x 102 482 = 4.82 x 102

10 = 1 x 101 89 = 8.9 x 101 (not usually done)

1 = 100

1/10 = 0.1 = 1 x 10-1 0.32 = 3.2 x 10-1 (not usually done)

1/100 = 0.01 = 1 x 10-2

0.053 = 5.3 x 10-2

1/1000 = 0.001 = 1 x 10-3 0.0078 = 7.8 x 10-3

1/10000 = 0.0001 = 1 x 10-4

0.00044 = 4.4 x 10-4


5/174

9

SI Prefixes List

Prefix Symbol 10n Long/Short Scale

yotta (gr. okto - eight) Y 1024 Quadrillion/Septillion

zetta (lat. septem - seven) Z 1021 Trilliard/Sextillion

eksa (gr. ex- six) E 1018 Trillion/Quintillion

peta (gr.penta - five) P 1015 Billiard/Quadrillion

tera (gr. teras monster) T 1012 Billion/Trillion

giga (gr. gigas giant) G 109 Milliard/Billion

mega (gr. megas great) M 106 Million/Million

kilo (gr. khilioi thousand) k 10 Thousand/Thousand

hecto (gr. hekaton hundred) h 10 Hundred/Hundred

deca (gr. deka ten) da 101 Ten/Ten

100 One/Onedecy (lat. decimus tenth) d 10-1 Tenth/Tenth

centy (lac. centum hundredth) c 10-2 Hundredth/Hundredth

milli (lac. mille thousand) m 10-3 Thousandth/Thousandth

mikro (gr. mikros small) 10-6 Millionth/Millionth

nano (gr. nanos dwarf) n 10-9 Milliardth/Billionth

pico (it.piccolo small) p 10-12 Billionth/Trillionth

femto (den. femten fifteen) f 10-15 Billiardth/Quadrillionth

atto (den. atten eighteen) a 10-18 Trillionth/Quintillionth

zepto (lat. septem - seven) z 10-21 Trilliardth/Sextillionth

yokto (gr. okto - eight) y 10-24 Quadrillionth/Septillionth

10

L-003 Measurement Uncertainties

Numbers taken in measurement may have different degrees of worth

depending on the error(s) in the method, and of the device(s) used as well as

the stability of the system being measured. The scientist must express theerror or uncertainty in any number measured.

The accuracy of a measurement is the degree of closeness ofmeasurements, as in target A

The precision of a measurement, is the degree ofreproducibility orrepeatability in the measurements, as in target B.

Target A is more accurate but less precise than target B

Target A Target B

Uncertainty in a measurement is due to a physical limit in the measurement,

e.g. a ruler.


6/174

11

The space between the smallest vertical lines on a ruler has no precise value.

Here is an estimate or uncertain value. The number 1 + 7/8 inch is accurate.

The last bit indicated by the arrow is a part of 1/8 unit of the ruler. The

uncertainty is therefore is centered about the estimated extra length of (

)

of 1/8 inch, with an uncertainty of +/- 1/16 inch. The arrow finds a value of

1 + 7/8 + (1/2)(1/8) inch, +/- 1/16 inch.

The centimeter scale for the ruler has less uncertainty in measurement (+/- 1

mm) and it is more precise than the inch scale.

In general, the uncertainty in a single measurement from an instrument

is half the smallest unit of the instrument.

Significant figures:

1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS

significant.

2) ALL zeroes between non-zero numbers are ALWAYS significant.

3) ALL zeroes which are SIMULTANEOUSLY to the right of thedecimal point AND at the end of the number are ALWAYS

significant.

4) ALL zeroes which are to the left of a written decimal point and are

in a number >= 10 are ALWAYS significant.

Examples:

12

A). 30.02 has 4 sig. Figs.

B). 50.0 has 3 sig. Figs.

C). 0.9003 has 4 sig. Figs. (0 left of decimal not sig.)

Rounding numbers:

Round number down if the value to truncate is = 5.

1232 -> 1230,

1239 -> 1240,

1235 -> 1240

Calculations with significant figures:

The rule for addition and subtraction is when quantities are added or

subtracted; the number of decimal places in the answer is equal to the

number of decimal places in the quantity with the smallest number of

decimal places

1.76 + 6.854 = 8.624 = 8.62 is correct answer

1.76 - 6.854 = -5.254 = -5.25 is correct answer

When you multiply or divide, keep as many significant

digits as found in the least accurate number.

1.76 * 6.854 = 12.06304 = 1.20 is the correct number.

1.76 / 6.854 = 0.233566 = 0.234 is the correct number.


7/174

13

L-004 Graphing Data

Once the scientist has collected a data set it becomes necessary to review thecollective data visually, i.e. graphically.

In mathematics and computer science, a tuple is an ordered element,(x, y). In principle each value the independent variable, x is related tothe dependant variable, y. Often y = f(x), that is y is the result of somefunction, f acting on x.

The invention of Cartesian coordinates in the 17th century by RenDescartes revolutionized mathematics by providing the firstsystematic link between Euclidean geometry and algebra.

A Cartesian coordinate system specifies each point uniquely ina plane by a pair ofnumericalcoordinates, which are the signeddistances from the point to two fixed perpendiculardirected lines,measured in the same unit of length. This x, y space is also called 2-dimensional Euclidean Space.

14

The x, y graph shows the placement of tuples [ (2,3), (0,0), ( -3,1),

(-1,5,-2.5)] in Cartesian 2-Space.

Graphing should not be done by hand. Use common software such as Excelby Microsoft.

Graphing in Excel.

The easiest way to begin is to gather some existing work sheet with data and

graphs to use as a template for your own data.

Open Excel and enter the date pairs in a columnar fashion as shown.


8/174

15

High light the data to be graphed by holding the mouse button and dragging

the over the numbers.

16

Find and click on the chart wizard to display the menu window. Click on theoption of choice.

Here we choose and select the charttype from the subsequent option menu .

After clicking on FINISH your graph is placed in the spreadsheet.


9/174

17

18

L-005 Vectors

Part A

A displacement is the shortest distance from an initial to the finalposition.

A dog takes a walk (along the dotted path). He begins at the green square

and ends at the red dot. The black vector mathematically represents his

displacement with its head at the end point and its tail at the start.

The vector, gives the resultant displacement of the dog and thedirection to his concluding position. The length of the arrow is a measure of

the magnitude of the displacement.

The displacement value itself is a scalar quantity has no sense of direction in

and of itself. The displacement is the same for any direction. It is the addedsense of direction that constitutes a vectored value.


10/174

19

Such a vector is used for operations in the plane. This displacement can be

described by Cartesian values, (x, y), where x = (xo xf) and y =(yo-

yf).

The dog begins at location (xo yo) and ends at location ( x f , yf).

Vectors are named by a bold letter ( A) symbol, or a letter with arrow

overstrike, .

Vectors are equal if their lengths and direction are the same.

Figure shows equal vectors.

Vectors are unequal if their lengths and or directions are not the same

Figure shows unequal vectors. No two are the same.

20

Vectors are added by translation of vectors keeping length and directionunchanged while placing one end (tail B) of the vector to the other (head A).

Vectors are subtracted by inverting the one (B) then adding (-B)as before.

An arbitrary vector, V has some angle, with respect to a Cartesian axis.Therefore the vector can be mathematically decomposed into Cartesian x-y

components by sine and cosine operations. Note the vertical and horizontal


11/174

21

resulting components are vectors in their own rights and point along the x, y-axes respectively.

22

Part B

Adding vectors graphically is accomplished by drawing them head to tail.Simply translate the vectors without modifying angles.

The sum (resultant) is the red vector from the tail of the first to the head of

the last vector.

Many vectors are so added to form a polygon when the resultant vector is added.


12/174

23

Subtraction is done as addition after the subtrahand (the vector to be

subtartced) is reversed in direction.

Consider 1-dimensional (x-axis, or y-axis) addition/subtraction:

Adding vectors by decomposition, follow the rule that the sum of vector

components is equal to the sum of vectors.

This method uses the theorem of Pythagoras for magnitudes and angle. Any vector canbe decomposed into basic x-axis, and y axis components

24

Here = 90o

and, A2

= Ay2

+ Ax2

For arbitrary , A2

= Ay2+ Ax

2+ 2 Ay Ax cos[]

The sum of components for two or more vectors proceeds by summing the

all x-components to a resultant x-vector, and similarly for a resultant y-

vector.


13/174

25

= tan-1

[Ry/ Rx]

For several vectors, the resultants Rx, Ry -components will be the sum of terms for eachvector being added.

Example: Find the resultant vector for the sum of three vectors;30 units at 45

o,

50 units at 160o70 units at 90

answer 41 units, 231o

26

L-006 Velocity and Acceleration 1-D

Kinematics is the branch ofclassical mechanics that describesthe motion of objects without consideration of force.

Consider the motion of a car traveling at 55 mph. Its speed is 55 mphand this is a scalar quantity. The same car traveling 55 mph northward has

its motion now given as a displacement vector.

A car or other body can change its rate and or direction of motion. Rate is

distance over time, eg. mile over hour. The change in displacement vector,

D will occur over a unit of time t. Time is always a scalar. We consider 1-

dimension only.

The cars velocity V, is a vector given as

V = D / t.

Where t = 0, the velocity is instantaneous and not measurable. A cars

speedometer indicates an instantaneous speed but in actuality this value is

determined by the rotation of the wheels for a finite time interval and it is

not truly an instantaneous measurement.

Where t > 0 (this value is always positive), the velocity is an average

value.

Vavg = D / t =

In detail D = Df Do, where Do = initial position and Df = finialposition.

So


14/174


15/174

29

L-007 Random Walk

Any continuously moving object such as a deer feeding or a gas molecule in

a bottle proceeds in successive random steps. The result can be

mathematically formalized and the topic is used in science, economics,

psychology, and computer science.

The resultant distance traveled does not indicate the amount of random steps involved inthe final outcome (see above illustration).

Suppose however the placement off the random steps is quantized. Let a particle enter abox divided, as is the following illustration. The number of collisions along the path ofthe object through the container is recorded for the right versus the left vertical sections(see following graph).

30

Graph of the collision number versus location in container.

exit

start

0

1

2

3

4

5

6

7

Series1


16/174

31

L-008 Dimensional Analysis

Any calculation involving units must have dimensions that match. There for

a unit must have a conversion factor! This also referred to a unit analysis.

12 inches = 1 foot

1 = 1/12 ft/in

so 3 in * 1/12 ft/in = ft

Consider the number of seconds in a day.

60 secs = 1 min , so 1 = 60 secs/min,

60 min = 1 hr, so 1 = 60 min/hr,

24 hr = 1 day, so 1 = 24 hr/day, combine

1* 1*1 = 60* 60 * 24 * sec/min * min/* hr * hr/day

1 = 86,400 secs /day, or

1 day = 86,400 secs

Any unit can be converted in a like manner

32

L-009 Gyroscope

The gyroscopes wheel spins around its the holding axis for rotation. The entire systemand will spin around the axis of gravity after a tilt from the vertical for precession(wobble). The wobble starts when one side of the wheel is pulled down and the opposite

side rises upward.

The rotation of the gyroscopes wheel creates angular momentum that acts along therotational axis. Gravity creates a torque that acts perpendicularly to the axis of angularmomentum. The resulting action is a displacement of the rotational axis that must bemutaully perpendicular to the directions of momentum and torque. Instead of falling the

rotational axis processes. When the ration of the gyroscope ends, the angular momentumgoes to zero and the rotational axis falls on its side..

Axes of precession


17/174

33

L-010 Drawing with Power Point

Open Power Point for a new blank presentation. Note the bottom bar menu

with pop up menus for graphing objects.

34

From the top menu view choose ruler, guides.

To add a guide, hold down CTRL while dragging an existing guide (Lift the

drag then CRTL). To delete a guide, drag it off the slide. To hide the guides

without deleting them, clickGuides on the View menu

To place and object, locate the cursor over the item on the tool bar then

Click once. Move the cursor over the graphing area and click once.

Drag a guide to position it where you want to align the objects. Drag each

object near the guide so its center or edge automatically aligns with the

guide.


18/174

35

L-011 Motion Graphs

Reading graphs is vital in science.I. Consider the following example of position versus time graph.

For any time belonging to the horizontal (x-axis) interval of (0, 7) seconds, the positionof a moving object can be found displaced along the vertical (y-axis) interval of (+2, -2)cm and vice versa.A slope in a graph is given (y final - y initial)/ (x final x initial)

1. For the first two seconds (0, 2) on the x-axis the object moves vertically on the y-axis from (0, -2) cm in a negative direction with the slope:

-2/2 cm /sec = -1 cm/sec, a velocity vector.2. For time interval (2, 3) seconds the object the object moves vertically from

(-2. 1) cm in a positive direction with a slope3/1 cm /sec = 3 cm/sec, a velocity vector

3. For time interval (3, 6) seconds the object does not move vertically which gives

a zero velocity vector. The area under the curve shown as hatched rectangle is(6-3)^(1-0) = 3 cm*sec, not of physical interest

Y cm

X secs

36

4. For the last seconds (6 ,7) the object moves vertically from (1 , 0) cm with aslope

-1 /1 cm /sec = -1 cm/sec, a velocity vector

II Consider the following example of velocity versus time graph. The same shape as

the previous graph but the y-axis is now cm/sec.

5. For the first two seconds (0, 2) on the x-axis the object changes velocity on the y-axis from (0, -2) cm/sec in a negative direction with the slope:

-2/2 cm /sec2

= -1 cm/ sec2, an acceleration vector.

6. For time interval (2, 3) seconds the object the object moves vertically from (-2,1)in a positive direction with a slope

3/1 cm / sec2

= 3 cm/ sec2, an acceleration vector.

7. For time interval (3, 6) seconds the object does not move vertically which givesa zero acceleration vector. The area under the curve shown as a rectangle is

(6-3)^(1-0) = 3 (cm/sec)*sec which is displacement8. For the last seconds (6, 7) the object moves vertically from (1 , 0) cm with a

slope

-1 /1 cm / sec2

= -1 cm/ sec2, an acceleration vector.

Y cm/sec

X secs


19/174

37

II Consider the following example of acceleration versus time graph. The same shapeas the previous graph but the y-axis is now cm/sec2.

9. For time interval (3, 6) seconds the area under the curve shown as hatchedrectangle is (6-3)^(1-0) = 3 (cm/sec2)*sec which is velocity.

Note areas under triangular section would then be Y*X.

Y cm/sec2

X secs

38

L-012 Bernoullis Theorem

Bernoulli's theorem; as the speed of a fluid increases the pressure decreases.

Bernoulli's principle can be derived directly from Newton's 2nd law. If asmall volume of fluid is flowing horizontally from a region of high pressureto a region of low pressure, then there is more pressure behind than infront. This gives a net force on the volume, accelerating it along thestreamline.

K is the total energy and it is conserved in the flow of a fluid in the pipe

above. V is velocity and represents kinetic energy, while P is pressure andrepresents potential energy. In the section of system where flow (kinetic

energy) increases then pressure (potential energy) must decrease.

The argument of Bernoulli holds where gravity is constant, i.e. motion in

horizontal direction. This concept is the basis of an airfoils lift and a

moving balls curved path.


20/174

39

Tangent line to a curve at a point

In geometry, the tangent line (or tangent) to a curve at a givenpoint is

the straight line that "just touches" the point.

In the figure, the straight line (red) is the tangent to the (large) point in the graph.

If the graph is position vs. time then the tangent is the instantaneous

velocity at that point. If the graph is velocity vs. time then the tangent is the

instantaneous acceleration at that point.

40

L-013 Constant Acceleration / Free Fall.

Between two locations in space the velocity of a moving object may vary

wildly, e.g. a bee buzzing among the flowers.

The average acceleration (aavg) is the difference in the velocity (v) of thetwo points over the time (t) of transit between these points.

aavg = v/t

I. If the acceleration is constant for all time in the period (t), then theacceleration is constant, (ak). The final velocity, (vf) can be found

vf = vo+ ak*tBoth of the above equations work for average or constant acceleration but

the second is typically used with the constant form

Example: what is the final velocity of a ball starting at rest, moving withconstant acceleration of 20 m2/sec for 10 seconds?

Ans vo = 0 m/sec

t = 10 secak= 20 m/sec

2

vf = 0 m/sec + 20 m/sec2 * 10 secvf = 200 m/sec


21/174

41

II. If the acceleration is constant for all time in the period (t), then theacceleration is constant, (ak). The final displacement (df) using the sum of

terms with (vo, vf ,t) can be found.df = vo *t+ * (vf - vo,) *t

Example: What is the final distance a bird moving with a constant

acceleration covers after 10 secs of flight, if its initial velocity is 0 and itsfinal velocity is 30 m/sec?

Ans vo = 0 m/sec

t = 10 secvf = 30 m/sec

ak = 0 m/sec2

df = 0 m/sec + * (30 - 0) m/sec * 10 sec

df = 150 m/sec

42

III. If the acceleration is constant for all time in the period (t), then theacceleration is constant, (ak). The final displacement (df) using the sum of

terms with (d0,vo, ak ,t) can be found.df = d 0 + vo*t + *ak*t 2

Example: What is the final position a bird moving with a constant

acceleration of 10 m/sec2after 10 secs of flight, if its initial velocity is

2 m/sec and its initial position is 10 m from a tree?

Ans vo = 2 m/sec

t = 10 secd0 = 10 m

ak = 10 m/sec2

df = 10 m+ 2m * 10 sec + * 10 m/sec2 * (10 sec)2

df = 10 m + 20 m + 500 m = 530 m

IV. If the acceleration is constant for all time in the period (t), then theacceleration is constant, (ak). The final velocity, (vf) using the sum of terms

(d0 ,df,vo, ak) can be found.

vf2 = vo

2 + 2* ak*(df - d0)


22/174

43

Example: What is the final velocity a bird moving with a constantacceleration of 5 m/sec2

when its initial velocity is 2 m/sec, its initial

position is 10 m, and its final position is 100 m from a tree?

Ans vo = 2 m/secak= 5 m/sec2

d0 = 10 m

df = 110 m

vf2 = (2m/sec )2 + 2 * 5 m/sec2 * (110 10) m

vf = [4 (m/sec )2 + 1000 (m/sec )2 ]0.5 = 320 m/sec

44

Free fall under gravity:

Gravitational acceleration, (g) had the value of g is 9,8m/s or

approximately 10 m/ s for simplified calculations. One uses the above

equations for constant acceleration with

ak = g = -9.8 m/sec2

Figure shows a stone falling and its velocity change per second under

the acceleration by gravity.

Example;


23/174

45

A ball falls off the roof of the house. It takes 6 seconds to hit the ground.

What is the velocity before the ball crashes to the

ground?

Ans

vf

= vo+ g*t

vo = 0 m/sec

t = 6 secg= 10 m/sec

2

vf = 0 m/sec + 10 m/sec2 * 6 sec

vf = 60 m/sec

46

L-014 Newtons Laws of Motion

Three laws of mechanics describe the motion of objects. These Laws were

first describe d by Sir Isaac Newton in the 1600s

The Three Laws of Motion

1. First law of motion: An object at rest will remain at rest, and an object

in motion will remain in motion, at a constant velocity unless or untiloutside forces act upon it.

2. Second law of motion: The net force acting upon an object is aproduct of its mass multiplied by its acceleration.

3. Third law of motion: When one object exerts a force on another, the

second object exerts on the first a force equal in magnitude butopposite in direction.

Newtons First Law is normally taken as the definition of inertia. If there

is no net force acting on an object, then the object maintains a constantvelocity. If velocity is zero, then the object remains at rest.

An object only changes velocity in either magnitude or direction when acted

upon by an outside force.

Example: If a car is traveling with a helium-filled balloon in the back seat,

and the car suddenly speeds up, what happens to the balloon? What happensto you?


24/174

47

The answer for the balloon has to do with the motion of the air in your car.When the car accelerates forward all of the air rushes to the back of the car

and you have created an area in the back of your car of higher air pressure.

Since helium is lighter than air it wants to 'float' away from that area and movesforward while your body moves backward.

2. Newtons Second law is given by the vector equation that the resultant (net) force,

Fris the product of the mass,mof the object times the resultant (net)

acceleration,Ar.

Fr = m * ArThe units of force are the Newton,

N = kg * m/sec2.

The stone block (above) accelerates as a force is applied.

Example is a 10 kg cannon ball falling from a wagon. What is the force on the

ball if gravity is the only acceleration?

Answer Fr = m * Ar

Fr = 10 kg * 10 m/sec2.

Fr = 100 N , downward

Fm

A

48

3. Newtons Third law is stated For every action, there is an equaland opposite reaction. This is to say a pair of objects that interact

exert equal and opposite forces one upon the other.

Example is a heavy ball supported by a rope attached to an overhead

beam. Gravity pulls the ball downward but the tension (force) in the ropeequals the force of gravity but in the opposing direction.

Tension

Gravity


25/174

49

L-015 Periodic Motion

An object that moves in regular (equal) intervals of time displays periodic

motion. Common examples are the movement of a porch swing, repetition of

ocean waves coming ashore, and the rotation of the earth.

In addition to linear motion and rotational motion there is another kind ofmotion that is common in physics. This is the back and forth action of an

object in oscillation (vibrating).

I. Simple Harmonic Motion, SHM is an important type of periodicmotion. This is motion in which the acceleration is alwaysdirected towards an equilibrium position. The magnitude of arestorative acceleration is proportional to the displacement of theobject from rest or equilibrium.

Case 1. Mass spring systemconsists of a spring attached to a hanging

mass. As the mass is displaced a distance from its equilibrium position, work

is done and potential energy is stored in the spring.

50

The spring (considered mass less) holds a ball of some weight. Atequilibrium (no motion), the pull downward by gravity on the ball is

balanced by the pull up by the stretched spring.

Here (T) is the period in time of the motion, (m) is the mass of the ball, and

(k) is the spring constant.

Case 2. The swing pendulum is a weight suspended from a pivot point and

moving freely once displaced from its (resting) equilibrium position. There

is a restoring force by gravity that will accelerate the mass back toward theequilibrium position.


26/174

51

A string or rod (considered mass less) holds a hanging weight, which is,

displaced side ways. Gravity acts to restore the system.

Here (T) is the period in time, (L) is the length of the string (rod), and (g) is

the acceleration of gravity. There is no mass (m) term in the equation.

Case 3. Circular motion, describes the motion of a body traversing in

a circularpath at constant speed. However, the velocity is not constant as its

direction is tangential. This is the effect of centripetal acceleration, which is

constant in magnitude and directed towards the axis of rotation.

52

A solid object travels at constant speed in a circle about a midpoint.

T = 2 r / VHere (T) is the period in time, (r) is the radius of the circle and (V) is the

tangential velocity.

Case 4. Transverse motion of a vibrating string is periodicoscillation a perpendicular the direction the wave is traveling. A good

example is a moving wave in which the fluid oscillates up and downwhile traveling horizontally.


27/174

53

Vibration of string moving left to right reflects at a fixed end.

Vibration of string moving left to right reflects at an open end.

T = 1/ f

Here (T) is the period in time, and (f) is the frequency of oscillation.

54

L016 Forces in 2-d

Consider the following examples;

a.) **Perpendicular forces on wire, rope, et,

b.) **Normal force and friction,

c.) Inclined plane, pulley, lever

d.) Projectile motion

**The static (mechanical) equilibrium for an object is the condition in which

the object has a net force of zero acting upon it.

Type a.) Horizontal wire, rope, etc.

#1. In the figure below, a ball is suspended from two ropes. The system is

motionless. If the magnitude of T2 is 10.0 N, what is the magnitude of T1?

The ball is motionless, so the net force is zero. The component of T1 pullingto the left must be equal and opposite of the component of T2 to the right.

The component to the right is T2 sin 60. The component to the leftis T1 sin 30. With these components, we can solve for T1.

T1 = (T2 sin 60)/(T1 sin 30) = 17

T1

T2

mg

30o 60o


28/174

55

#2-

Half the 20.0 N force will pull on the left the other half on the right. The small angle of

distortion ();

= tan-1[ 0.1/50] = 0.115 degreesTr = 0.5 Fd / sin

= 10 N / 2.0*e-5

= 5000 N

Type b & c.) Normal force, friction, inclined plane, lever and pulley

The normal force Fnis the force component that is perpendicular to a surface

of contact such a floor or wall, which prevents the object from penetratingthe surface.

56

In the figure above, the normal force is perpendicular to the incline plane and in the

direction of cos. Note the movement of the crate of mass (m) is in the direction of sin.

The frictional force, Ff always opposes the direction of motion.

Friction is caused by the attractive forces between ruff surfaces as depicted in the aboveillustration

#1. The mass of a crate lying flat upon the floor is 50 kg and the Ff is

Fn

mg cos

mg

mg sin

FF


29/174

57

#1. The mass of a crate lying flat upon the floor is 50 kg and the Ff is250 N. What will be the acceleration of the crate for a 300 N pulling force?

a = (300 - 250)N/50kg = 10 m/sec2

#2 A20 kg box sits on a frictionless 30 inclined plane. What is the normalforce? What is the acceleration of the box?

Fn = mg cos 30o

= 20 kg * 10 m/sec2 * 0.866

= 177 N

ma = mg sin 30o

a = g sin 30o

a = 10 m/sec2 * 0.500

a = 5 m/sec2

#3 Consider the lever in the following illustration It offers a mechanicaladvantage in lifting heavy loads. The lever can increase the applied force by

the following ratio.

FL= Fe *E/ L

Where Fe is the applied force

E is the length of the force (effort) arm

FL is the load force

L is the length of the load arm

58

The length of the force arm is 10 meters and the load arm is 1 meter. The

applied force downward is 20 kilograms. This lever will lift a load of 200kilograms.

#4 Consider the pulley system in the following illustration.

In the standard pulley system the masses, [m < M] are connectedbya mass less and frictionless pulley and rope. The system is initially atrest. The acceleration may be calculated.

The Tension, acceleration, and velocity will be the same everywhere due to the rope.


30/174

59

T - M*g = M*a larger M moves downward

sign (-a)

T - m*g = + m*a smaller m moves upwardsign (+a)

where (g) is the acceleration by gravity and (a) is theresultant acceleration

a = g*(M -m)/ (M + m)

a = 10*(50-10)/(50+10) m/sec2

a = 6.7 m/sec2

upward form, and downward for M

Type d.) Projectiles

The motion in which a body is thrown or catapulted is called projectilemotion. The resulting path followed is called its trajectory and proceeds

under the influence of gravity. See illustration below.

The path of a projectile is illustrated with vertical velocities in red, and bluefor the horizontal velocity.

60

#1.) A rocket has a speed of 300 m/sec as it is launched horizontally from cliff64 meters above the ocean. How far does the bullet travel before striking the

water?

Part one of the answer is to use free fall equations to get the time the rocketwill fall vertically to the ground.

Dov= O m, vertical

Dfv = 64 m, verticalG = -10 m/sec2, vertical

Vov = 0 m/sec, vertical

Dfv = Vov*t +1/2 g*t2

64 m = 0*t (10/2 )* t2, (m/sec

2) * sec

2= m

t = Square root of [64 / 5]

= 3.6 sec, time is always positive

Part two is to use this time of flight to calculate how far the rocket will travel

under constant horizontal velocity.

Doh = 0 m/sec, horizontalDfh = ? m/sec, horizontal

t = 3.6 sec

Voh= 300 m/sec, horizontalDfh = Voh *t , (m/sec * sec) = m, horizontal

= 300*3.6 = 1080 m, horizontal

#2) Consider the same problem but the rocket leaves at 450

of elevation. Thusthe vertical part becomes a free fall calculation up then a free fall calculation

down for total time.Dov = O m, vertical upDfv= 64 m, vertical down

g = -10 m/sec2, vertical

Vov = 300 * Sin 450m/sec = 212 (m/sec), vertical up

Vfv = 0 m/sec, vertical upVfv = Vov + g*t, vertical up

0 m/sec = 212 (10 )* t, m/sec + (m/sec2) * sec

= m/sec , vertical up

t = [-212 /-10] = 21.2 sec, vertical up

Dfv = Vov*t +1/2 g*t2Dfv = 212*21.2 +(-10/2)*(21.2)2 m, vertical upDffv = 2247 + 64 = 2311 m, total distance above water


31/174

61

Part two is the horizontal distance.Doh = 0 m/sec, horizontalDfh = ? m/sec, horizontalt = 21.2 up + 21.2 down + 3.6 down sec = 46 sec, total

Voh = 300 cos 450m/sec, horizontal

Dfh = Voh *t , (m/sec * sec) = m, horizontal= 212 *46 = 9752 m, horizontal

62

Addendum: Uniform Circular motion / Torque

Uniform circular motion describes the motion of a body traveling in

a circular path with a constant speed (see figure following). The radius (r) of

rotation remains constant. The body's velocity is not constant however and it

remains tangential to the circumference, and orthogonal to the radius at all

times. Changing velocity denotes acceleration (ar) , which is constant andcentripetal at all times. This acceleration is always directed along the radius

toward the axis of rotation.

An object moving about a circle travels one circumference, 2r distance in

one revolution in time, (T). The tangential (linear) velocity, (v) is given.

V = 2r/T


32/174

63

The centripetal acceleration (ar) is given.

ar = V2/ r

Angular velocity, () is given in radians/sec, (rad/sec) where one radian is

the ratio between the length of one radius of arc to the length of the radius,

(see following figure).

One revolution is therefore, 2r/r = 2 rad = 360o

= V/r

r

r

64

Eample #1. A DVD has a scratch 3 cm from the center which skips 45 time

a minute. What is the linear velocity of the rotating DVD? What is the

angular velocity of the DVD?

Answer: T= (60/45) (secs/rev) = 1.3 s

R = 3 cm

linearV = 2r/T (cm/sec)linearV = 2 (3)/(1.3) = 14.5cm/secangular = V/r = 14.5/3 = 4.8 rad/sec

Example #2 A small plane circles an airport at a distance of 10,000 m

every 3000 secs. What is the centripetal acceleration of the plane?

Answer:

linearV = 2r/T (m/sec)linearV = 2(10000)/(3000) (m/sec)linearV = 20.9 m/sec

ar = V2/ r (m

2/sec)

2*(1/m)

ar = 20.92/ 10000 (m/sec

2)

ar = 0.044 m/sec2


33/174

65

Torque, () is a turning force that rotates an object about an axis or pivot.For example, moving a wrench connected to a nut or bolt produces a torque

that loosens or tightens the nut or bolt. The force acts at right angle to the

axis of rotation (see the following figure).

= force x length arm (N*m)Example #3. A force of 120 N acts perpendicularly upon an arm 0.5 m

from the axis. What torque is achieved?

= force x length arm (N*m) = 120 x 0.5 (N*m) = 60 N*m

rotation force

66

L-017 Vector Space

Consider the Cartesian plane, or 2-space where any point is uniquely defined

by a pair of coordinate, (x, y). Any point (a, b) in this space can be located

by another pair of coordinates, (P, ) with (P), the position vectors tail atthe origin and its head landing on the point (a, b). Theta, () is the positive(ccw) angle from the horizontal axis to (P).

Typically any position vector can be written in component vector termsthatare parallel to the two axes, P = px+ py. Thus any point in the space is

further identifiable by coordinates (px, py).

The algebra of adding or subtracting vectors proceeds by adding andsubtracting component vectors of the same axes. See the previous section L-

005. The resultant vectors magnitude |R| is obtained by the Pythagorean

theorem and the resultant angle (r) by the tangent inverse. The positionvector is said to span 2-space, i.e. all points in the plane may be so mapped.

See following figure.

(x, y), (|R|, )

R


34/174

67

This concept is readily extended to 3-space. In an attempt to simplify

further algebraic operations, a set of mutually orthogonal, linear

independent, unit base vectors (i,j, k) are chosen. That is to say each of theunit base vectors is positioned 90o from the other base vectors, and the

magnitude of one component vector |ai|, or |bj|, or |dk| does not affect thevalue of any other component vectors.

Further |i| = |j| = |k| =1.

j

i

k

ik

j

68

So let H =ai +bj +dk , and G =mi +nj +wkthen H + G = (a + m)i + (b + n)j + (d+ w)kand H - G = (a - m)i+ (b - n)j + (d- w)k

The magnitude | H + G |2

= (ai + mi)2 + (bj + nj)

2 + (dk+ wk)2 , the direction

of

H + G is given by the set of direction cosines for (i , j , k);cos2 (i ) = (ai + mi)2 /| H + G |2cos2 (j ) = (aj + mj)2 /| H + G |2

cos2 (k) = (ak+ mk)2 /| H + G |2where the identity cos2 (i ) + cos2 (j ) + cos2 (k) = 1 obtains bysubstitution.

Example #1. Find the magnitude and direction for the given vector

F =100i +153.2j + 80.8kAnswers | F |

2= (100)2 + (153.2)2 + (80.8)2

| F |2

= 40,000

| F | = 200 isthe magnitudeThe resultant direction is the set of cosines;cos

(i ) = 100/200 =0.500,

cos(j ) = 153.2/ 200 = 0.766,cos(k) = 80.8/ 200 = 0.404with a check, 0.500

2 + 0.7662 + 0.4042 = 1

Example # 3 Find the Dot product A *B given


35/174

69

Example #2. Find the resultant vector, H - G given

H =3i + 0j +1k and G =2i +3j +0kThe answer:

H - G = (3 - 2)I + (0 - 3)j + (1- 0)kH - G = i - 3j + k

Now consider the multiplication of two vectors;

H =ai +bj +dk , and G =mi +nj +wk

I. The Dot Product of two vectors is given;H * G = a*b + b*n +d*w, and it is also given as

H * G = (| H |*| G |)*cos(),where () is the included angle between the two vectors. Theresultant product is a scalar quantify and not a vector.

The factor, | G |*cos(), is called the projection of G onto H thatis the vector componentofGparallel to the direction of H. The

reverse is also true | H |*cos() forthe projection of H onto Gat the same time.

70

Example # 3. Find the Dot product, A *B given

A =0i +3j +-7k and B =2i +3j +1k , and find the angle, () between thetwo vectors.Answer: A *B = 0*2 + 3*3 +(-7*1) = 2

| A |2

= (0)2 + (3)2 + (-7)2 = 7.62 2

| B |2

= (2)2 + (3)2 + (1)2 = 3.74 2

= inverse cos[(2) /(7.62 * 3.74)] = inverse cos(0.070) = 86o

II. The Cross Product, A X B of two vectors A, B is a multiplication

where the resultant is also a vector. Given

Cross Product, C is calculated,

The vectorC is orthogonal to both vectors A and B. according to the righthand rule ofA moving onto B while the thumb points inthe direction ofC.

It is noted that the magnitude and direction cosines are obtained as shown

previously.

Vector space problems


36/174

71

Example # 4. Find the cross product of vectors A = i+j and B = 2j + 2k as

well as the resultant vectors direction. Using above equations.

Answer C = 2i 2j +2k with direction cosines

cos(i ) = 2/3.46 = 0.578,cos(j ) = 2/ 3.46 = 0.578,cos(k) = 2/3.46 = 0.578with a check, 0.5782 + 0.5782 + 0.5782 = 1

Note the precession of a gyroscope, () sideways is related by the crossproduct with gravitational downward torque, (), and horizontal(mutually perpendicular) angular momentum, (L) resulting from the angularspin.

= X L

72

Vector space problems

1. Add the vectors

A = i + 3kB = 3j + 1k

C = 2i + 2j + 1k

Ans A+B+C = 3i +5j +5k

2. subtract vectors

A = 235i + 35k - 77k

B = -235i + 123j - 10kAns A-B = 470i +-88j +67k

3. Find the magnitude and the direction cosines for A+B where

A = i + 3k

B = 3j + 1kAns | A+B |= 5.01

cos(i ) = 0.200,

cos(j ) = 0.60

cos(k) = 0.80

4. Find the dot product of vectors A*C and C*A and find the included

angle

B = 3j + 1k

C = 2i + 2j + 1k

Ans A*C = 7

= 42.4o

5. Find the cross product H = AXB with resultant angle for

A = i + 3kB = 3j + 1k

Ans H = -9i + -j + 3k

cos(i ) = 0.95

cos(j ) = 0.10

cos(k) = 0.32

L 018 U i l G it ti


37/174

73

L-018 Universal Gravitation

I. Planetary MotionKeplers Laws.

Law # 1. All planets revolve in an elliptic path with a star at one focus.

For any point on an ellipse, the value of R1+ R2 is constant where (F1, F2)

are the foci of the ellipse.

The equation for an ellipse in x-y space is (x-h)2/a

2+ (y-k)

2/b

2=1

where intercepts |a| >0, and |b| >0 and the elliptic center is located at [h, k].

Law # 2. A line drawn from a star to a revolving planet will sweep out

equal areas in equal periods of time. Thus a planet moves faster when it

transits the closer to the star. The two shaded areas in the figure below are

equal valued.

F1 F2

R1 R2

74

The Area, (A) under the arc length (r*d) is given asA = (r* r*) where r is the radius, (in measure of radians) is thesubtended angle of the arc.

dA/dt = (r* r*d/dt)but d/dt = v/r is just angular velocityso dA/dt = (r* r* v/r ) = (r*v)

In order for a constant dA/dt the term (r*v) must be also be held constant.

Therefore r 1/v.

Law # 3. The square of the ratio of two plants periods of revolution isequal to the cube of the ratio of the two plants radii from the star.

(Ta/Tb)2

= (ra/rb)3

The period (T) is the time for one orbit around a star. The planets distance

form the star(r) is the semi-major axis.

The semi-major axis runs from the center, through a focus of the elliptical

path to the edge of the orbit. It is the measure of the radius taken at the

orbit's two most distant points.

r*

(Ta2)/(ra

3) is constant for all objects which orbit a star. Newton's law of Universal Gravitation


38/174

75

(Ta )/(ra )is constant for all objects which orbit a star.

Example #1. One moon and a large asteroid orbit the center of a galaxy. The

orbital radius of the moon is 10, 000 parsec. The orbital radius of the

asteroid is 1000 parsec. The period of revolution of the asteroid is 101millennia. What is the period of orbit for the moon?

Answer:

Tmoon = Tast * (rmoon/rast)3/2

= 101 * (10000/ 1000 )

1.5

= 3194 millennia

check 31942/ 10000

3= 1.04 * 10

-5millennia

2/parsec

3

1012/ 1000

3= 1.02 * 10

-5millennia

2/parsec

3

76

Newton s law of Universal Gravitation

Every object in the universe attracts every other object by a force

proportional to the product of the masses (m, M) and inversely proportional

to the square of their separation distance, (d). This force, (F) acts equally onboth objects.

F = G*m*M/d2

The attractive force is measured in Newtons, (N) and the masses,

(m, M) in kilograms, and distance, (d center to center) in meters, and the

constant G is approximately equal to 6.6741011 N m2 kg

2.

Let (Me = mass of the Earth), and (m) be mass of an object on the Earths

surface. Then the radius, (re) of the Earth is distance from its center to its

surface. Force, F = mg for acceleration of gravity, (g = 9.8m/sec2). So

M m

d

attractive

. force

b.) dg= -1.63 m/sec2


39/174

77

F = G* m*Me /re2

= mg

g = G* Me / re2

Example #1. A 100 kg gram person stands of the surface of the earth. If the Earth weighs6.4*1024 kg, then find the radius of the Earth both analytically and numerically.

Answer: re2

= (G* Me)/g, and re = 6.602 106

m

Variation of acceleration due gravity.

The equation for Newtons law of Universal Gravitation indicates a change

of gravity with distance from the Earths center.dF/dre = G*m*Me*d(re

-2 )

where d(re-2

)/dre = -2re-3

= -2/re3

so, dF/dre = -2*G*m*M/re3

dF = -2*G*m*M *(1/re3

)*dre

now divide above equation by F = G* m*Me /re2

therefore dF/ F = dg/g = -(2/re) dre

example #2 Given g = 9.806 m/sec2 and re = 6.02 *106 find the change in (g)

a.) dg at dre = 1000m ,and

b.) dg at dre = 500,000m above the Earth.

Answer a.) dg= - g (2/re) dre = -9.806*2 * dre /(6.02 106) m/sec

2

dg = -0.003 m/sec2

78

) gThe case of an object in circular motion above the Earth

F = G* m*Me /re2

= mac

Where centripetal acceleration, ac = v2/ re

So m* v2/ re = G* m*Me /re2

Then velocity, (v) of the orbiting object is

v2 = G*Me /re

example #3 An astronaut orbits the earth at 200,000 above. What is astronauts speed?

v2 = G*Me /(re + 20000)v = 22,280 m/sec

Centripetal acceleration is also given for period of revolution,(T)

ac = 42re /T

2

by substitution G* m*Me /re2 = m42re /T2

then period of revolution just above the Earth is

T2 = 42 re3/(G*Me )

example #4 If a satellite is 40000 m above the Earth what is its period of

revolution. 1 sec = 3.12 * 10-8

year

Answer T2

= 42 (re + 40000)3/(G*Me )T = 1.44 hr


40/174

Example #3 What is the radius, and the center of the circle for theti i


41/174

81

III. Conic Sections

A conic section is curve obtained by intersecting a right circular conical

figure with a plane surface. The resulting curve is described by a specificform of a 2nd degree equation. The curve can be defined as all the loci of

points whose distances are in a fixed relation to a fixed point, (focus), and afixed line, (directrix). There are four, (4) types of conic sections.

1.) A circle is set of all points at a distance fixed by the line segment

called the radius, (r) and the focus called the center point, (h, k).

The circle has a standard equation.

(x - h )2

+ (y - k)2

= r2

82

equation given

(x -8 )2 + (y + 2)2 = 49

Answer (h. k) = (8,-2), and r = 7

Example #4 Given the center of a circle at (-3, 7), and the radius =6, find the equation of the circle.

Answer. (x + 3 )2 + (y - 7)2 = 36

Example# 5 graph the equation (x )2 + (y )2 = 9

answer

2.) The ellipse is the set of all points found by keeping the sum of thedistances from two foci a constant The midpoint of the line


42/174

83

distances from two foci a constant. The midpoint of the line

segment connecting the foci is called the center, (h, k) of the ellipse.

The ellipse has a standard equation which is always eaual to (1). Thevalues of the constants are (a > 0, b > 0) are distances from the

center point to the intercepts of the ellipse, (horizontal, vetical)

respectively.

((x h)/a )2 + ((y k)/b)2 = 1

F1 F2

(h, k) (h+a, k)(h-a, k)

(h, k+b)

(h, k-b)84

Example #7 Find the center point and intercepts of the ellipse

((x 5)2 /9 + (y 6)2 /4 = 1answer (h, k) = (5, 6) and a = 3, and b = 2

Example #7 Write in standard form an ellipse with center at the origin andintercept at (-7, 0), and intercept at (0,11).

Answer x 2 /49 + y2 /121 = 1

Example #8 Graph x 2 /49 + y2 /25 = 1

3.) The parabola is a conic section formed with a point, focus and a line,

directrix. The loci of all points equidistant from the focus to the dirtectrix

is the parabola.


43/174

85

The equation of a parabola comes in two forms. The point (h, k) is the vertexof the parabola, i.e. the closet point to the origin of the xy-graph. The axis of

symmetry for a parabola in given as the line;

-b/(2a)

Axis of symmetry value of (a) direction parabola opens

y for x2 > 0 vertically up

< 0 vertically down

x for y2 > 0 horizontally right

< 0 horizontally left

Example #9 For the parabola y = (x-2)2 + 0 find the vertex and the axis

of symmetry.

Answers: in std for y = x2

4y +4, so the axis of symmetry is the liney = -(-4)/(2*1) = 2

from the vertex form h = 2, k = 0 so the vertex is the point (2, 0).

86

Example #10. A satellite dish is a parabola of the form y = ax2. If a chosen

dish is of the form y = x2/32 and the distance, (p) from any dishs center to

the focal p0int is given as a = 1/(4p). Find the value of (P) for the chosen

satellite dish.

Answer p =8 (typically in inches)

Example #11. graph the parabola y = 4 x2.

4). The hyperbola is a conic section that has two pieces called brancheswhich are mirror images of each other and resemble parabolas. A hyperbola

is centered on a point, (h. k). The locus on each branch closest to the center

is a branch's vertex.The line going from one vertex, through the center to other vertex is called the transverseaxis. The foci of the hyperbola are located inside each branch.The hyperbola is the locus of all points whose difference in distances from the two foci isconstant.


44/174

87

There are two standard forms of the hyperbolic equation((x h)/a )

2- ((y k)/b)

2= 1 for horizontal, vertices = (h-a, k), (h+a, k)

((y k)/b)2 -((x h)/a )2 = 1 for vertical, vertices = (h, k-b), (h, k+b)

There are two asymptotes y = bx/a, and y =-bx/a.These asymptotes are the diagonals of the fundamental rectangle with

corners

(a, b), (a, -b), (-a, b ), (-a, -b) which is centered on (h, k).

88

Example #12. Given the hyperbola x2 /25 - y2 /16 = 1, find the equations for

the asymptotes, the center point, and the vertices of its fundamental

rectangle.Answer: y = +/- 4x/5

centered on (h, k) = (0, 0)

(5, 4), (5, -4), (-5, 4 ), (-5, -4)

Example #13. Given the hyperbola x2 /25 - y2 /16 = 1, find the vertex of the

hyperbola.Answer: vertices = (h-a, k), (h+a, k) = (-5, 0), (5, 0)

Example #14, Graph the hyperbola y2 /121 - x2 /169 = 1,

Answer


45/174

89 90

Practice questions

1. Find quadratic F(x) for roots x = 5, and x = -11

Answer f(x) = (x5)*(x+ 11)

2. Find the roots of the 7x2+ 4x + -3 = 0

Answer x = -1 and x = -3/7

3. Find the equation of a circle for r2

= 3 with center at (3/8, 0).

Answer(x - 3/8 )2

+ y2

= 3

4. Given x2

+ (y + 1. 7)2

= 25 find the center and radius.

Answer center = (0,-1.7), and r = 5

5. Plot (x + 1 )2 + (y - 1)2 = 9

answer

6 For what value of (z) does the following ellipse pass through the

i

9. For the parabola x = (y-3)2 5, find the vertex and the axis of

symmetry.2


46/174

91

point

(-4, 4)? ((x + 5)2

/4 + (y + z)2

/9 = 1answer z = - 1.41

7 What is the standard form of the ellipse with vertices;(2, 3), (6, 3), (4, 7), (4, -1).

answer ((x - 4)2

/4 + (y 3)2

/16 = 1

8. Graph x2

/25 + y2

/49 = 1

92

Answers: in standard for x = y2 -6y +25, so the axis of symmetry is the line

y = -(-6)/(2*1) = 3, and from the vertex form h = -5, k = 3 so vertex is at(-5, 3) .

10. A satellite dish is a parabola of the form y = ax2. A dish has the

distance, (p) from a dishs center to its focal point given as, a =

1/(4p).

Find the equation of a particular satellite with a value of p = 9.Answer y = x2 /36

11. Graph x = -y2

+2

12. Given the hyperbola y2 /121 - x2 /169 = 1, find the equations for the

asymptotes, the center point, and the vertices of its fundamental rectangle.

Answer: y = +/- 11x/13centered on (h, k) = (0, 0)

(13, 11), (-13, 11), (13, -11), (-13, -11)

13. Given the hyperbola y2 /121 - x2 /169 = 1, find the vertices the

hyperbola.

Answer: vertices = (h, k-b), (h, k+b) = (0, 11), (0, -11)

14. Graph the hyperbola x2

/9 - y2

/36 = 1

L-020 Momentum


47/174

93

answer

94

L 020 Momentum

Impulse

An impulse, (J) happens when a force, (F) occurs during

a time interval, (t).

A force, (F) of constant direction is zero for all times except for the time period from

initial time, (ti) until the final time, (tf) in duration, (t). See the figure above. The areaunder curve is an impulse, (J) = force per time,

J = (F) (t) = m*a*t.Where mass, (m) mass, and (a) is acceleration.

Example #1. An 80 kg box is accelerated to velocity of 5 m/sec in0.2 seconds. What is the impulse? What is the average force, (Favg)?

Answer: Impulse = 80 5 kg*m/sec *(sec/sec) = 400 N*sec

Favg = 400/0.2 N*sec

= 2000 N

F

t t

Linear momentum

A l t


48/174

95

Linear momentum

The linear momentum, (p) is the product of the object's mass times its

velocity. p = mv

Example #2 A 0.59 kg ball flies at 110 m/sec. What is the momentum of theball?

Answer: p = 0.59 * 110 kg*m/sec

= 65 N*sec

An impulse creates a change in momentum, (p).

J = mvfinal mvintial

The area under the force function, F(t) vs. time curve is the impulse value.

Consider the force functionF(t) = -2t2

+ 4x +1 on the time interval [0, 2]

then

J = -2/3(t3) + 2t

2+ t evaluated on [0, 2]

= -2/3(23) +2

2+ 2 - (-2/3(0

3) +2*0

2+0 ) = +16/3 + 8 +2 0

J = 4.67 N*sec

J = F(t)*dt

t

96

Angular momentum

A particle of mass, (m) traveling in uniform circular motion at a radius, (r)experiences a tangential velocity, (vt), and a tangential linear momentum,

(mvt). However the rotation induces a magnitude of angular momentum,(|L|).

|L| = |r*mvt |

Example # 3. The angular momentum is 4.6 N*m*sec for a 1 kg object inuniform circular motion of radius 0.5 m. What is the mass tangential

velocity?

Answer: vt = L/( r*m) = 4.6/(1*0.5) = 9.2 m/sec

N*m*sec = kg * m*m*sec/(sec2* m*kg) = m/sec

r

mvt

Conservation of linear momentum in 1 d and 2 d

The center of mass


49/174

97

Conservation of linear momentum in 1-d and 2-d

Momentum is conserved before and after a force acts upon a system. So the

vector sum ofmivi for all (i) is constant.

Example #4. Consider two balls each weighing (m =1.5 kg) then let the two

objects collide. Ball labeled A is initially traveling at 4 m/sec while ball

labeled B is at rest. Ball A strikes B and the two continue in the same

direction after with the identical velocity. What is their final velocity?

Answer: before a = mva = 1.5 * 4 kg*m/sec = 6 N*sec

b = mvb = 0

after a + b = 2*m* vf = 6 N*sec

vf = 6/(2*1.5) kg*m/(sec*kg) = 6/3 = 2 m/sec for both

Now let the two balls leave in different directions.

Example #5 A ball labeled A with mass = 2.0 kg is initially traveling at 4m/sec while ball labeled B with mass = 3.0 kg is at rest. Ball A strikes B and

Ball A leaves at an angle of 60o

to the left of its initial direction but ball B

moves in a direction 30o

to the right of the incidence. What is the finalmomenta of A, and B?

Answer before a = mva = 2 * 4 kg*m/sec

after a = 8 * sin 30o

= 8 * 0.5 = 4 N*sec

b = 8 * cos 30o = 8 * 0.866 = 6.9 N*sec

98

The center of mass

The center of mass, (cm) is a location that represents the average of several

numbers, (i) of masses, (ms) with respect to the distances, (xs) of each mass

from a reference point (usually zero). The concept is likened to a see-sawbalanced at a pivot point.

Above a mass less rod holds a 2m object at one end and a 3m object at theother. Any force acting upon the system can be analyzed as if the force acts

solely upon the center of mass. Here s = 2.

xcm = (2m*x1 + 3m*x2) / ( 5m) kg*m/kg = m

xcm = (2*x1 + 3*x2) / ( 5) since m is a constant

Consider the above illustration; Let x1 = 2 cm and x2 = 10 cm, then thecenter of mass is located;

xcm = (2*2 + 3*10) / ( 5) = 34/5 = 6.8 cm

Note there does not need to be any mass at the center of mass position.

2m cm 3m

x1

xcm

x2

There may be (3) masses so (s = 3) located in a plane (nonlinear positions).


50/174

99

There may be (3) masses so (s 3) located in a plane (nonlinear positions).

In this case the position vectors, (r) in can be used to find the

resultant position vector for the center of mass, (rcm).

rcm = (1/M) * ms*rs

In example let each object have the same mass = 1 kg so

rcm = (1/3) * 1*rs

and 1*r1 =

1* r2 = ,

1*r3 = , so

rcm = ( 1/3)* (1* + 1* + 1*)

rcm = ( 1/3)*

rcm = again there is no mass at this location

The object may be a continuous mass such as a length, (L cm) of solid rod with a variablemass per unit length, (x grams/cm). A piece of rod say at 2 cm would have a mass of 2cm*2 gram/cm = 4 grams. At a position x = 4 cm the mass is 4* 4 = 16 grams, and 25grams at position x = 5 cm, etc.

Note the mass is not in uniformly distributed along the points of theThe rod has no volume or area. It gets more mass along its length from x =o cm until x =

L cm. The rod is magical for reasons of simplifying the analysis by integral calculus.

xcm = (L3/3)/(L2/2) = 2L/3 cm

xcm x*x*dx x dx

100

L-021 Solid Geometry

I. Disk

The disk is a solid within a circle of radius, (r), The circumference of the

circle, C = 2**r, and area A = *r2 and arc length, L = d*r where d is therange of angle along the circumference.

We wish to look from point, (p) at a distance, (x) away from a disk of radius,(r). Let the disk contain some uniform surface property, ( per unit area).What is the disks accumulated effect, (V) at location (p)?

The answer is given by integral calculus.

Consider the inscribed circular strip of circumference 2**y, and width dy.This strip contains an area 2**y*dy. Hence the total surface of the strip has**2**y*dy where the constant of proportionality, () is introduced with

appropriate units. The value of d2

= x2

+y2

.

Px

r

yd

II. Toroid

A common form is


51/174

101

The strips contribution to the effect at (p) is given as dV.

dV = **2**y*dy/(x2 + y2)

The integration will go on the interval [0, r]. The answer is given.

V = ( **/2)*[(x2 + r2) - x]

Example #1 Find the value of (V) for a disk of radius (r = 88) with = 4 at apoint x =200 units away.

Answer V = 237 *

V **2**y*dy/(x2 + y2)

102

A torus, a type of torrid is a 3-dimensional ring, e.g. a donut. It is usually

made of a metal for scientific applications. It ha a cross sectional radius, (r)

and a radius, (R) of the ring.

The general formula

For a material of uniform density, () the total mass of the torus is given.

M = *2*2*r2*R

R

r

Area

Volume

Example #2. Find the total mass, (M) of a torus with minor radius, r = 0.5

One steradian is1 sr =r2, the maximum solid (conical) angle that can be

subtended by an area of 1 m2. There are 4*, or approximately 12.6

steradians in a complete sphere A solid angle of 1sr encompasses about


52/174

103

p ( )

cm, and major radius of R = 1.5 cm. The density = 2.47 g/cc.

Answer M = 18.3 g

III. Sphere

A sphere is a perfectly round object in 3-space. A sphere is symmetric with

respect to its center, with all points on the surface lying the same

distance from the center point.

Sphere of radius, (r) has a volume = (4/3) **r3, and a surface area = 4**r

2.

The standard form of the equation of a sphere is given with center point as

(h, k, s).

(x - h)2

+ (y - k)2

+ (z - s)2

= r2

The solid angle, () is a two-dimensional angle in 3-space. It is a measure of

how large an object appears. A small object nearby may subtend the same

solid angle as a larger object farther away. The moon and sun both have the

same size of appearance but are not the same distance from the observer on

Earth.

The steradian, (sr) is the dimensionless unit of a solid angle.

2

104

steradians in a complete sphere. A solid angle of 1sr encompasses about

1/12.6, or 7.9577 percent, of the space surrounding the center point of a

sphere.

To subtend is to be opposite to and extend from one side to the other, e.g. a

hypotenuse subtends a right angle.

Example #3. The view of the stars in the heavens above subtends a

maximum of 4 sr. If on a cloudy night only 45% of the sky is clear then

what solid angle results for star watching.

Answer 5.65 sr

The annulus is the region lying between two concentric circles with unequal

radii (r1 < r2). The area of the annulus is A = (r1- r2)2.

A spherical shell is a generalization of an annulus to three dimensions. A

spherical shell is therefore the region between two concentric spheres ofunequal radii.

r2

r1

Now consider an annulus of total mass, (M) with radius, (y). Consider the

The circumference of the ring is 2*y*sin(). Its arc length is y*d islinear for very small (d).


53/174

105

point, (P) a distance, (x) away.

A gravitational field, (Ffield) is Newtons Universal Law with the secondmass term removed. The force is not exerted until a second mass (M)

appears at position, (r) distant from the first mass.

Ffield = G*m/r2

There is a component of the field in the x-direction but

G*m/r2 *cos() = G*m/r2*(x/r)

and r2 = x2 + y2

therefore along the x-axis Ffield = G*m*x/r3.

Now let the ring have an arbitrary mass density, ( per unit area). Let the ring be at angle,() with an angular width, (d). The radius of the ring becomes y*sin().

py x

r

106

Thus the area of the ring is [ 2*y*sin()* y*d] which renders the mass as*[ 2*y*sin()* y*d].

therefore the field due to the ring along the x-axis is

Ffield = 2G*x**[*y2*sin()*d]/r

3.

The field due to the entire shell is found by integration over () from [0, ]which will yields

Ffield = 4G* **y2/s2

p

y

x

r

d

s

Example #4. What is the Ffield for the spherical shell if (s = y)

L 022 Bi i l C ffi i t


54/174

107

Answer Ffield = 4G* *

Problems in solid geometry

#1. For the value of (V) = 42 for a disk of radius (r = 8) with = 4, = 1. Find thedistance to point (p).

Answer x = 40.4 units

#2. If the total mass, M = 750g of a torus with minor radius, r = 0.5 cm, and major radiusof R = 1.5 cm. Find the density, ().

Answer = 101 g/cm3

#3.

answer 0.2 sr 4.4 sr

#4 For the spherical shell, if the Ffield = 16G* *then what is the relation between y and s.

Answer y/s = 2

108

L-022 Binomial Coefficients

I. Factorial

Take any integer (n >= 0), then the product of all positive integers less than

or equal to (n) is given as the factorial (n!).n! = 1*2*3*...(n-1)*n

e.g. 0! = 1

1! = 1

2! = 1*2 = 23! = 1*2*3 = 6

4! = 1*2*3*4 = 24

10! = 3628800

Example #1. Find 5!

Answer 120

Some factorial arithmetic:

11! + 12! = 11! + (12*11!) = (1 + 12)*11! = 13*11! = 13!/12

(7! + 8!)/(2!*3!*4!) =7!*( 1+8)/(2*2*3*4!) = (7!/4!)* 9/12

= ( 4!*5*6*7/4!)*3/4 = (4!/4!)*630/4

= 157.5

Example #2. Find (5! + 6!)/(3!*4!)

II. Binomial Coefficient

Example #4. Expand (x + y)2

where n =2 , k =0, 1, 2


55/174

109

The binomial coefficient is used to find the coefficients in the expansion of

a polynomial, and combinations of elements in a set.

Example #3. Find

Answer 5!/3!(5-3)! = 5!/ 2!*3! = (3!*4*5)/ 2*3! = 20/2 = 10

1.) polynomial expansion (the binomial theorem)

The algebraic expansion of the powers of (x + y)n

given by a binomialformula.

Cop

5

3

110

Answer * x0*y

(2-0)+ *x

1*y

(2-1)+ *x

2*y

(2-2)

= 2!/0!(2-0)!*x0*y

2+ 2!/1!(2-1)!*x*y + 2!/2!(2-2)!*x

2*y

0

= y2

+ 2*x*y + x2

2.) Permutation on a set of objects

Permuting refers to ordering in rearrangement of object. For example, there

are six permutations of the set {1,2,3} where the number of objects, n = 3.

(1, 2, 3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).

The number of permutations of (n) distinct objects is n-factorial (n!).

Example #5. Find the number of permutations for the set [a, b, c, d]

Answer since n = 4, there are 4! = 24 permutations

For illustration the 24 permutations for the set [a, b, c, d] are shown.

2

1

2

2

2

0

Problem set for the binomial coefficient


56/174

111

3.) Combinations

A combination is a way of selecting (k) objects out of asset of (n).

There are [ n!/(n-k)!] permutations of (k) objects taken out of a set of (n).

The number of combinations is the binomial coefficient.

n!/((n-k)!*k!)

Example #6. Find the number of combination of (5) cards drawn one at a

time from a deck of (52) but the same card cannot be drawn more than once.

Answer 2598960

112

#1. Find 6!

Answer 720

#2. Find (10! + 11!)/(2!*3!*4!)

Answer 20

Example #3. Find

Answer 7

Cop

#4. Expand (x + y)3 where n = 3

Answer x3

+ 3x2y + 3xy

2+ y

3

#5. Find the number of permutation s for the set [A, B, C, D, E].

Answer since n = 5, there are 5! = 120 permutations

#6. If 4 people are chosen randomly from a group of 10.

How many ways can this be done?Answer 210

7

6

L-023 Sequences / progressions / LimitsThe (a) is the initial value, and (d) is called the constant of commondifference.

Example #2 Find the arithmetic progression for the sequence


57/174

113

I. Sequence

A function, yn = f(n) on positive integers, n > = 0 is called a sequence.The values of the function are real numbers. The sequence is defined as

the set of all yn generated by f(n).

y0 = f(0),

y1 = f(1),

y2 = f(2), .

An example of a sequence is 0, 2, 4, 6, 8,.2*n,..

Example #1. Construct a sequence for the squares of odd integers.

Answer 1, 9, 25, 49,.

f(0) = 1 = 12

= (2*0+ 1)

2

f(1) = 9 = 32 = (2*1 + 1)2

f(2) = 25 = 52 = (2*2 + 1)2

f(3) = 49 = 72

= (2*3+ 1)

2

etc

f(n) = (2*n

+ 1)

2

II. Arithmetic progression

An arithmetic progression is a sequence in which each term is obtained

by adding a constant to the preceding term, e.g.,.

(a +i*d) = a, a + d, a + 2d, a + 3d, a + 4d, .

114

Example #2. Find the arithmetic progression for the sequence

13, 15, 17, 19.107, 109

answer ( a + i*d) = (13 + i*2) so 109 = 13 + (48) *2

The sum of an arithmetic progression (a + i*d) is

(a + i*d) = (n+1)*a +d*(n *(n+1))/2

Example #3. Given the arithmetic progression (13 + i*2), find the sum

of terms from 13 to 109.

Answer 13*(48 + 1) + 2*48*49/2 = 2989

III. Geometric Progression

A geometric progression is a sequence in which each term is obtained by

from its predecessor through multiplication by a constant, e.g.

a ar, ar2, ar3, ar4,

The (a) is the initial value, and (r) is called the common ratio.

Example #4 For geometric progression is given find (a , r)

3, 32, 33, 34, 35,.

Answer 3, 3*31, 3*3

2, 3*3

3, 3*3

4, ., so (a , r) = (3, 3)

n

i=0

The sum of a geometric progression is given Example #6 Does sequence 1, 2, 3, 4, 5, 6.n converge?Answer NO, since as n , f(n).


58/174

115

Example #5 Find the sum of the first n = 11 terms of the geometric

progression

2, 3/8, 9/128,

answer r = 3/8/2 = 3/16, (a, r ) = (2, 3/16)

sum = 2*( 1 (3/16)12

)/ (13/16) = 2.46

IV Limit of a sequence

A sequence of numbers, f(0), f1), f(3), f(4)..f(n) is said to have a limit if to

converges to a number (L) as the number (n) gets arbitrarily large.

Limit f(n) L as n

If a sequence does not have a limit then it is said to diverge.

116

Example #7. Does the sequence 1, , 1/3, ,..1/n converge?

Answer YES as n, 1/n 0

Example #8. Find the limit of the sequence with

f(n) = (n3 3n)/ (n4 +8)

answer multiply f(n)*( n4 / n4 ) = (1/n 3/n3)/ (1+8/ n4 )n, f() 0/1 = 0 = L

Consider the value x = 0.1428571 then we construct a sequence, a(n) from decimalapproximations.

N x - a(n)

0 0.1428571 = 0.0

1 0.0428571


59/174

115

Example #5 Find the sum of the first n = 11 terms of the geometric

progression

2, 3/8, 9/128,

answer r = 3/8/2 = 1/16, (a, r ) = (2, 1/16)

sum = 2*( 1 (1/16)12

)/ (15/16) = 2.13

IV Limit of a sequence

A sequence of numbers, f(0), f1), f(3), f(4)..f(n) is said to have a limit if to

converges to a number (L) as the number (n) gets arbitrarily large.

Limit f(n) L as n

If a sequence does not have a limit then it is said to diverge.

116

Example #7. Does the sequence 1, , 1/3, ,..1/n converge?

Answer YES as n, 1/n 0

Example #8. Find the limit of the sequence with

f(n) = (n3 3n)/ (n4 +8)

answer multiply f(n)*( n4 / n4 ) = (1/n 3/n3)/ (1+8/ n4 )n, f() 0/1 = 0 = L

Consider the value x = 0.1428571 then we construct a sequence, a(n) from decimalapproximations.

N a(n)

0 0.1428571 = 0.0

1 0.0428571


60/174

117

#1 Find the sequence for 0, 1, 8, 27, 64,..answer f(n) = n3

#2. Given the arithmetic progression (a + i*d) = (135 + I*3) find the sequence

answer 135, 138, 141, 144, .

#3. Given the arithmetic progression of 38 terms begins with 1 and ends

with 112. find progression and the sum of terms.

Answer (1 + i*3), 2147

#4 For geometric progression is given find (a , r)

210, 211, 212, 213,.

Answer (a , r) = (210

, 2)

#5 Find the sum of the first n = 11 terms of the geometric

progression 2, 8, 32, answer sum = 11.2 million

Example #6 Does sequence 12, 22, 32, 42, .n2 converge?

Answer NO

118

#8. Find the limit of the sequence with

f(n) = 1 + (-1)n /n

answer L = 1

L-024 Work Energy PowerWork can be done by the force function (Hooks law), F(x) = - k*x when a spring isstretched by a mass to an extreme point. The sign indicates the force is opposite to thedirection the spring is stretched. The value of (k) is the spring constant.

L


61/174

119

Energy is the ability to exert a force through a distance, i.e. create a physical change. Thesystems net energy changes. Work, (W) is the quantity of force, (F) acting through a

distance, (d).

W = F*d kg*m2/sec2 = Joule

Kinetic energy, (K) of a system changes due to work.

K = W = F*d kg*m2/sec2 = Joule

K = mv2

Thus a force must act in the same direction as the velocity in order to do work. So force

acting at an angle, (

) to an objects direction of motion does work by only its parallelcomponent.

W = F*d*cos ()

Example # 1. A box moves 4.0 m horizontally by a 6 N force acting 37

o

from the horizon.What work is done?

Answer W = 6*4*cos (37o) = N*m = kg*m*m/sec2 = Joule19.2 J

v

F

120

W = -k* x2/2 = -k*[L2/2 - 02/2] = -k *L2/2

Example # 2. Given a spring with an unknown, (k) = - 400 N/m that has been stretched 3cm. Find the value of (W).

Answer W = -kL2/2 = - 400*(0.03)2 /2 = (kg*m /sec2/m) *m = JouleW = - 0.18 J

The work done per unit time is power, (P)P = W/t = J/sec = watt

Example #3. A 1500 kg car accelerates from 0 to 25 m/s in 10 secs. What power isrequired?

Answer P = W/t = K/t = * m ( vf2 vi

2) = (1500/2)*(252 0

2)/10P = 4,690 watt = kg*(m/sec)

2/ sec = J/sec

W =

L

0

-k*x dx

Law of the conservation of energy applies to a closed system in which particles do notleave or enter. The system must also be isolated from outside forces. In such conditions,the total energy content of the system is constant.

The definition of total energy is called mechanical energy, (E). This is the sum of theki i ( ) d i l ( ) i f f l d i l d

L-025 Fluids

A force, (F) is applied perpendicularly to all surface areas, (A) surrounded

by a fluid. The result is pressure, (P).


62/174

121

kinetic, (K) and potential, (U) energies for any state of a closed, isolated system.

E = K + U

For a system involving gravity, U = Ug = m*g*h where (m) is mass, (g) is gravitationalacceleration, and (h) is vertical height.

Example # 4. A 10kg object falls to the ground. It has k = 400 J of kinetic energy. Fromwhat height was it dropped?

Answer K (max) = Ug (max) = m*g*h = 400 N

h = 400/(10* 9.8) = (kg*m*m/sec2)/(kg*m/sec2) = 4.08 m

122

y p , ( )

P = F / A = N/m2

= Pa (pascal)

The force is attributed to the kinetic motion of particles composing the fluid

when they collide with the surface.

Example #1. A 100 kg ball rests on a 1 cm2 spot. What is the pressure on

the spot?

Answer; 100 kg*10 (m/sec2)/ 0.01 m

2= 1*10

6Pa

The Earths atmosphere exerts 10 N on every cm2 of surface;

1 atmosphere = 10 N/cm2

= 100 kPa

Any fluid that has a change in pressure at a given point transmits the

pressure uniformly through out, Pascals Principal. Consider hydrostatic

pressure change with depth in the ocean.

P = *g*h

Where () is density, (g) is gravitational acceleration, and (h) is depth. Thepressure is uniform in all horizontal directions.

Example #2. Mercury has a density, ( = 13.6 g/cm3). What is the pressure at

100 cm below the surface?

Ans P = 13,600*10*0.100 = (kg/m3 )*(m/ sec2 )*m = kg*m/(m2 *sec2) = Pa

P = 13.6 kPa

This confirms a difference in vertical pressure for an immersed object. If thepressure above is less than the pressure below an object will float,

Archimedes Principle. The upward force, (Fb) is then proportional to an

objects volume, (V).

L-026 Thermodynamics


63/174

123

Fb = *g*V

Example #3. Aluminum block weighs 0.1 kg and displaces 4*105 m3 ofwater when submerged. What is the buoyant pressure on the block?

Answer; Fb = (0.1/4*105

) *10 * 4*105

= (kg/m3

)*(m

/sec2)*m

3= N

Fb = 1 N

Bernoullis Equation for the vertical flow of air that suspends a floating

sphere is given.

P1 +1/2 v12

+ gh1 = P2 +1/2 v22

+ gh2

Where (P) is pressure, = 1.2 kg/m3

is the density of air, (v) is velocity, g =

9.8 m/sec2, and (h) is height respectively in regions 1 and 2 that differ in

crossectional area.

Example #4. Let a small ball be held suspended by a pressure of 120 kPa at

the bottom side. Ifv1 = 14 m/s and v2 = 20 m/s, then find P2.

Answer Let gh1 gh2 so the terms cancel SoP1 +1/2 v1

2= P2 +1/2 v2

2or

P2 = v22*P1 /v1

2

P2 = 202 *120/14

2

P2 = 245 kPa

124

0th law refers to Temperature, oT

1st law refers to internal energy, E2nd law refers to entropy, S

3rd law refers to absolute, zero motion, 0 oK

Temperature is a measurement of the average kinetic energy of the

molecules in a system that may be measured. Temperature is the internal

energy contained within the system.

Heat, (Q) is a measure of the energy transferred from one system to another.

The greater the heat absorbed by a material, the more rapidly the atoms

within the material begins to move and thus the greater the rise intemperature.

Calorimetry is the method used to determine the heat released or absorbed in

a chemical reaction. The specific heat, (c) is the amount of heat per unitmass required to raise the temperature by one degree Celsius, (C). One (1)

kcal is the heat needed to rise one (I) kg of water one (1 ) C. The specificheat of water is c =1 calorie/(gram* C) = 4.186 joule/(gram* C).

The quantity of heat, (Q) lost or gained when an object changes temperature

is given.

Q = m*c*T

QTf

Ti

c dT

Example # 1. Some 100.0 g of water is required to be exactly 38.0oC. It is

presently 40.0oC. A piece of aluminum originally at a temperature (24.0oC)

is added to the bath. Of what mass should the Al chunk be? Specific heat of

Al = 0.215 cal/(g* oC) and specific heat of water = 1.00 cal/(g* oC).

Example #2. Copper has a thermal conductivity of 0.0092 kcal/(sec m C).If the crossectional area of a copper slab is 1 m2 what will be the heat

conducted for a temperature gardient of 10 C /m ?

Answer dQ/dt = - 0.0092*1 * 10 kcal/(sec m Co))*( m

2)* (C /m )


64/174

125

Answer. Qwater= QAl

|mw*cw*Tw | = |mAl*cAl*TAl | = g*cal/(g*oC) *(

oC )= calories

|100 *0.215*(38.0 40.0) | = |mAl*1.00*(38.0 24.0) |

mAl = 3.12 g

0. Zeroth Law of thermodynamics

The zeroth law of thermodynamics says that if two systems are in thermal

equilibrium with a third, then they are also in thermal equilibrium with each

other.

Let a piece of cold metal be placed into a warm liquid contained in a beaker.The liquid and beaker lose heat to the metal object. The closed system will

settle into equilibrium with the temperature becoming equal in all three. The

internal energy, (E) of the system is constant.

The transfer of heat between regions of different temperatures is called heat

conduction. Consider a slab of material of crossectional area, (A) andthickness, (dx) with difference in temperatures, (dT). The heat, (dQ) flows

perpendicularly between faces of the material.

dQ/dt = -k*A*dT/dx

Here (k) with units kcal/(sec*m*C) is the constant of thermal conductivity,and (dT/dx) is the temperature gradient. The larger the value of (k) the

better heat is condcuted in a material.

126

Q ( )) ( ) ( )

= -0.092 kcal/sec

Heat and work; 1 calorie = 4.186 joule.

Heat is a form of energy that is transferred between regions of different

temperatures. Work involves a transfer of energy between states of a system.

A system that changes equilibrium states by processes that involves Q, Ware called thermodynamic. Consider a piston of crossectional area, (A)

filled with an idea gas at pressure, (p). The base is a temparature, (T)

resevoir of high thermal conductivity. The walls are thermally insulating.

dW = F*dx = p*A*dx = p*dV

The area under the p_V graph is the work done by the piston compressing

the gas.

dx

A

P

T

F

WVf

Vi

p dV

I. 1st law of thermodynamics

The first law of th

Madonna H S Physics Tutorials

Documents

Transcript of Madonna H S Physics Tutorials