Chapter 1-7

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Chapter 1-7

MASTER CLASS NOTES FOR Chapter 1,2,3,4, & 7

MASTER NOTES



Chapter 1

Table of Contents

Chapter 1 The Science of PhysicsChapter 2 Motion in One DimensionChapter 3 Two- Dimensional Motion and VectorsChapter 4 Forces and the Laws of MotionChapter 7 Rotational Motion and the Law of Gravity

The Science of Physics



Section 1 What is PhysicsChapter 1

The Topics of Physics

• Physics is simply the study of the physical world.




The areas of Physics1. Mechanics - The study of motion and its causes.

– Falling objects, friction, weight, spinning objects.

2. Thermodynamics – The study of heat and temperature.– Melting and Freezing processes, engines,

refrigerators.

3. Vibration and Wave Phenomena – The study of specific types of repetitive motion.– Springs, pendulums, sound




The areas of Physics (cont)

4. Optics – The study of light.– Mirrors, lenses, color, astronomy

5. Electromagnetism – The study of electricity, magnetism, and light.– Electrical charge, circuitry, permanent magnets,

electromagnets.

6. Relativity – The study of particles moving at any speed, including very high speed.– Particle collisions, particle accelerators, nuclear energy.




The areas of Physics (cont.)

7. Quantum Mechanics – The study of submicroscopic particles.– The atom and its parts



Types of observations• Qualitative- descriptive, but not true measurements

– Hot– Large

• Quantitative- describe with numbers and units– 100C– 15 meters

Chapter 1




The Scientific Method

• The scientific method is a way to ask and answer scientific questions by making observations and doing experiments.

• Steps of the scientific : – Observation (Ask a Question)– Collect Data (Do Background Research)– Construct a Hypothesis (Educated guess)– Test Your Hypothesis by Doing Experiments – Analyze Your Data and Draw a Conclusion

• The conclusion is only valid if it can be verified by other people.

– Communicate Your Results



Chapter 1

The Scientific Method

Section 1 What is Physics?




The Scientific Method (cont)

• System – A set of items or interactions considered a distinct physical entity for the purpose of study.– Decide what to study and eliminate everything else that

has minimal or no effect on the problem.– Draw a diagram of what remains (Model)

• Models – A replica or description designed to show the structure or workings of an object, system, or concept. – Models help guide experimental design



Chapter 1

The System


http://en.wikipedia.org/wiki/File:System_boundary.svg



Chapter 1

The Scientific Model


http://en.wikipedia.org/wiki/File:Atmosphere_composition_diagram.jpg




The Scientific Method (cont)

• Hypothesis – A reasonable explanation for observations, one that can be tested with additional experiments.– The hypothesis must be tested in a controlled

experiment.• Controlled Experiment- Only one variable at

a time is changed to determine what influences the phenomenon you are observing.



Section 2 Measurements in ExperimentsChapter 1

Numbers As Measurements

• Numerical measurements in science contain the value (number) and Dimension.

• Dimension is the physical quantity being measured (length, mass, time, temperature, electric current)

• Each dimension is measured using units and prefixes from the SI system.

• The dimension must match the unit. (ex. If you are measuring length, use the meter(m), not the kilogram(kg)




• SI is the standard measurement system for science.• Used so that scientists can communicate with the

same language.• There are seven base units. They are:

– Meter(m) – length– kilogram(kg) – Mass– Second(s) – Time– Kelvin(K) – Temperature– Ampere(A) – current– Mole(mol) – amount of substance– Candela(cd) – luminous intensity




• Common Metric Prefixes:-See handout or visit reference section of website-Be able to convert between any prefix and another.



How good are the measurements?

• Scientists use two word to describe how good the measurements are:

• Accuracy- how close the measurement is to the actual value.

• Precision- how well can the measurement be repeated.



Differences

• Accuracy can be true of an individual measurement or the average of several.– Problems with accuracy are due to error

• Precision requires several measurements before anything can be said about it.– Precision describes the limitation of the measuring

instrument.



Percent Error

• Percent error = (Experimental Value – Accepted value) x 100

Accepted Value

• Percent error can be negative.



Let’s use a golf analogy



Accurate? NoPrecise? Yes



Accurate? YesPrecise? Yes



Precise? No

Accurate? Maybe?



Accurate? Yes

Precise? We cant say!



Significant FiguresScientific Notation

Accuracy and Precision



Significant figures (sig figs)

• How many numbers mean anything.• When we measure something, we can (and do)

always estimate between the smallest marks.

21 3 4 5



Significant figures (sig figs)

• The better marks the better we can estimate.• Scientist always understand that the last number

measured is actually an estimate.

21 3 4 5



Significant figures (sig figs)• The measurements we write down tell us about the

ruler we measure with• The last digit is between the lines• What is the smallest mark on the ruler that measures

142.13 cm?

141 142



Sig figs.

• How many sig figs in the following measurements?• 458 g• 4085 g• 4850 g• 0.0485 g• 0.004085 g• 40.004085 g

405.0 g4050 g0.450 g4050.05 g0.0500060 g



Rounding rules

• Look at the number behind the one you’re rounding.• If it is 0 to 4 don’t change it.• If it is 5 to 9 make it one bigger.• Round 45.462 to four sig figs.• to three sig figs.• to two sig figs.• to one sig figs.

45.4645.54550



Scientific notation

• All non-zero digits in scientific notation are significant figures.

• Any ending zero will be after the decimal point to be significant

• 1.20 x 103

• Sometimes you must write in scientific notation to use the correct sig figs.



Watch the Sig Figs

• When rounding, you don’t change the size of the number.

• You should end up with a number about the same size.

• Use place holders- they’re not significant.– Round 15253 to 3 sig figs– Round 0.028965 to 3 sig figs

153000.0290



Pacific Atlantic

Present Absent

If the decimal point is absent, start at the Atlantic (right), find the first non zero, and count all the rest of the digits

230000 1750



Pacific Atlantic

Present Absent

If the decimal point is PRESENT, start at the Pacific (left), find the first non zero, and count all the rest of the digits

0.045 1.2300



Using your calculatorwith scientific notation

• EE and EXP button stand for x 10 to the• 4.5 x 10-4 • push 4.5• push either EXP or EE• push 4 +/- or -4• see what your display says.



Practice these problems

(4.8 x 10 5 ) x (6.7 x 10-6)

(6.8 x 10 -6)

(3.2 x 10 4)• Remember when you multiply you add exponents

• 106 x 10-4

• When you divide you subtract exponents.



Adding and Subtracting

• You can’t add or subtract numbers until they are to the same power of ten.

• Your calculator does this automatically.• (4.8 x 10 5 ) + (6.7 x 106)

• (6.8 x 10 -6) - (3.2 x 10-5)• Remember- standard form starts with a number

between 1 and 10 to start.



Adding and subtracting with sig figs

• The last sig fig in a measurement is an estimate.• Your answer when you add or subtract can not be

better than your worst estimate. • have to round it to the least place of the

measurement in the problem.



For example

27.93 6.4+ First line up the decimal places27.936.4+

Then do the adding..

34.33Find the estimated numbers in the problem.

27.936.4

This answer must be rounded to the tenths place.



Practice

• 4.8 + 6.8765• 520 + 94.98• 0.0045 + 2.113• 500 -126• 6.0 x 103 - 3.8 x 102 • 6.0 x 10-2 - 3.8 x 10-3 • 5.33 x 1022 - 3.8 x 1021



Multiplication and Division

• Rule is simpler• Same number of sig figs in the answer as the least in

the question• 3.6 x 653• 2350.8• 3.6 has 2 s.f. 653 has 3 s.f.• answer can only have 2 s.f.• 2400



Multiplication and Division

• Same rules for division.• practice • 4.5 / 6.245• 4.5 x 6.245• 9.8764 x .043• 3.876 / 1980• 16547 / 710



The Metric System



Measuring• The numbers are only half of a measurement.• It is 10 long.• 10 what?• Numbers without units are meaningless.• How many feet in a yard?• A mile?• A rod?



The Metric System• Easier to use because it is a decimal system.• Every conversion is by some power of 10.• A metric unit has two parts.• A prefix and a base unit.• prefix tells you how many times to divide or multiply by

10.



Base Units• Length - meter - more than a yard - m• Mass - grams - about a raisin - g• Time - second - s• Temperature - Kelvin or ºCelsius K or ºC• Energy - Joules- J• Volume - Liter - half of a two liter bottle- L• Amount of substance - mole - mol



Prefixes• kilo k 1000 times• deci d 1/10• centi c 1/100• milli m 1/1000• micro μ 1/1000000• nano n 1/1000000000• kilometer - about 0.6 miles• centimeter - less than half an inch• millimeter - the width of a paper clip wire



Volume

• calculated by multiplying L x W x H • Liter the volume of a cube 1 dm (10 cm) on a side• 1L = 1 dm3

• so 1 L = 10 cm x 10 cm x 10 cm• 1 L = 1000 cm3 • 1/1000 L = 1 cm3 • 1 mL = 1 cm3



Volume

• 1 L about 1/4 of a gallon - a quart• 1 mL is about 20 drops of water or 1 sugar cube



Mass

• 1 gram is defined as the mass of 1 cm3 of water at 4 ºC.

• 1000 g = 1000 cm3 of water• 1 kg = 1 L of water



Mass

• 1 kg = 2.5 lbs• 1 g = 1 paper clip• 1 mg = 10 grains of salt



Converting

k h D d c m• how far you have to move on this chart, tells you how

far, and which direction to move the decimal place.• The box is the base unit, meters, Liters, grams, etc.



Conversions

• Change 5.6 m to millimeters

k h D d c m

starts at the base unit and move three to the right.move the decimal point three to the right

56 00



Conversions

• convert 25 mg to grams• convert 0.45 km to mm• convert 35 mL to liters• It works because the math works, we are dividing or

multiplying by 10 the correct number of times.

k h D d c m



What about micro- and nano-?

• The jump in between is 3 places• Convert 15000 μm to m• Convert 0.00035 cm to nm

k h D d c m μ n3 3



Measuring Temperature

• Celsius scale.• water freezes at 0ºC• water boils at 100ºC• body temperature 37ºC• room temperature 20 - 25ºC

0ºC



Measuring Temperature• Kelvin starts at absolute zero (-273 º C)• degrees are the same size• C = K -273• K = C + 273• Kelvin is always bigger.• Kelvin can never be negative.

273 K



Temperature is different

• from heat.• Temperature is which way heat will flow. (from hot to

cold)• Heat is energy, ability to do work.• A drop of boiling water hurts,• kilogram of boiling water kills.



Units of energy are

• calories or Joules• 1 calorie is the amount of heat needed to raise the

temperature of 1 gram of water by 1ºC.• A food Calorie is really a kilocalorie.• 1 calorie = 4.18 J



Conversion factors

• “A ratio of equivalent measurements.”• Start with two things that are the same.

1 m = 100 cm• Can divide by each side to come up with two ways of

writing the number 1.



Conversion factorsConversion factors

100 cm1 m =100 cm 100 cm




11 m =100 cm




11 m =100 cm

100 cm=1 m1 m 1 m




11 m =100 cm

100 cm=1 m

1



Conversion factors

• A unique way of writing the number 1.• In the same system they are defined quantities so

they have unlimited significant figures.• Equivalence statements always have this

relationship.• big # small unit = small # big unit• 1000 mm = 1 m



Write the conversion factors for the following

• kilograms to grams• feet to inches• 1.096 qt. = 1.00 L



What are they good for?What are they good for? We can multiply by one creatively to We can multiply by one creatively to

change the units .change the units . 13 inches is how many yards?13 inches is how many yards? 36 inches = 1 yard.36 inches = 1 yard. 1 yard = 11 yard = 1

36 inches 36 inches 13 inches x 1 yard 13 inches x 1 yard ==

36 inches 36 inches



Conversion factors

• Called conversion factors because they allow us to convert units.

• Really just multiplying by one, in a creative way.• Choose the conversion factor that gets rid of the unit

you don’t want.



Dimensional Analysis

• Dimension = unit• Analyze = solve• Using the units to solve the problems.• If the units of your answer are right, chances are you

did the math right.




• Using with metric units• Need to know equivalence statements• If it has a prefix, get rid of it with one conversion

factor• To add a prefix use a conversion factor



Practice• 25 mL is how many L?

• 5.8 x 10-6 mm is how many nm?



Dimensional Analysis• In the same system, unlimited sig figs• From one system to another. The conversion factor

has as many the most sig figs in the measurements.

• 1 inch is 2.54 cm• 3 sf

1 inch

2.54 cm




• A race is 10.0 km long. How far is this in miles? – 1 mile = 1760 yds– 1 meter = 1.094 yds




• Pikes peak is 14,110 ft above sea level. What is this in meters?– 1 mile = 1760 yds– 1 meter = 1.094 yds



Multiple units• The speed limit is 65 mi/hr. What is this in m/s?– 1 mile = 1760 yds– 1 meter = 1.094 yds

65 mihr

1760 yd1 mi 1.094 yd

1 m 1 hr60 min

1 min60 s



Multiple units

• Lead has a density of 11.4 g/mL. What is this in pounds per quart?– 454 g = 1 lb– 1 L = 1.06 qt



Units to a Power

• How many m3 is 1500 cm3?

1500 cm33 1 m100 cm

1 m100 cm

1 m100 cm

1500 cm33 1 m

100 cm

33



Units to a Power

• How many cm2 is 15 m2?• 36 cm3 is how many mm3?



• A European cheese making recipe calls for 2.50 kg of whole milk. An American wishes to make the recipe has only measuring cups, which are marked in cups. If the density of milk is 1.03 g/cm3 how many cups of milk does he need?

1 qt = 2 pints1 yd = 3 ft.

1 mile = 1.61 km1 m = 1.094 yds1 L = 1000 cm3

1 gal = 4 qt1 L = 1.06 qt1 lb = 454 g

1 mi =1760 yds1 pint = 2 cups



• A barrel of petroleum holds 42.0 gal. Empty it weighs 75 lbs. When it is filled with ethanol it weighs 373 lbs. What is the density of ethanol in g/cm3?

1 qt = 2 pints1 yd = 3 ft.

1 mile = 1.61 km1 m = 1.094 yds1 L = 1000 cm3

1 gal = 4 qt1 L = 1.06 qt1 lb = 454 g

1 mi =1760 yds1 pint = 2 cups



Chapter 2

Table of Contents

Section 1 Displacement and Velocity

Section 2 Acceleration

Section 3 Falling Objects

Motion in One Dimension



Section 1 Displacement and VelocityChapter 2

Objectives• Describe motion in terms of frame of reference,

displacement, time, and velocity.

• Calculate the displacement of an object traveling at a known velocity for a specific time interval.

• Construct and interpret graphs of position versus time.




One Dimensional Motion

• To simplify the concept of motion, we will first consider motion that takes place in one direction.

• One example is the motion of a commuter train on a straight track.

• To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.




Displacement

x = xf – xi displacement = final position – initial position

• Displacement is a change in position.• Displacement is not always equal to the distance

traveled.• The SI unit of displacement is the meter, m.



Chapter 2

Positive and Negative Displacements

Section 1 Displacement and Velocity




Average Velocity• Average velocity is the total displacement

divided by the time interval during which the displacement occurred.

f iavg

f i

x xxvt t t

average velocity = change in position

change in time =

displacementtime interval

• In SI, the unit of velocity is meters per second, abbreviated as m/s.




Velocity and Speed

• Velocity describes motion with both a direction and a numerical value (a magnitude).

• Speed has no direction, only magnitude.

• Average speed is equal to the total distance traveled divided by the time interval.

distance traveledaverage speed = time of travel




Interpreting Velocity Graphically

– Object 1: positive slope = positive velocity

– Object 2: zero slope= zero velocity – Object 3: negative slope = negative

velocity

• For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph.

• If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity.




Interpreting Velocity Graphically, continued

The instantaneous

velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path.



Section 2 AccelerationChapter 2

Objectives

• Describe motion in terms of changing velocity.

• Compare graphical representations of accelerated and nonaccelerated motions.

• Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration.



Chapter 2

Changes in Velocity• Acceleration is the rate at which velocity changes

over time.


f iavg

f i

v vvat t t

change in velocityaverage acceleration =

time required for change

• An object accelerates if its speed, direction, or both change.

• Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.



Chapter 2

Changes in Velocity, continued

• Consider a train moving to the right, so that the displacement and the velocity are positive.

• The slope of the velocity-time graph is the average acceleration.


– When the velocity in the positive direction is increasing, the acceleration is positive, as at A.

– When the velocity is constant, there is no acceleration, as at B.

– When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.



Chapter 2

Velocity and Acceleration




Chapter 2

Motion with Constant Acceleration• When velocity changes by the same amount during each time interval, acceleration is

constant. • The relationships between displacement, time, velocity, and constant acceleration are

expressed by the equations shown on the next slide. These equations apply to any object moving with constant acceleration.

• These equations use the following symbols:x = displacementvi = initial velocityvf = final velocityt = time interval




Chapter 2

Equations for Constantly Accelerated Straight-Line Motion




Sample Problem

Final Velocity After Any DisplacementA person pushing a stroller starts from rest, uniformlyaccelerating at a rate of 0.500 m/s2. What is thevelocity of the stroller after it has traveled 4.75 m?

Section 2 AccelerationChapter 2



Sample Problem, continued


1. DefineGiven:

vi = 0 m/s a = 0.500 m/s2

x = 4.75 mUnknown:

vf = ?Diagram: Choose a coordinate system. The most convenient one

has an origin at the initial location of the stroller, as shown above. The positive direction is to the right.

Chapter 2



Chapter 2



2. PlanChoose an equation or situation: Because the initial

velocity, acceleration, and displacement are known, the final velocity can be found using the following equation:

2 2 2f iv v a x

2 2f iv v a x

Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf .



Chapter 2



Tip: Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity. So, the final velocity must be positive.

3. CalculateSubstitute the values into the equation and solve:

4. EvaluateThe stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

2 2(0 m/s) 2(0.500 m/s )(4.75 m)fv

2.18 m/sfv



Section 3 Falling ObjectsChapter 2

Objectives

• Relate the motion of a freely falling body to motion with constant acceleration.

• Calculate displacement, velocity, and time at various points in the motion of a freely falling object.

• Compare the motions of different objects in free fall.



Chapter 2

Free Fall

• Free fall is the motion of a body when only the force due to gravity is acting on the body.

• The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration.

• Free-fall acceleration is denoted with the symbols ag (generally) or g (on Earth’s surface).




Chapter 2

Free-Fall Acceleration• Free-fall acceleration is the same for all objects,

regardless of mass.• This book will use the value g = 9.81 m/s2.• Free-fall acceleration on Earth’s surface is –9.81 m/s2

at all points in the object’s motion. • Consider a ball thrown up into the air.

– Moving upward: velocity is decreasing, acceleration is –9.81 m/s2

– Top of path: velocity is zero, acceleration is –9.81 m/s2

– Moving downward: velocity is increasing, acceleration is –9.81 m/s2




Sample Problem

Falling ObjectJason hits a volleyball so that it moves with an initialvelocity of 6.0 m/s straight upward. If the volleyballstarts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?

Section 3 Falling ObjectsChapter 2





1. DefineGiven: Unknown: vi = +6.0 m/s t = ?

a = –g = –9.81 m/s2 y = –2.0 m

Diagram: Place the origin at the Starting point of the ball (yi = 0 at ti = 0).

Chapter 2



Chapter 2

Sample Problem, continued2. Plan Choose an equation or situation:

Both ∆t and vf are unknown. Therefore, first solve for vf using the equation that does not require time. Then, the equation for vf that does involve time can be used to solve for ∆t.


2 2 2f iv v a y f iv v a t

2 2f iv v a y f iv vt

a

Rearrange the equation to isolate the unknown: Take the square root of the first equation to isolate vf. The second equation must be rearranged to solve for ∆t.



Chapter 2


Tip: When you take the square root to find vf , select the negative answer because the ball will be moving toward the floor, in the negative direction.


2 2 22 (6.0 m/s) 2(–9.81 m/s )(–2.0 m)f iv v a y

2 2 2 2 2 236 m /s 39 m /s 75 m /s –8.7 m/sfv

3. Calculate Substitute the values into the equation and solve: First find the velocity of the ball at the moment that it hits the floor.



Chapter 2


4. EvaluateThe solution, 1.50 s, is a reasonable amount of time for the ball to be in the air.


2 2

–8.7 m/s 6.0 m/s –14.7 m/s–9.81 m/s –9.81 m/s

f iv vt

a

1.50 st

Next, use this value of vf in the second equation to solve for ∆t.



Multiple Choice

Standardized Test PrepChapter 2

Use the graphs to answer questions 1–3.

1. Which graph represents an object moving with a constant positive velocity?

A. I C. III B. II D. IV



Multiple Choice



1. Which graph represents an object moving with a constant positive velocity?




Multiple Choice, continued



2. Which graph represents an object at rest?

F. I H. III G. II J. IV






2. Which graph represents an object at rest?

F. I H. III G. II J. IV






3. Which graph represents an object moving with a constant positive acceleration?







3. Which graph represents an object moving with a constant positive acceleration?






4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of 73 km/h to the south.What is the bus’s displacement?

F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/h to the south





4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of 73 km/h to the south.What is the bus’s displacement?

F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/h to the south





5. What is the squirrel’s displacement at time t = 3.0 s?

A. –6.0 mB. –2.0 m C. +0.8 m D. +2.0 m

Use the position-time graph of a squirrelrunning along a clothesline to answer questions 5–6.





5. What is the squirrel’s displacement at time t = 3.0 s?

A. –6.0 mB. –2.0 m C. +0.8 m D. +2.0 m







6. What is the squirrel’s average velocity during the time interval between 0.0 s and 3.0 s?

F. –2.0 m/sG. –0.67 m/s H. 0.0 m/sJ. +0.53 m/s





6. What is the squirrel’s average velocity during the time interval between 0.0 s and 3.0 s?

F. –2.0 m/sG. –0.67 m/s H. 0.0 m/sJ. +0.53 m/s






7. Which of the following statements is true of acceleration?

A. Acceleration always has the same sign as displacement.

B. Acceleration always has the same sign as velocity.C. The sign of acceleration depends on both

the direction of motion and how the velocity is changing.D. Acceleration always has a positive sign.

•





7. Which of the following statements is true of acceleration?

A. Acceleration always has the same sign as displacement.

B. Acceleration always has the same sign as velocity.C. The sign of acceleration depends on both

the direction of motion and how the velocity is changing.D. Acceleration always has a positive sign.

•





8. A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move during this time?

F. 12 m G. 93 mH. 120 m J. 190 m





8. A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move during this time?

F. 12 m G. 93 mH. 120 m J. 190 m





9. Which of the following statements is true for a ball thrown vertically upward?

A. The ball has a negative acceleration on the way up and a positive acceleration on the way

down.B. The ball has a positive acceleration on the way

up and a negative acceleration on the way down.C. The ball has zero acceleration on the way up and

a positive acceleration on the way down.D. The ball has a constant acceleration throughout

its flight.





9. Which of the following statements is true for a ball thrown vertically upward?

A. The ball has a negative acceleration on the way up and a positive acceleration on the way

down.B. The ball has a positive acceleration on the way

up and a negative acceleration on the way down.C. The ball has zero acceleration on the way up and

a positive acceleration on the way down.D. The ball has a constant acceleration throughout

its flight.



Short Response


10. In one or two sentences, explain the difference between displacement and distance traveled.



Short Response


10. In one or two sentences, explain the difference between displacement and distance traveled. Answer: Displacement measures only the net change in position from starting point to end point. The distance traveled is the total length of

the path followed from starting point to end point and may be greater than or equal to the displacement.



Short Response, continued


11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity:

a. for the time interval from t = 0.0 min to t = 10.0 min

b. for the time interval from t = 10.0 min to t = 20.0 min

c. for the time interval from t = 20.0 min to t = 30.0 min

d. for the entire run





11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity. Answers will vary but should be approximately as follows:

a. for t = 0.0 min to t = 10.0 minAnswer: +2400 m, +4.0 m/s

b. for t = 10.0 min to t = 20.0 minAnswer: +1500 m, +2.5 m/s

c. for t = 20.0 min to t = 30.0 minAnswer: +900 m, +2 m/s

d. for the entire runAnswer: +4800 m, +2.7 m/s





12. For an object moving with constant negative acceleration, draw the following:

a. a graph of position vs. timeb. a graph of velocity vs. time

For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin.





12. For an object moving with constant negative acceleration, draw the following:

a. a graph of position vs. timeb. a graph of velocity vs. time

For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin.Answers:





13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s.

a. If the snowmobile accelerates at a rate of +0.50 m/s2 for 7.0 s, what is its final velocity?b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop?





13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s.

a. If the snowmobile accelerates at a rate of +0.50 m/s2 for 7.0 s, what is its final velocity?b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop?

Answers: a. +6.5 m/s b. 5.0 s



Extended Response


14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s.

a. What is the car’s average acceleration?b. What is the car’s average velocity?c. How far did the car move while accelerating?

Show all of your work for these calculations.



Extended Response


14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s.

a. What is the car’s average acceleration?b. What is the car’s average velocity?c. How far did the car move while accelerating?

Answers: a. 1.6 m/s2 eastward b. 24 m/s c. 240 m



Extended Response, continued


15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m.

a. How long does it take the ball to reach its highest point?

b. How long is the ball in the air?






15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m.

a. How long does it take the ball to reach its highest point?b. How long is the ball in the air?


Answers: a. 2.55 s b. 5.18 s



Chapter 3 Two-Dimensional Motion and Vectors



Two-Dimensional Motion and VectorsChapter 3

Table of Contents

Section 1 Introduction to Vectors

Section 2 Vector Operations

Section 3 Projectile Motion

Section 4 Relative Motion



Section 1 Introduction to VectorsChapter 3

Objectives

• Distinguish between a scalar and a vector.

• Add and subtract vectors by using the graphical method.

• Multiply and divide vectors by scalars.



Chapter 3

Scalars and Vectors• A scalar is a physical quantity that has magnitude

but no direction.– Examples: speed, volume, the number of pages

in your textbook

• A vector is a physical quantity that has both magnitude and direction.– Examples: displacement, velocity, acceleration

• In this book, scalar quantities are in italics. Vectors are represented by boldface symbols.




Chapter 3

Graphical Addition of Vectors

• A resultant vector represents the sum of two or more vectors.

• Vectors can be added graphically.


A student walks from his house to his friend’s house (a), then from his friend’s house to the school (b). The student’s resultant displacement (c) can be found by using a ruler and a protractor.



Chapter 3

Triangle Method of Addition

• Vectors can be moved parallel to themselves in a diagram.

• Thus, you can draw one vector with its tail starting at the tip of the other as long as the size and direction of each vector do not change.

• The resultant vector can then be drawn from the tail of the first vector to the tip of the last vector.




Chapter 3

Properties of Vectors

• Vectors can be added in any order.

• To subtract a vector, add its opposite.

• Multiplying or dividing vectors by scalars results in vectors.




Section 2 Vector OperationsChapter 3

Objectives• Identify appropriate coordinate systems for solving

problems with vectors.

• Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector.

• Resolve vectors into components using the sine and cosine functions.

• Add vectors that are not perpendicular.



Chapter 3

Coordinate Systems in Two Dimensions• One method for diagraming

the motion of an object employs vectors and the use of the x- and y-axes.

• Axes are often designated using fixed directions.

• In the figure shown here, the positive y-axis points north and the positive x-axis points east.




Chapter 3Determining Resultant Magnitude and Direction• In Section 1, the magnitude and direction of a

resultant were found graphically.

• With this approach, the accuracy of the answer depends on how carefully the diagram is drawn and measured.

• A simpler method uses the Pythagorean theorem and the tangent function.




Chapter 3Determining Resultant Magnitude and Direction, continued

The Pythagorean Theorem

• Use the Pythagorean theorem to find the magnitude of the resultant vector.

• The Pythagorean theorem states that for any right triangle, the square of the hypotenuse—the side opposite the right angle—equals the sum of the squares of the other two sides, or legs.


c2 a2 b2

(hypotenuse)2 (leg 1)2 (leg 2)2



Chapter 3Determining Resultant Magnitude and Direction, continued

The Tangent Function

• Use the tangent function to find the direction of the resultant vector.

• For any right triangle, the tangent of an angle is defined as the ratio of the opposite and adjacent legs with respect to a specified acute angle of a right triangle.


tangent of angle = opposite legadjacent leg



Chapter 3

Sample ProblemFinding Resultant Magnitude and Direction An archaeologist climbs the Great Pyramid in

Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30 102 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?




Chapter 3



1. DefineGiven: y = 136 mx = 1/2(width) = 115 mUnknown:

d = ? = ?Diagram: Choose the archaeologist’s starting

position as the origin of the coordinate system, as shown above.



Chapter 3



Rearrange the equations to isolate the unknowns:2 2 –1 tan

yd x yx

d 2 x2 y2 tan yx

2. Plan Choose an equation or situation: The

Pythagorean theorem can be used to find the magnitude of the archaeologist’s displacement. The direction of the displacement can be found by using the inverse tangent function.



Chapter 3



2 2 –1

2 2 –1

tan

136 m(115 m) (136 m) tan 115

178 m 49.8

yd x yx

d

d

3. Calculate

4. Evaluate Because d is the hypotenuse, the archaeologist’s

displacement should be less than the sum of the height and half of the width. The angle is expected to be more than 45 because the height is greater than half of the width.



Chapter 3

Resolving Vectors into Components• You can often describe an object’s motion more conveniently by

breaking a single vector into two components, or resolving the vector.

• The components of a vector are the projections of the vector along the axes of a coordinate system.

• Resolving a vector allows you to analyze the motion in each direction.




Chapter 3

Resolving Vectors into Components, continued

Consider an airplane flying at 95 km/h.• The hypotenuse (vplane) is the resultant vector

that describes the airplane’s total velocity.

• The adjacent leg represents the x component (vx), which describes the airplane’s horizontal speed.

• The opposite leg represents the y component (vy), which describes theairplane’s vertical speed.




Chapter 3

Resolving Vectors into Components, continued

• The sine and cosine functions can be used to find the components of a vector.

• The sine and cosine functions are defined in terms of the lengths of the sides of right triangles.


sine of angle = opposite leghypotenuse

cosine of angle = adjacent leghypotenuse



Chapter 3

Adding Vectors That Are Not Perpendicular


• Suppose that a plane travels first 5 km at an angle of 35°, then climbs at 10° for 22 km, as shown below. How can you find the total displacement?

• Because the original displacement vectors do not form a right triangle, you can not directly apply the tangent function or the Pythagorean theorem.

d2

d1



Chapter 3Adding Vectors That Are Not Perpendicular, continued


• You can find the magnitude and the direction of the resultant by resolving each of the plane’s displacement vectors into its x and y components.

• Then the components along each axis can be added together.

As shown in the figure, these sums will be the two perpendicular components of the resultant, d. The resultant’s magnitude can then be found by using the Pythagorean theorem, and its direction can be found by using the inverse tangent function.



Chapter 3

Sample ProblemAdding Vectors Algebraically A hiker walks 27.0 km from her base camp at 35°

south of east. The next day, she walks 41.0 km in a direction 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement




Chapter 3

1 . Select a coordinate system. Then sketch and label each vector.


Given: d1 = 27.0 km 1 = –35°

d2 = 41.0 km 2 = 65°

Tip: 1 is negative, because clockwise movement from the positive x-axis is negative by convention.

Unknown: d = ? = ?




Chapter 3

2 . Find the x and y components of all vectors.


For day 1 :x1 d1 cos1 (27.0 km)(cos –35 ) = 22 kmy1 d1 sin1 (27.0 km)(sin –35 ) = –15 km

Make a separate sketch of the displacements for each day. Use the cosine and sine functions to find the components.

For day 2 :x2 d2 cos2 (41.0 km)(cos 65 ) = 17 kmy2 d2 sin2 (41.0 km)(sin 65 ) = 37 km




Chapter 3

3 . Find the x and y components of the total displacement.


xtot x1 x2 22 km + 17 km = 39 kmytot y1 y2 –15 km + 37 km = 22 km

4 . Use the Pythagorean theorem to find the magnitude of the resultant vector.d 2 (xtot )

2 (ytot )2

d (xtot )2 (ytot )

2 (39 km)2 (22 km)2

d 45 km




Chapter 3

5 . Use a suitable trigonometric function to find the angle.


–1 –1 22 kmtan = tan 39 km

29 north of east

yx




Section 3 Projectile MotionChapter 3

Objectives

• Recognize examples of projectile motion.

• Describe the path of a projectile as a parabola.

• Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion.



Chapter 3

Projectiles• Objects that are thrown or launched into the air and are subject to

gravity are called projectiles.

• Projectile motion is the curved path that an object follows when thrown, launched,or otherwise projected near the surface of Earth.

• If air resistance is disregarded, projectiles follow parabolic trajectories.




Chapter 3

Projectiles, continued

• Projectile motion is free fall with an initial horizontal velocity.

• The yellow ball is given an initial horizontal velocity and the red ball is dropped. Both balls fall at the same rate.

– In this book, the horizontal velocity of a projectile will be considered constant.

– This would not be the case if we accounted for air resistance.




Chapter 3

Kinematic Equations for Projectiles• How can you know the displacement, velocity, and acceleration of

a projectile at any point in time during its flight?

• One method is to resolve vectors into components, then apply the simpler one-dimensional forms of the equations for each component.

• Finally, you can recombine the components to determine the resultant.




Chapter 3

Kinematic Equations for Projectiles, continued

• To solve projectile problems, apply the kinematic equations in the horizontal and vertical directions.

• In the vertical direction, the acceleration ay will equal –g (–9.81 m/s2) because the only vertical component of acceleration is free-fall acceleration.

• In the horizontal direction, the acceleration is zero, so the velocity is constant.




Chapter 3

Kinematic Equations for Projectiles, continued

• Projectiles Launched Horizontally– The initial vertical velocity is 0.– The initial horizontal velocity is the initial velocity.

• Projectiles Launched At An Angle


– Resolve the initial velocity into x and y components.

– The initial vertical velocity is the y component.

– The initial horizontal velocity is the x component.



Chapter 3

Sample ProblemProjectiles Launched At An Angle A zookeeper finds an escaped monkey hanging from a

light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole,which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s,will the dart hit the monkey, the banana, or neither one?




Chapter 3

1 . Select a coordinate system. The positive y-axis points up, and the positive x-

axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m.





Chapter 3

2 . Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis.

1 1 4.00 mtan tan 21.810.0 m

yx





Chapter 3

3 . Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-

axis to isolate the unknown t, which is the time the dart takes to travel the horizontal distance.

x (vi cos )t

t x

vi cos

10.0 m(50.0 m/s)(cos 21.8 )

0.215 s





Chapter 3

4 . Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases.

For the banana, vi = 0. Thus:yb = ½ay(t)2 = ½(–9.81 m/s2)(0.215 s)2 = –

0.227 m

The dart has an initial vertical component of velocity equal to vi sin , so:

yd = (vi sin )(t) + ½ay(t)2 yd = (50.0 m/s)(sin )(0.215 s) +½(–9.81 m/s2)(0.215 s)2

yd = 3.99 m – 0.227 m = 3.76 m





Chapter 3

5 . Analyze the results. Find the final height of both the banana and the dart.

ybanana, f = yb,i+ yb = 5.00 m + (–0.227 m)

ybanana, f = 4.77 m above the ground

The dart hits the banana. The slight difference is due to rounding.


ydart, f = yd,i+ yd = 1.00 m + 3.76 m

ydart, f = 4.76 m above the ground




Section 4 Relative MotionChapter 3

Objectives

• Describe situations in terms of frame of reference.

• Solve problems involving relative velocity.



Chapter 3

Frames of Reference• If you are moving at 80 km/h north and a car

passes you going 90 km/h, to you the faster car seems to be moving north at 10 km/h.

• Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north.

• This simple example demonstrates that velocity measurements depend on the frame of reference of the observer.




Chapter 3

Frames of Reference, continuedConsider a stunt dummy dropped from a plane.(a) When viewed from the plane, the stunt dummy falls straight

down. (b) When viewed from a stationary position on the ground, the

stunt dummy follows a parabolic projectile path.




Chapter 3

Relative Velocity• When solving relative velocity problems, write down the

information in the form of velocities with subscripts.

• Using our earlier example, we have:• vse = +80 km/h north (se = slower car with respect to

Earth)• vfe = +90 km/h north (fe = fast car with respect to Earth)• unknown = vfs (fs = fast car with respect to slower car)

• Write an equation for vfs in terms of the other velocities. The subscripts start with f and end with s. The other subscripts start with the letter that ended the preceding velocity: • vfs = vfe + ves




Chapter 3

Relative Velocity, continued• An observer in the slow car perceives Earth as moving south

at a velocity of 80 km/h while a stationary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/h. In equation form:• ves = –vse

• Thus, this problem can be solved as follows:• vfs = vfe + ves = vfe – vse

• vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n

• A general form of the relative velocity equation is:• vac = vab + vbc




Multiple Choice

1. Vector A has a magnitude of 30 units. Vector B is perpendicular to vector A and has a magnitude of 40 units. What would the magnitude of the resultant vector A + B be?

A. 10 unitsB. 50 unitsC. 70 unitsD. zero




Multiple Choice

1. Vector A has a magnitude of 30 units. Vector B is perpendicular to vector A and has a magnitude of 40 units. What would the magnitude of the resultant vector A + B be?

A. 10 unitsB. 50 unitsC. 70 unitsD. zero





2. What term represents the magnitude of a velocity vector?

F. accelerationG. momentumH. speedJ. velocity





2. What term represents the magnitude of a velocity vector?

F. accelerationG. momentumH. speedJ. velocity





Use the diagram to answer questions 3–4.

3. What is the direction of the resultant vector A + B?

A. 15º above the x-axisB. 75º above the x-axisC. 15º below the x-axisD. 75º below the x-axis






3. What is the direction of the resultant vector A + B?

A. 15º above the x-axisB. 75º above the x-axisC. 15º below the x-axisD. 75º below the x-axis






4. What is the direction of the resultant vector A – B?

F. 15º above the x-axisG. 75º above the x-axisH. 15º below the x-axisJ. 75º below the x-axis






4. What is the direction of the resultant vector A – B?

F. 15º above the x-axisG. 75º above the x-axisH. 15º below the x-axisJ. 75º below the x-axis





Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river

that flows toward the south at a speed of 5.0 m/s.

5. What is the resultant velocity relative to an observer on the shore ?

A. 3.2 m/s to the southeastB. 5.0 m/s to the southeastC. 7.1 m/s to the southeastD. 10.0 m/s to the southeast







5. What is the resultant velocity relative to an observer on the shore ?

A. 3.2 m/s to the southeastB. 5.0 m/s to the southeastC. 7.1 m/s to the southeastD. 10.0 m/s to the southeast







6. If the river is 125 m wide, how long does the boat take to cross the river?

F. 39 sG. 25 sH. 17 sJ. 12 s







6. If the river is 125 m wide, how long does the boat take to cross the river?

F. 39 sG. 25 sH. 17 sJ. 12 s





7. The pilot of a plane measures an air velocity of 165 km/h south relative to the plane. An observer on the ground sees the plane pass overhead at a velocity of 145 km/h toward the north.What is the velocity of the wind that is affecting the plane relative to the observer?

A. 20 km/h to the northB. 20 km/h to the southC. 165 km/h to the northD. 310 km/h to the south





7. The pilot of a plane measures an air velocity of 165 km/h south relative to the plane. An observer on the ground sees the plane pass overhead at a velocity of 145 km/h toward the north.What is the velocity of the wind that is affecting the plane relative to the observer?

A. 20 km/h to the northB. 20 km/h to the southC. 165 km/h to the northD. 310 km/h to the south





8. A golfer takes two putts to sink his ball in the hole once he is on the green. The first putt displaces the ball 6.00 m east, and the second putt displaces the ball 5.40 m south. What displacement would put the ball in the hole in one putt?

F. 11.40 m southeastG. 8.07 m at 48.0º south of eastH. 3.32 m at 42.0º south of eastJ. 8.07 m at 42.0º south of east





8. A golfer takes two putts to sink his ball in the hole once he is on the green. The first putt displaces the ball 6.00 m east, and the second putt displaces the ball 5.40 m south. What displacement would put the ball in the hole in one putt?

F. 11.40 m southeastG. 8.07 m at 48.0º south of eastH. 3.32 m at 42.0º south of eastJ. 8.07 m at 42.0º south of east





Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball

horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m.

9. What is the initial speed of the girl’s ball relative to the boy?A. 1.0 m/s C. 2.0 m/sB. 1.5 m/s D. 3.0 m/s







9. What is the initial speed of the girl’s ball relative to the boy?A. 1.0 m/s C. 2.0 m/sB. 1.5 m/s D. 3.0 m/s







10. If air resistance is disregarded, which ball will hit the ground first?F. the boy’s ball H. neitherG. the girl’s ball J. cannot be determined







10. If air resistance is disregarded, which ball will hit the ground first?F. the boy’s ball H. neitherG. the girl’s ball J. cannot be determined







11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground?A. the boy’s ball C. neitherB. the girl’s ball D. cannot be determined







11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground?A. the boy’s ball C. neitherB. the girl’s ball D. cannot be determined







12. What is the speed of the girl’s ball when it hits the ground?

F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s







12. What is the speed of the girl’s ball when it hits the ground?

F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s




Short Response


13. If one of the components of one vector along the direction of another vector is zero, what can

you conclude about these two vectors?



Short Response


13. If one of the components of one vector along the direction of another vector is zero, what can

you conclude about these two vectors?

Answer: They are perpendicular.





14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move

horizontally and vertically?





14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move

horizontally and vertically?

Answer: 31.5 m horizontally, 26.4 m vertically





15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second

ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically?





15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second

ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically?

Answer: 29.4 m/s



Extended Response


16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work.



Extended Response


16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work.

Answer: 10.8 m





Read the following passage to answer question 17.Three airline executives are discussing ideas for developing

flights that are more energy efficient.

Executive A: Because the Earth rotates from west to east, we could operate “static flights”—a helicopter or airship could begin by rising straight up from New York City and then descend straight down four hours later when San Francisco arrives below.

Executive B: This approach could work for one-way flights, but the return trip would take 20 hours. continued on the next slide





Executive C: That approach will never work. Think about it.When you throw a ball straight up in the air, it comes straight back down to the same point.

Executive A: The ball returns to the same point because Earth’s motion is not significant during such a short time.

17. State which of the executives is correct, and explain why.





17. State which of the executives is correct, and explain why.

Answer: Executive C is correct. Explanations should include the concept of relative velocity—when a helicopter lifts off straight up from the ground, it is already moving horizontally with Earth’s horizontal velocity. (We assume that Earth’s motion is constant for the purposes of this scenario and does not depend on time.)



Chapter 3

Graphical Addition of Vectors




Chapter 3

Adding Vectors That Are Not Perpendicular




Chapter 3

Projectiles




Chapter 3

Frames of Reference




Chapter 7 Circular Motion and Gravitation



Circular Motion and GravitationChapter 7

Table of Contents

Section 1 Circular Motion

Section 2 Newton’s Law of Universal Gravitation

Section 3 Motion in Space

Section 4 Torque and Simple Machines



Chapter 7

Objectives

• Solve problems involving centripetal acceleration.

• Solve problems involving centripetal force.

• Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the centripetal force.




Chapter 7

Tangential Speed

• The tangential speed (vt) of an object in circular motion is the object’s speed along an imaginary line drawn tangent to the circular path.

• Tangential speed depends on the distance from the object to the center of the circular path.

• When the tangential speed is constant, the motion is described as uniform circular motion.




Chapter 7

Centripetal Acceleration


http://www.youtube.com/watch?v=fSfVVz0eIis

http://www.youtube.com/watch?v=fSfVVz0eIis



Chapter 7


• The acceleration of an object moving in a circular path and at constant speed is due to a change in direction.

• An acceleration of this nature is called a centripetal acceleration.

CENTRIPETAL ACCELERATION

ac vt

2

r

centripetal acceleration = (tangential speed)2

radius of circular path




Chapter 7

Centripetal Acceleration, continued

• (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes.

• (b) For short time intervals, ∆v is directed toward the center of the circle.

• Centripetal acceleration is always directed toward the center of a circle.




Chapter 7

Centripetal Acceleration, continued

• You have seen that centripetal acceleration results from a change in direction.

• In circular motion, an acceleration due to a change in speed is called tangential acceleration.

• To understand the difference between centripetal and tangential acceleration, consider a car traveling in a circular track. – Because the car is moving in a circle, the car has a centripetal

component of acceleration. – If the car’s speed changes, the car also has a tangential

component of acceleration.




Chapter 7

Centripetal Force• Consider a ball of mass m that is being whirled in a

horizontal circular path of radius r with constant speed.

• The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force.

• This net force, which is is directed toward the center of the circle, is a centripetal force.




Chapter 7

Centripetal Force, continued

Newton’s second law can be combined with the equation for centripetal acceleration to derive an equation for centripetal force:

ac vt

2

r

Fc mac mvt

2

r

centripetal force =

mass (tangential speed)2

radius of circular path




Chapter 7


• Centripetal force is simply the name given to the net force on an object in uniform circular motion.

• Any type of force or combination of forces can provide this net force.

– For example, friction between a race car’s tires and a circular track is a centripetal force that keeps the car in a circular path.

– As another example, gravitational force is a centripetal force that keeps the moon in its orbit.




Chapter 7


• If the centripetal force vanishes, the object stops moving in a circular path.

• A ball that is on the end of a string is whirled in a vertical circular path.– If the string breaks at the position

shown in (a), the ball will move vertically upward in free fall.

– If the string breaks at the top of the ball’s path, as in (b), the ball will

move along a parabolic path.




Chapter 7

Describing a Rotating System

• To better understand the motion of a rotating system, consider a car traveling at high speed and approaching an exit ramp that curves to the left.

• As the driver makes the sharp left turn, the passenger slides to the right and hits the door.

• What causes the passenger to move toward the door?




Chapter 7

Describing a Rotating System, continued


• As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight path.

• If a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat.

• If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath.



Section 2 Newton’s Law of Universal GravitationChapter 7

Objectives

• Explain how Newton’s law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides.

• Apply Newton’s law of universal gravitation to solve problems.



Chapter 7

Gravitational Force

• Orbiting objects are in free fall.• To see how this idea is true, we can use a thought

experiment that Newton developed. Consider a cannon sitting on a high mountaintop.


Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.



Chapter 7

Gravitational Force, continued

• The centripetal force that holds the planets in orbit is the same force that pulls an apple toward the ground—gravitational force.

• Gravitational force is the mutual force of attraction between particles of matter.

• Gravitational force depends on the masses and on the distance between them.




Chapter 7

Gravitational Force, continued

• Newton developed the following equation to describe quantitatively the magnitude of the gravitational force if distance r separates masses m1 and m2:


1 22

2

mass 1 mass 2 gravitational force constant(distance between masses)

gmmF Gr

Newton's Law of Universal Gravitation

• The constant G, called the constant of universal gravitation, equals 6.673 10–11 N•m2/kg.



Chapter 7

Newton’s Law of Universal Gravitation


http://www.youtube.com/watch?v=Y50HeIUS4tk

http://www.youtube.com/watch?v=Y50HeIUS4tk



Chapter 7

Gravitational Force, continued• The gravitational forces that two masses exert on each other are always equal in magnitude and opposite in

direction. • This is an example of Newton’s third law of motion.• One example is the Earth-moon system, shown on the next slide. • As a result of these forces, the moon and Earth each orbit the center of mass of the Earth-moon system. Because

Earth has a much greater mass than the moon, this center of mass lies within Earth.




Chapter 7

Newton’s Law of Universal Gravitation




Chapter 7

Applying the Law of Gravitation• Newton’s law of gravitation accounts for ocean tides.

• High and low tides are partly due to the gravitational force exerted on Earth by its moon.

• The tides result from the difference between the gravitational force at Earth’s surface and at Earth’s center.




Chapter 7

Applying the Law of Gravitation, continued• Cavendish applied Newton’s law of universal gravitation to find the value of G and Earth’s mass.

• When two masses, the distance between them, and the gravitational force are known, Newton’s law of universal gravitation can be used to find G.

• Once the value of G is known, the law can be used again to find Earth’s mass.




Chapter 7

• Gravity is a field force.• Gravitational field strength,

g, equals Fg/m.• The gravitational field, g,

is a vector with magnitude g that points in the direction of Fg.

• Gravitational field strength equals free-fall acceleration.


The gravitational field vectors represent Earth’s gravitational field at each point.

Applying the Law of Gravitation, continued



Chapter 7

Applying the Law of Gravitation, continued

• weight = mass gravitational field strength• Because it depends on gravitational field

strength, weight changes with location:


weight = mg

g Fgm

GmmEmr2

GmEr2

• On the surface of any planet, the value of g, as well as your weight, will depend on the planet’s mass and radius.



Section 3 Motion in SpaceChapter 7

Objectives

• Describe Kepler’s laws of planetary motion.

• Relate Newton’s mathematical analysis of gravitational force to the elliptical planetary orbits proposed by Kepler.

• Solve problems involving orbital speed and period.



Chapter 7

Kepler’s Laws

Kepler’s laws describe the motion of the planets.• First Law: Each planet travels in an elliptical orbit

around the sun, and the sun is at one of the focal points.• Second Law: An imaginary line drawn from the sun to

any planet sweeps out equal areas in equal time intervals.

• Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun.




Chapter 7

Kepler’s Laws, continued

• Kepler’s laws were developed a generation before Newton’s law of universal gravitation.

• Newton demonstrated that Kepler’s laws are consistent with the law of universal gravitation.

• The fact that Kepler’s laws closely matched observations gave additional support for Newton’s theory of gravitation.




Chapter 7


According to Kepler’s second law, if the time a planet takes to travel the arc on the left (∆t1) is equal to the time the planet takes to cover the arc on the right (∆t2), then the area A1 is equal to the area A2.

Thus, the planettravels faster when itis closer to the sunand slower when it is farther away.




Chapter 7


• Kepler’s third law states that T2 r3.

• The constant of proportionality is 42/Gm, where m is the mass of the object being orbited.

• So, Kepler’s third law can also be stated as follows:

22 34T r

Gm




Chapter 7


• Kepler’s third law leads to an equation for the period of an object in a circular orbit. The speed of an object in a circular orbit depends on the same factors:

T 2r3

Gm vt G

mr

• Note that m is the mass of the central object that is being orbited. The mass of the planet or satellite that is in orbit does not affect its speed or period.

• The mean radius (r) is the distance between the centers of the two bodies.




Chapter 7

Planetary Data




Chapter 7

Sample ProblemPeriod and Speed of an Orbiting Object Magellan was the first planetary spacecraft to be launched from a

space shuttle. During the spacecraft’s fifth orbit around Venus, Magellan traveled at a mean altitude of 361km. If the orbit had been circular, what would Magellan’s period and speed have been?




Chapter 7


1. DefineGiven:r1 = 361 km = 3.61 105 mUnknown:T = ? vt = ?

2. PlanChoose an equation or situation: Use the equations for

the period and speed of an object in a circular orbit.

T 2r 3

Gm vt

Gmr




Chapter 7


Use Table 1 in the textbook to find the values for the radius (r2) and mass (m) of Venus.

r2 = 6.05 106 m m = 4.87 1024 kg

Find r by adding the distance between the spacecraft and Venus’s surface (r1) to Venus’s radius (r2).

r = r1 + r2

r = 3.61 105 m + 6.05 106 m = 6.41 106 m




Chapter 7


3. Calculate

4. EvaluateMagellan takes (5.66 103 s)(1 min/60 s) 94 min to

complete one orbit.

T 2r3

Gm =2

(6.41106 m)3

(6.67310–11N•m2 /kg2 )(4.87 1024 kg)

T 5.66 103 s

vt Gmr

(6.67310–11N•m2 /kg2 )(4.87 1024 kg)

6.41106 mvt 7.12 103 m/s




Chapter 7

Weight and WeightlessnessTo learn about apparent weightlessness, imagine that

you are in an elevator:– When the elevator is at rest, the magnitude of the

normal force acting on you equals your weight. – If the elevator were to accelerate downward at 9.81

m/s2, you and the elevator would both be in free fall. You have the same weight, but there is no normal force acting on you.

– This situation is called apparent weightlessness.– Astronauts in orbit experience apparent

weightlessness.




Chapter 7

Weight and Weightlessness




Section 4 Torque and Simple MachinesChapter 7

Objectives

• Distinguish between torque and force.

• Calculate the magnitude of a torque on an object.

• Identify the six types of simple machines.

• Calculate the mechanical advantage of a simple machine.




Rotational Motion

• Rotational and translational motion can be analyzed separately.– For example, when a bowling ball strikes the pins, the pins

may spin in the air as they fly backward.– These pins have both rotational and translational motion.

• In this section, we will isolate rotational motion.

• In particular, we will explore how to measure the ability of a force to rotate an object.




The Magnitude of a Torque

• Torque is a quantity that measures the ability of a force to rotate an object around some axis.

• How easily an object rotates on both how much force is applied and on where the force is applied.

• The perpendicular distance from the axis of rotation to a line drawn along the direction of the force is equal to d sin and is called the lever arm.

= Fd sin torque = force lever arm




The Magnitude of a Torque, continued

• The applied force may act at an angle.

• However, the direction of the lever arm (d sin ) is always perpendicular to the direction of the applied force, as shown here.



Chapter 7

Torque




Chapter 7

Torque and the Lever Arm


In each example, the cat is pushing on the door at the same distance from the axis. To produce the same torque, the cat must apply greater force for smaller angles.




The Sign of a Torque

• Torque is a vector quantity. In this textbook, we will assign each torque a positive or negative sign, depending on the direction the force tends to rotate an object.

• We will use the convention that the sign of the torque is positive if the rotation is counterclockwise and negative if the rotation is clockwise.

Tip: To determine the sign of a torque, imagine that the torque is the only one acting on the object and that the object is free to rotate. Visualize the direction that the object would rotate. If more than one force is acting, treat each force separately.



Chapter 7

The Sign of a Torque




Chapter 7

Sample ProblemTorque

A basketball is being pushed by two players during tip-off. One player exerts an upward force of 15 N at a perpendicular distance of 14 cm from the axis of rotation.The second player applies a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. Find the net torque acting on the ball about its center of mass.




Chapter 7



1. DefineGiven:F1 = 15 N F2 = 11 Nd1 = 0.14 m d2 = 0.070 m

Diagram:

Unknown:net = ?



Chapter 7



2. PlanChoose an equation or situation: Apply the definition of torque to each force,and add up the individual torques.

Tip: The factor sin is not included in the torque equation because each given distance is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. In other words, each given distance is the lever arm.

= Fdnet = 1 + 2 = F1d1 + F2d2



Chapter 7



4. Evaluate The net torque is negative,so the ball rotates in a

clockwise direction.

3. Calculate Substitute the values into the equation and

solve: First,determine the torque produced by each force.Use the standard convention for signs.

1 = F1d1 = (15 N)(–0.14 m) = –2.1 N•m2 = F2d2 = (–11 N)(0.070 m) = –0.77 N•mnet = 1 + 2 = –2.1 N•m – 0.77 N•m net = –2.9 N•m




Simple Machines• A machine is any device that transmits or modifies force,

usually by changing the force applied to an object.

• All machines are combinations or modifications of six fundamental types of machines, called simple machines.

• These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.



Chapter 7

Simple Machines





Simple Machines, continued

• Because the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force.

• This ratio is called mechanical advantage.• If friction is disregarded, mechanical advantage

can also be expressed in terms of input and output distance.

MA FoutFin

dindout





The diagrams show two examples of a trunk being loaded onto a truck.

• In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work.

• In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work.





• The simple machines we have considered so far are ideal, frictionless machines.

• Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat.

• The efficiency of a machine is the ratio of useful work output to work input.

eff Wout

Win

– The efficiency of an ideal (frictionless) machine is 1, or 100 percent.

– The efficiency of real machines is always less than 1.



Chapter 7

Mechanical Efficiency




Multiple Choice

1. An object moves in a circle at a constant speed. Which of the following is not true of the object?

A. Its acceleration is constant.B. Its tangential speed is constant.C. Its velocity is constant.D. A centripetal force acts on the object.




Multiple Choice

1. An object moves in a circle at a constant speed. Which of the following is not true of the object?

A. Its acceleration is constant.B. Its tangential speed is constant.C. Its velocity is constant.D. A centripetal force acts on the object.





Use the passage below to answer questions 2–3.A car traveling at 15 m/s on a flat surface turns in a

circle with a radius of 25 m.

2. What is the centripetal acceleration of the car?

F. 2.4 10-2 m/s2

G. 0.60 m/s2

H. 9.0 m/s2 J. zero







2. What is the centripetal acceleration of the car?

F. 2.4 10-2 m/s2

G. 0.60 m/s2

H. 9.0 m/s2 J. zero







3. What is the most direct cause of the car’s centripetal acceleration?

A. the torque on the steering wheelB. the torque on the tires of the carC. the force of friction between the tires and the roadD. the normal force between the tires and the road







3. What is the most direct cause of the car’s centripetal acceleration?

A. the torque on the steering wheelB. the torque on the tires of the carC. the force of friction between the tires and the roadD. the normal force between the tires and the road





4. Earth (m = 5.97 1024 kg) orbits the sun (m = 1.99 1030 kg) at a mean distance of 1.50 1011 m. What is the gravitational force of the sun on Earth? (G = 6.673 10-11 N•m2/kg2)

F. 5.29 1032 N G. 3.52 1022 NH. 5.90 10–2 N J. 1.77 10–8 N





4. Earth (m = 5.97 1024 kg) orbits the sun (m = 1.99 1030 kg) at a mean distance of 1.50 1011 m. What is the gravitational force of the sun on Earth? (G = 6.673 10-11 N•m2/kg2)

F. 5.29 1032 N G. 3.52 1022 NH. 5.90 10–2 N J. 1.77 10–8 N





5. Which of the following is a correct interpretation of the expression ?

A. Gravitational field strength changes with an object’s distance from Earth.

B. Free-fall acceleration changes with an object’s distance from Earth.

C. Free-fall acceleration is independent of the falling object’s mass.

D. All of the above are correct interpretations.


ag g GmEr2




5. Which of the following is a correct interpretation of the expression ?

A. Gravitational field strength changes with an object’s distance from Earth.

B. Free-fall acceleration changes with an object’s distance from Earth.

C. Free-fall acceleration is independent of the falling object’s mass.

D. All of the above are correct interpretations.


ag g GmEr2




6. What data do you need to calculate the orbital speed of a satellite?

F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbitH. mass of satellite and radius of orbit onlyJ. mass of planet and radius of orbit only





6. What data do you need to calculate the orbital speed of a satellite?

F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbitH. mass of satellite and radius of orbit onlyJ. mass of planet and radius of orbit only





7. Which of the following choices correctly describes the orbital relationship between Earth and the sun?

A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle.C. The sun orbits Earth in an ellipse, with Earth

at one focus.D. Earth orbits the sun in an ellipse, with the sun

at one focus.





7. Which of the following choices correctly describes the orbital relationship between Earth and the sun?

A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle.C. The sun orbits Earth in an ellipse, with Earth

at one focus.D. Earth orbits the sun in an ellipse, with the sun

at one focus.







8. The three forces acting onthe wheel have equal magnitudes. Which force willproduce the greatest torque on the wheel?

F. F1

G. F2

H. F3

J. Each force will produce the same torque.






8. The three forces acting onthe wheel have equal magnitudes. Which force willproduce the greatest torque on the wheel?

F. F1

G. F2

H. F3

J. Each force will produce the same torque.






9. If each force is 6.0 N, the angle between F1 and F2 is 60.0°, and the radius of the wheel is 1.0 m, what is theresultant torque on the wheel?

A. –18 N•m C. 9.0 N•mB. –9.0 N•m D. 18 N•m






9. If each force is 6.0 N, the angle between F1 and F2 is 60.0°, and the radius of the wheel is 1.0 m, what is theresultant torque on the wheel?

A. –18 N•m C. 9.0 N•mB. –9.0 N•m D. 18 N•m




10. A force of 75 N is applied to a lever. This force lifts a load weighing 225 N. What is the mechanical

advantage of the lever?

F. 1/3G. 3H. 150J. 300





10. A force of 75 N is applied to a lever. This force lifts a load weighing 225 N. What is the mechanical

advantage of the lever?

F. 1/3G. 3H. 150J. 300





11. A pulley system has an efficiency of 87.5 percent. How much work must you do to lift a desk

weighing 1320 N to a height of 1.50 m?

A. 1510 JB. 1730 JC. 1980 JD. 2260 J





11. A pulley system has an efficiency of 87.5 percent. How much work must you do to lift a desk

weighing 1320 N to a height of 1.50 m?

A. 1510 JB. 1730 JC. 1980 JD. 2260 J





12. Which of the following statements is correct?

F. Mass and weight both vary with location.G. Mass varies with location, but weight does not.H. Weight varies with location, but mass does

not.J. Neither mass nor weight varies with location.





12. Which of the following statements is correct?

F. Mass and weight both vary with location.G. Mass varies with location, but weight does not.H. Weight varies with location, but mass does

not.J. Neither mass nor weight varies with location.





13. Which astronomer discovered that planets travel in elliptical rather than circular orbits?

A. Johannes KeplerB. Nicolaus CopernicusC. Tycho BraheD. Claudius Ptolemy





13. Which astronomer discovered that planets travel in elliptical rather than circular orbits?

A. Johannes KeplerB. Nicolaus CopernicusC. Tycho BraheD. Claudius Ptolemy




Short Response

14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as

shown below.




Short Response

14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as

shown below.


Answer: The water remains in the pail even when the pail is upside down because the water tends to move in a straight path due to inertia.




15. Explain why approximately two high tides take place every day at a given location on Earth.





15. Explain why approximately two high tides take place every day at a given location on Earth.

Answer: The moon’s tidal forces create two bulges on Earth. As Earth rotates on its axis once per day, any given point on Earth passes through both bulges.





16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an

example.





16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an

example.

Answer: You would have to apply the input force over a greater distance. Examples may include any machines that increase output force at the expense of input distance.




Extended Response


17. Mars orbits the sun (m = 1.99 1030 kg) at a mean distance of 2.28 1011 m. Calculate the length of

the Martian year in Earth days. Show all of your work. (G = 6.673 10–11 N•m2/kg2)



Extended Response

17. Mars orbits the sun (m = 1.99 1030 kg) at a mean distance of 2.28 1011 m. Calculate the length of

the Martian year in Earth days. Show all of your work. (G = 6.673 10–11 N•m2/kg2)


Answer: 687 days



Chapter 7





Chapter 7

Centripetal Force




Chapter 7

Kepler’s Laws





The Magnitude of a Torque




Simple Machines

Documents

Chapter 1-7