аЯрЁБс>ўџ Y[ўџџџXџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџьЅСM јП?bjbjт=т= "bWW>7џџџџџџl МЄ\ \ \ 8 А ЄјЖи ю "
L\
\
;;;GIIIIII$Ў ЮNmE;7;;;mдд\
\
лВ;Ід\
\
G;GОGZ6G\
Ь №ЄШczХЄИ\ сGGШ0јGuGЄЄддддйTopic #7(. Motion in Two Dimensions
1. Equilibrium of Forces
2. Inclined Planes
3. Projectiles and Projectile Motion
4. Projectiles dropped or thrown sideways (Cliff Problems)
5. Projectile launched upwards at an angle to the horizontal
6. Uniform Circular Motion
7. Placing a Satellite in Orbit
8. Simple Harmonic Motion
9. Torque
Notes should include:
Equilibrium of Forces: An object experiences a balance of forces when the net force on the object is zero (Newtons in the Metric system; Pounds in the British system). When the forces are balanced the object is said to be in a state of equilibrium. Consider a situation where a commercial sign is hanging between two vertical supports such as two posts. The sign is suspended between two posts and hangs by two cables attached to opposite ends of the sign. The forces acting on the sign are the tension forces in the cables and the force of gravity acting downwards. The two cables are hung at angles measured from the horizontal direction of the sign. Together the two cables tension forces add up to a resultant force (vector) pointing upwards, which cancels the gravitational force (vector) pointing downwards. If the combined force of the two cable forces can be thought of as a resultant then the force of gravity can be thought of as an equilibrant. So long as the resultant force and the gravitational force cancels each other out the sign remains in place. The sign is said to be in a state of equilibrium.
Inclined Planes: Often an objects motion is not limited to only horizontal and/or vertical scenarios. Now when an object is on an inclined plane or ramp, certain forces are always involved. Gravitational force is of course present and pulls directly downwards on the object that is positioned on the ramp. Normal force is present as well, but on a ramp it does not equal the magnitude of the force of gravity. The definition of the normal force is The contact force exerted by a surface on an object perpendicular and away from the surface. This definition tells us that on a ramp the normal force is perpendicular to the surface the object is on, but the surface is no longer parallel to the horizontal, because the surface is an inclined plane. On an inclined plane the normal force is always less than the gravitational forces magnitude. In fact as the angle of inclination (the amount of tilt) increases the value of the normal force decreases. Another force to consider in the study of the inclined plane is the parallel force. This is the force acting on the object parallel to the surface of the inclined plane. In some scenarios the force of friction will be a factor to consider, while in others it will not play a role in the solution to a problem.
Projectiles and Projectile Motion: Projectile motion is the motion of an object that is projected or thrown into the air. Unless launched straight up without a cross wind, the projectile follows a curved path. This kind of motion involves two-dimensional motion because there is both an x and a y component to the motion. A projectile is anything that you can launch into the air. Unless it too is under study, the means by which you launch the projectile is unimportant. The path followed by the projectile is called its trajectory, and the shape of the path is that of a parabola. As always, when discussing the motion of an object, a frame of reference is important. For example, how does the motion of a ball being thrown up and down by a boy sitting on a (moving bus appear differently to some observers on a sidewalk as the bus drives by the observer compared to his friend sitting across the aisle on the same bus? In simple terms his friend would see only vertical motion with respect to the floor of the bus. The observers outside the bus would see a parabolic curve being followed by the ball with respect to the ground. It is always necessary to be clear about the frame of reference in solving a problem. It affects your perspective. Also the question of how is time affected by the motion?. If both the friend on the bus and an observer standing on the sidewalk had stopwatches, would there be a difference in the time each observed for the ball to rise up and fall back to the thrower's hand?
Projectiles dropped or thrown sideways (Cliff Problems): Consider time in a different situation. Suppose two identical balls are resting in a device sticking off the edge of a high cliff. Assuming there is no significant air friction because both are relatively heavy objects. Would both balls reach the ground at the same time when one was simply allowed to fall at exactly the same moment one was thrown exactly sideways? Think, don't rush to a hasty answer. If you consider what we said about vectors acting at right angles being independent to one another, you will predict that the vertical motion of the two balls is independent of any sideways motion. Since the time to fall to the ground is only affected by height, would not the two balls reach the ground at the same time, with one landing at the base of the cliff while the other would land some distance away from the base of the cliff? Think about it. That second ball is given a sideways (horizontal) velocity is not relevant to the time for it to reach the ground. Both balls would reach the ground at the same time, because neither ball was given a vertical velocity. The equation that we encountered earlier in our studies said that vertical distance was found by dy = vyi t + 0.5 g t2. If you know the height of the cliff, its initial velocity and the gravitational acceleration, you can calculate the time for an object to fall to the ground. In such a case of free fall the initial velocity would be 0 m/s. If an object such as the second ball is given a horizontal velocity, the horizontal distance it travels while falling can be found from the equation dx = vx t once the time of the fall is known.
Projectile launched upwards at an angle to the horizontal: When an object is launched at an angle with respect to the ground (the horizontal), at some initial launch velocity, the object follows a trajectory that is truly parabolic in shape. The initial launch velocity has an x component and a y component respectively. Now is a good time to review the process of resolving vector quantities into x and y components. Once, you identify or calculate the vertical and horizontal velocity components, use the vertical velocity component and the equation vf = vi + gt to find the time of the flight of the projectile. At this point you will have to have decided whether downwards motion is negative or positive, because it affect the vector values of acceleration, velocity, and displacement. Once you have determined the time for the object to travel through the air, you can focus on how high did it travel (maximum height), and how far did it travel from the launch point. This horizontal distance from the launch point is called the range. In the military launching projectile down range on a firing range is common terminology.
Uniform Circular Motion: Have you ever made some object go round and round as you held onto some tethering device like a piece of string, etc. The object was probably experiencing circular motion. For an object to experience circular motion there has to be an unbalanced force pulling the object towards the center of its circular path. This force produced by the string in our example is called a centripetal force, and it produces a centripetal acceleration towards the center of the circular path causing the object's velocity's direction to change continually even though the magnitude of the object's velocity, its speed, remains constant. This means an object following a circular path is doing so, because a force (centripetal force) causes it to accelerate (centripetal (acceleration) towards the center of the path at right angles to its forward constant speed. If the centripetal force were suddenly removed, the centripetal acceleration would stop and the object would follow a straight-line path that would be a tangent line to the curved path at the moment the force was stopped.
To find centripetal force you use the equation that requires you multiply the mass of the object times the centripetal acceleration it is experiencing. This equation is Fc = m ac. The equation for the centripetal acceleration is ac = v2 / r, where v is the object's velocity and r is the radius of the circular path the object is traveling. To find the velocity of the object you need to calculate the length of the circular path, which is the circumference, and divide it by the period of the object, which is the time for the object to complete the circular path once. To better understand period, you need to study frequency.
As stated above, the time for an object to complete one trip around a circular path is defined as the period of the event. Related to period is another measurement called frequency. Frequency is defined as the number of times an event occurs in a specified unit of time, such as the second, and is often expressed in the unit hertz (hz) which is the number of cycles per second. A cycle is one completion of an event. For example an object traveling once around a circular path is one cycle. Let us say that a ball goes around your head while attached to a string you are holding in your hand. It moves with a frequency of 2 hz. This means it travels twice around your head in one second of time. On the other hand the period of an event is the time for the event to take place. In the situation just described the ball would have a period of 0.5 s, because the time for the ball to circle your head once is the reciprocal of the frequency. The reciprocal of 2 Hz is 0.5 s. So 0.5 seconds is the time required for the ball to circle your head once.
Placing a Satellite in Orbit: The motion of satellites in orbit, while often elliptical, can be close to if not actually circular in shape. To get a reasonable value for the question what velocity must a satellite have to be in a certain orbit and essentially stay there, you would use the following equation. v = ( g r )0.5. The value g, of course, is the acceleration due to gravity, r is the radius of the orbit, and the exponent 0.5 means that you are raising the value in the parentheses to the one-half power, or take the square root of the value.
Simple Harmonic Motion: Have you ever watch something vibrate? A guitar string vibrates. A weight oscillating up and down on the end of spring looks to be vibrating. The motion of a swinging pendulum looks similar to vibrating motion. The characteristics associated with vibrating motion, is the moving of an object back and forth over the same path, reversing its direction of motion at the end of each cycle. A special kind of vibrating motion is called simple harmonic motion (SHM). A common way to describe it is in terms of a mass "vibrating" up and down on a spring. Whenever the spring is either stretched or compressed, there is a restoring force, which resists the distortion experienced by the spring due to the up and down vibration, that works towards returning the mass to the position of equilibrium (the rest position).
The motion of a pendulum at least in terms of motion involving small angles of less than 15o, is cited as an example of SHM. The amplitude of the swing is the distance from the equilibrium position. The period of a pendulum is the time for the hanging mass to make one complete swing. This is equivalent to one vibration or cycle. The equation to find the period of a pendulum is the following. T = 2 p№ ( L / g )0.5. The L, which stands for the length of the pendulum, is a vertical displacement of the pendulum mass from the pivot point of the pendulum.
(Torque: Often this topic is described in terms of the lever arm. Have you ever tried to accomplish a task using a lever arm? A common definition of a lever arm is that it is the perpendicular distance from the axis of rotation of an object to a line along which a force acts. The torque is the multiplication product of the force and the lever arm. The larger the torque is, the greater the change in the rotational motion of an object. Torques can stop, start, or change the direction of rotational motion. As an example, consider a child's seesaw. The see saw has a pivot point, often called a fulcrum. To balance two children of unequal weight so they may use the seesaw effectively, you would use the equation W1 d1 = W2 d2. W represents weight and d represents the distance from the fulcrum. The symbol for torque is t№, where t№ = F d. F is the force perpendicular to the torque arm and d is the distance the force is from the pivot point of the torque arm. A simple example is a socket wrench being used to tighten a bolt. The socket is the pivot point. The person using the wrench then applies a certain amount of force to the wrench handle at a certain distance from the socket.
Vocabulary: Resultant, Equilibrant, Component Vector, Inclined Plane, Normal Force, Parallel Force, Projectile, Trajectory, Maximum Height, Range, Flight Time, uniform Circular Motion, Centripetal Force, Centripetal Acceleration, Centrifugal Force, Rigid Rotating Object, Lever Arm and Torque
Skills to be learned:
Solve Balance Force Problems (hanging signs)
Solve Inclined Plane Problems (objects on ramps)
Solve Projectile Motion Problems (launched off cliffs)
Solve Projectile Motion Problems (launched from the ground)
Solve Uniform Circular Motion Problems (objects in circular paths)
Solve Satellite Velocity in Orbit Problems
Solve Simple Harmonic Motion Pendulum Problems (Pendulums)
*Solve Torque Problems
Assignments:
Textbook: Read / Study / Learn Chapter 7 applications of two dimensional motion and force using vectors correctly
Workbook Exercise(s): PS# 7-1, PS#7-2, PS#7-3 (Inclined Planes)
Activities TBA
Resources:
This Handout and the Overhead and Board Notes discussed in class
Textbook: Chapter 7
WB Lessons and Problem Sets
HYPERLINK "http://www.physicsphenomena.com" www.physicsphenomena.com - Motion in Two Dimensions
( HYPERLINK "http://www.physicsphenomena.com" www.physicsphenomena.com / Motion in Two Dimensions
( HYPERLINK "http://www.physicsphenomena.com" www.physicsphenomena.com / Motion in Two Dimensions
( HYPERLINK "http://www.physicsphenomena.com" www.physicsphenomena.com / Motion in Two Dimensions
( HYPERLINK "http://www.physicsphenomena.com" www.physicsphenomena.com / Motion in Two Dimensions
PAGE
PAGE 3
kШ и ИкЕЭгыєўБВзoptvњћџ(b>IP[RSWXЌ#!4!H!`!" "$$$$Т$У$Ч$Ш$И&О&''l**Ў+Б+,А,90:0222283:3в6д6и6к6т6ф6ш6ъ6Ј7Њ7§ѕ§я§я§я§я§я§я§я§ѕ§я§ы§ы§ч§ы§ы§я§я§я§ы§ы§я§я§я§ѕ§ы§ы§ы§ч§я§я§я§ч§я§ч§р§ч§ѕ§ы§ы§ы§ы§ж5CJOJQJ\CJOJQJCJH*CJH*
5CJ\ jу№0JCJCJT$%?SxГ№-HSTjkЧ Ш ЗИ'(л#м#Q&R&§§§§їїї§§§§§§§§§§§§§§§§§§§§§а`аЯ=l?
??ўўўўR&k*l*,,н/о/6383@9A9f:g:h:~:Ќ:о:;S;;У;џ;<<%<<и<ш<щ<§§§§§§§§§§§§§§§§§§§§§§§§ї§§§а^аЊ7К7М7Т7Ф7A9L9h:~:<%<щ<ѓ<h=i====А=Б=Я=а=б=в=џ=>>>>6>7>8>9>f>g>h>>>>>> >Э>Ю>Я>ч>ш>????4?5?6?N?O?l?m?s?t?§ѓ§э§ф§ф§ф§ф§н§гнЮн§ШУЛУИУШУАУИУШУЈУИУШУ УИУ0J
j0JUjЄUjЛUjвU0JjщU jU
jу№0J0JCJjCJU
jCJU56CJ\]
5CJ\5CJOJQJ\CJ;щ<є<6=K=h=Я=6>>?k?l?u?v?w????
???????§§§§їѕѕѕѕ§ьц§ьц§§§§§§§їh]hјџ&`#$а`аt?u?w?x?~???????§і§іюі§ыCJ0JmHnHu
j0JU0J
1hАа/ Ар=!А"А# $ %АщDаЩъyљКЮЊKЉwww.physicsphenomena.comрЩъyљКЮЊKЉBhttp://www.physicsphenomena.com/щDаЩъyљКЮЊKЉwww.physicsphenomena.comрЩъyљКЮЊKЉBhttp://www.physicsphenomena.com/щDаЩъyљКЮЊKЉwww.physicsphenomena.comрЩъyљКЮЊKЉBhttp://www.physicsphenomena.com/щDаЩъyљКЮЊKЉwww.physicsphenomena.comрЩъyљКЮЊKЉBhttp://www.physicsphenomena.com/щDаЩъyљКЮЊKЉwww.physicsphenomena.comрЩъyљКЮЊKЉBhttp://www.physicsphenomena.com/
i8@ёџ8NormalCJ_HaJmH sH tH <A@ђџЁ<Default Paragraph Font.U@Ђё. Hyperlink>*B*phџ, @,Footer
ЦрР!&)@Ђ&Page Number6@"6
Footnote TextCJaJ8&@Ђ18Footnote ReferenceH*Б
.њ8gЮ5њ8џџџџџџџџџџџџџџџџџџџџџџњ8bџџџџ$%?SxГ№-HSTjkЧШЗ
И
'(лмQ"R"k&l&((н+о+
..Џ2А2е3ж3з3э34M4
4Т4525n55556G6W6X6c6Ѕ6К6з6>7Ѕ78s8к8л8ф8х8ц8ё8ђ8ѓ8є8ѕ8і8ї8ј8ћ80000000000000000000000000000000000000000000000000000000000000@0@0@0@0
@0@0@0@0@0@0
00000Њ7t??*.0R&щ<?+-/?,з677њ8Xџ!!Tџ1JiБаџ7fXџXџXџXџ№8№@ёџџџї№№№0№( №
№№B
№S№ПЫџ ?№њ8жнnpsvљћўQS[]
#Ё#Ї$Ќ$У$Х$>7л8ѓ8є8ћ8
авGMnpQSЁ#Ј#Є$Ч$Ѓ'Є'q-t-Ъ3Э3>7л8ѓ8є8ћ8333333333 И
И
Б
В
P"P"U"["a"b"ђ#ѓ#є# $>$?$k$q$Ђ$Ќ$(+)+++++н+о+),:,o-p-
----Г-Д-Ю- ...//з0з0у0ф0)1ж3ь3L4щ45(515a5m556666F6R6V6b6=7>7@7Є7Ѕ7к8л8у8ц8№8є8є8ѕ8ј8ћ8џџTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychonC:\Documents and Settings\Administrator\Application Data\Microsoft\Word\AutoRecovery save of Topic7-MSword.asdTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychonC:\Documents and Settings\Administrator\Application Data\Microsoft\Word\AutoRecovery save of Topic7-MSword.asdTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychonC:\Documents and Settings\Administrator\Application Data\Microsoft\Word\AutoRecovery save of Topic7-MSword.asdTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docTychoTC:\Documents and Settings\Administrator\Desktop\My Physics Lessons\Topic7-MSword.docџ@Ќ$Ќ$ tЌ$Ђ$@o-o.o0o1њ8`@`2`h@`6`p@џџUnknownџџџџџџџџџџџџGz џTimes New Roman5Symbol3&z џArial"q№аhJвЄ§-a!№ ДД20d№72№џџTopic #7TychoTychoўџр
ђљOhЋ+'Гй0lЌИШдр№
(4
@LT\dф Topic #7 opiTycho#7ychychNormal7Tycho717hMicrosoft Word 9.0@и^Ќ:@њчЧMіФ@wЦczХ§-ўџеЭе.+,љЎDеЭе.+,љЎDhpЄЌДМФ
ЬсфBraheEnterprizesya№7э Topic #7TitleЄ 8@_PID_HLINKSфA\XS!http://www.physicsphenomena.com/XS !http://www.physicsphenomena.com/XS!http://www.physicsphenomena.com/XS!http://www.physicsphenomena.com/XS!http://www.physicsphenomena.com/
!"#$%&'()*+,-./01ўџџџ3456789ўџџџ;<=>?@ABCDEFGўџџџIJKLMNOўџџџQRSTUVWўџџџ§џџџZўџџџўџџџўџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџRoot Entryџџџџџџџџ РFPБЎШczХ\Data
џџџџџџџџџџџџ21Tableџџџџџџџџ:WordDocumentџџџџ"bSummaryInformation(џџџџџџџџџџџџHDocumentSummaryInformation8џџџџџџџџPCompObjџџџџjObjectPoolџџџџџџџџџџџџPБЎШczХPБЎШczХўџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџџўџ
џџџџ РFMicrosoft Word Document
MSWordDocWord.Document.8є9Вq