phys109-mechanics - middle east technical...

Phys109-MECHANICS

Altug Özpineci

METU

PHYS109

Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 1 / 175

First Week Your Lecturer

Your Lecturer

Title and Name: Prof. Dr. Altug ÖzpineciAdministrative Duty: Vice to Dept. Chair. responsible for thecourses given by the departmentSpecialization: High Energy Physics, Hadron Physics (propertiesof quarks and gluons)e-mail: [email protected] method to contact: through e-mailWeb Page: http://www.metu.edu.tr/~ozpineciCourse Web Page: Course Material -> Phys109


mailto://[email protected]

http://www.metu.edu.tr/~ozpineci

First Week Syllabus

Syllabus

Text Book: “Physics for Scientists & Engineers,” 4th Edition, C.GiancoliSubjects to be covered:

Physics as a Science (≈1. week)Describing Motion-Kinematics (≈1. and 2. weeks)Causes for Changes in Motion - Dynamics (≈2. and 3. weeks)Applications of Newton’s Laws (≈3.-10. weeks)(?)Probability in Physics - Thermodynamics (≈ 11. week till the endof semester)

20 Chapters in 14 weeks (?)


First Week Syllabus

Applications of Newton’s Laws:FrictionCircular MotionGravitationWork and EnergySystems of ParticlesCollisionsRotationStaticsFluidsOscillations and Waves...


First Week Supplamentary Material and References

Supplamentary Material and References

H. C. Ohanian, “Physics”R. P. Feynman “Lectures on Physics, Vol. 1”http://www.feynmanlectures.info

Open Courseware Project (OCW)MIT OCW (http://ocw.mit.edu) (in particular see Physics I:Classical Mechanics, Prof. Walter Lewin, Turkish translation is alsoavailable)TÜBA OCW (in Turkish) (http://www.acikders.org.tr)METU OCW (http://ocw.metu.edu.tr)

“Feynman’s Lost Lecture: The Motion of Planets Around the Sun ”Any other related book in the library


http://www.feynmanlectures.info

http://ocw.mit.edu

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/

http://www.acikders.org.tr

http://ocw.metu.edu.tr

First Week Grading

Grading

Grades are NOT the aim of studying! They are only means ofmeasuring how much you have learned!


First Week Grading

Grading

Grades are NOT the aim of studying! They are only means ofmeasuring how much you have learned!

15% Lab5% Pop Quizes20% each midterm (two midterms in total)20% Final Exam5% Term Report15% HomeworkBonus: Can increase your grade upto half a letter

Translate 5 items from English wikipedia to Turkish wikipedia(oryour mother tongue if different). You have to let us know in a monthwhich items you are planning to translateAsk good questions on piazza! (signup athttp://piazza.com/metu.edu.tr/fall2013/phys109)


http://www.piazza.com

http://piazza.com/metu.edu.tr/fall2013/phys109

First Week Grading

Piazza

Piazza is an open platform to manage class Question andAnswersSimilar to LMS or METUOnline but students can ask questionsanonymously!Can collaborate to find answers!


http://lms.metu.edu.tr

http://online.metu.edu.tr

First Week Grading

Pop Quizes

The purpose is to encourage youto attend the lecturesto study regularly,

Will be simple question that can be easily done if you have payedattention to the lectureTheir time and number will not be announced! They can be anyday, and any time during the lecture


First Week Grading

Reports:

The purpose of the report isEncourage you to work on your ownEncourage you to work in groupsPractice presenting your findings to your colleagues


First Week Grading

Reports:

Reports should be prepared by a group of three studentsYou are allowed to determine your own group until October 3,2013.If you do not inform me about your group until the deadline, I willform groups from the remaining students


First Week Grading

Reports:

Reports should be about any subject related with the coursematerialSubject of the report should be chosen before October 24, 2013.Preferable, your group should choose your own report subjectbased on your own interest. After the deadline, the instructor orthe assistants will assign your group a subjectIt should be aimed at teaching somebody else who does not knowanything about the subjectA mid report should be handed in before November 24, 2013.


First Week Grading

Reports:

At the end of the semester, Each group will present their reportThe instructor/assistants will choose which member of the groupwill present which part of the report(PARTIALLY) HANDWRITTEN REPORTS WILL NOT BEACCEPTED


First Week Grading

HOMEWORK

Their purpose is to encourage you to study regularly and topracticeYou can discuss the solutions in groupsThe homework you hand in should be what YOU understand


First Week Grading

Final Exam

Conditions under which you will NOT be allowed to take the final exam:

Failing the labBeing absent in more than 20% of Pop QuizesNot taking any of the midterms without any excuseNot completing the term report


First Week Ethics

Ethics

You should never claim the work to be yours if you have not doneit

Handing in somebody else’s (from your friends, from internet, orfrom some other source) solutions in homework/midterms(cheating/plagiarism)Claiming that your data/solution is correct even if you know thatthey are not (falsification/data fabrication)

If you quote somebody else’s work, make sure that you citehim/her so that the reader understands that you do not claim to bethe owner of your workAccording to discipline regulation of committee of highereducation, cheating is punished by sending the student away forat least one semester (YOK Ögrenci Disiplin Yönetmeligi)


http://www.yok.gov.tr/web/guest/icerik/-/journal_content/56_INSTANCE_rEHF8BIsfYRx/10279/17960

First Week Prerequisite Courses

Prerequisite Courses

Phys109/110 and MATH119/120 are prerequisites to the higherlevel coursesIf you fail them, most probably, you can not graduate in four years.

!

Prerequisite!Chart!of!Must!Courses!

!

MATH%119% MATH%120%

PHYS%209% PHYS%210%

PHYS%311% PHYS%334%

PHYS%110%

%

PHYS%202% PHYS%203% PHYS%221%

PHYS%300%

PHYS%222%

PHYS%430%

%

PHYS%431%

PHYS%109%

%


First Week Special Physics Group

Special Physics Group

Students are required to work harderThey learn more subjects in more detailA small group (15-20) of students selected at the beginning ofsecond yearSelected based on the success in the first year


First Week Learning/Lecturing Physics

Learning/Lecturing Physics

Watch: Confessions of a Converted Lecturer, by Eric MazurMain lessons:

”Traditional lecturing is nothing but the transfer of lecture notes fromthe notes of the lecturer to the notes of the student without passingthrough the brains of neither” by XXX YYYIn a traditional lecture, students only learn 22% of what they knowno matter what the lecturer does/doesn’t doIt can increase to 44% on average if the students participate in theclass


http://www.youtube.com/watch?v=WwslBPj8GgI


Learning/Lecturing Physics

Watch: Confessions of a Converted Lecturer, by Eric MazurMain lessons:

”Traditional lecturing is nothing but the transfer of lecture notes fromthe notes of the lecturer to the notes of the student without passingthrough the brains of neither” by XXX YYYIn a traditional lecture, students only LEARN 22% of what theyknow no matter what the lecturer does/doesn’t doIt can increase to 44% on average if the students participate in theclass


http://www.youtube.com/watch?v=WwslBPj8GgI


“LECTURING”

Lecture: late 14c., "action of reading, that which is read," fromM.L. lectura "a reading, lecture," from L. lectus, · · ·lesson: l early 13c., "a reading aloud from the Bible," · · ·

Source: http://www.etymonline.com



“LECTURING”

Lecture: late 14c., "action of reading, that which is read," fromM.L. lectura "a reading, lecture," from L. lectus, · · ·lesson: l early 13c., "a reading aloud from the Bible," · · ·

Source: http://www.etymonline.comThe way we lecture did not change since 14th century.



Learning

When do you think that you have LEARNED a subject?I think that I have learned a subject if I can

carry out derivations without looking at any other referencerepeat the reasonings

starting from the first principlesLearning physics is NOT about memorizing formulas and applyingthem (you will be given formulas in all the exams)



Pay Attention To Answers That are Not Answers

Not all answers are really answers.Why the leaves are green?Because they reflect green light.Rephrase the question: Why leaves reflect green light?Because the contain chlorophyll.Rephrase the question: Why does chlorophyll reflect light?...



Best Teachers:

Your best teachers are (in order of importance):1 Yourself: learn to learn on your own. (use the library!) You are the

only person that will know when you have really learned a subject.2 Your friends: Collaborate. Your professors will not have a clue why

you do not understand things that are obvious for them.3 Your professor. Even if you think that a professor does not know

anything, s/he still knows more physics than you, and has moreexperience.


First Week Physics

What is Physics?

Student participation is crucial for students to learn.Not so easy in large classrooms with ≈ 150 studentOptimum number of students in a class ≈ 17 (PHED graduatestudents)Try the SMS system: send your answers to 4660 (Turkcell, Avea,or Vodafone free of charge)


http://echoice.mobilogren.net

First Week Physics

Physics

Physics is about everything around you!Physics tries to find relationship between observations.Measurement is a crucial part of physics


First Week Physics

Physics

Physics is about everything around you! Look around yourself!Physics tries to find relationship between observations.Measurement is a crucial part of physics


First Week Physics

MEASUREMENT

Has to be repeatable by anybody (that has necessary equipment)Units and errors are a crucial part of the measurement!


First Week Physics

Precision: how well repeated measurements yield similar resultsAccuracy: how close the measurement is to the real value


First Week Physics

Precision: how well repeated measurements yield similar resultsAccuracy: how close the measurement is to the real value Howcan one know the real value without measurement? How is itpossible to be sure that a given measurement is accurate?


First Week Physics

Certain quantities are defined, not measured!1 m: Length that light travels in 1/299,792,458 second1 s: Time required for 9,192,631,770 periods of radiation emittedby cesium atoms

Speed of light is exactly 299,792,458 m/s!


First Week Physics

Certain quantities are defined, not measured!1 m: Length that light travels in 1/299,792,458 second1 s: Time required for 9,192,631,770 periods of radiation emittedby cesium atoms

Speed of light is exactly 299,792,458 m/s!Platinum cylinder in International Bureau of Weights andMeasures, Paris

These are the fundamental units (in SI system). Everything else ismeasured relative to these units.


First Week Physics

Q: How to measure learning?


First Week Physics

Learn LATEX


First Week Kinematics of Motion

nature

model

MathematicalRepresentation

model

Mathematical DerivationMATH119 & MATH120

nature

Interpretation



Assume all the motion is along a given line.

The position can be specified by a unique number: distance fromorigin O.One side is denoted as "+" , the other side "-"The choice of O and the "+" side is completely arbitrary

A B

O

xi = +3.0 cm

xf = −1.0cm



Assume all the motion is along a given line.The position can be specified by a unique number: distance fromorigin O.

One side is denoted as "+" , the other side "-"The choice of O and the "+" side is completely arbitrary

A BO

xi = +3.0 cm

xf = −1.0cm



Assume all the motion is along a given line.The position can be specified by a unique number: distance fromorigin O.One side is denoted as "+"

, the other side "-"The choice of O and the "+" side is completely arbitrary

A BO

xi = +3.0 cm

xf = −1.0cm



Assume all the motion is along a given line.The position can be specified by a unique number: distance fromorigin O.One side is denoted as "+" , the other side "-"

The choice of O and the "+" side is completely arbitrary

A BO

xi = +3.0 cm

xf = −1.0cm



Assume all the motion is along a given line.The position can be specified by a unique number: distance fromorigin O.One side is denoted as "+" , the other side "-"The choice of O and the "+" side is completely arbitrary

A BO

xi = +3.0 cm

xf = −1.0cm



Definitions:

Displacement: the change in the position of an object ∆x .

∆x = (final position)− (initial position)= (3.0 cm)− (−1.0 cm) = 4.0 cm (1)

Average velocity: If ∆t is the time that an object moves by ∆x ,average velocity is

v =∆x∆t

(2)



Definitions:

Displacement: the change in the position of an object ∆x .

∆x = (final position)− (initial position)= (3.0 cm)− (−1.0 cm) = 4.0 cm (1)

Average velocity: If ∆t is the time that an object moves by ∆x ,average velocity is

v =∆x∆t

(2)


Greek Letters∆: Finite differences of any sizeδ: Finite differences of small sized : Infinitesimal difference (smallerthan anything else)


x(t)

O t

A

1.0 s0.2 m

B

4.0 s

3.2 m

|∆x |

|∆t |α

vAB =(3.2 m)− (0.2 m)

(4.0 s)− (1.0 s)= 1.0 m/s = tanα

(3)



x(t)

O t

A

1.0 s0.2 m

B

4.0 s

3.2 m

|∆x |

|∆t |α

vAB =(3.2 m)− (0.2 m)

(4.0 s)− (1.0 s)= 1.0 m/s

= tanα

(3)



x(t)

O t

A

1.0 s0.2 m

B

4.0 s

3.2 m

|∆x |

|∆t |α

vAB =(3.2 m)− (0.2 m)

(4.0 s)− (1.0 s)= 1.0 m/s = tanα (3)



As the final time moves closer to the initial time, i.e. the point Bmoves towards point A, we obtain the instantaneous velocity:

vinst = limB→A

vAB = limtf→ti

∆x∆t

= limtf→ti

xf − xi

tf − ti=

dxdt

≡ v

(4)

If δt is a sufficiently small amount of time, the displacement duringthis time is δx = vδt

xf = xi + vδt (5)



As the final time moves closer to the initial time, i.e. the point Bmoves towards point A, we obtain the instantaneous velocity:

vinst = limB→A

vAB = limtf→ti

∆x∆t

= limtf→ti

xf − xi

tf − ti=

dxdt≡ v (4)

If δt is a sufficiently small amount of time, the displacement duringthis time is δx = vδt

xf = xi + vδt (5)



QuestionIf v(t) is know for all t ∈ (ti , tf ), and a particle is at the positionx(ti) = x0 initially, how can we find x(t) for any t ∈ (ti , tf )?

A: Assume δt is sufficiently small and tf = ti + Nδt .

x(ti + δt)− x(ti) = v(ti)δtx(ti + 2δt)− x(ti + δt) = v(ti + δt)δt

x(ti + 3δt)− x(ti + 2δt) = v(ti + 2δt)δt· · ·

x(ti + Nδt = tf )− x(ti + (N − 1)δt) = v(t2 + (N − 1)δt) (6)

x(tf )− x0 =N−1∑k=0

v(ti + kδt)δt

δt→0−→∫ tf

tiv(t)dt

(7)

Read Zeno’s paradox!








x(tf )− x0 =N−1∑k=0

v(ti + kδt)δt

δt→0−→∫ tf

tiv(t)dt

(7)

Read Zeno’s paradox!Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 32 / 175







x(tf )− x0 =N−1∑k=0

v(ti + kδt)δt δt→0−→∫ tf

tiv(t)dt (7)

Read Zeno’s paradox!Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 32 / 175


Special Case: Motion with constant velocity v0:In this case

x(tf )− x0 =N−1∑k=0

v(ti + kδt)δt =N−1∑k=0

v0δt = v0Nδt = v0(tf − ti) (8)

x(t) = v0(t − ti) + x0 (9)

Note that for motion with constant velocity v = v0. Hence ∆x = v0∆t



The same steps can be repeated for the change of velocity.a = ∆v

∆t . The unit of acceleration is m/s2

ainst = lim∆t→0∆v∆t ≡ a

v(t) = v(t0) +∫ tf

t0a(t ′)dt ′



The same steps can be repeated for the change of velocity.a = ∆v

∆t . The unit of acceleration is m/s2

ainst = lim∆t→0∆v∆t ≡ a

v(t) = v(t0) +∫ tf

t0a(t ′)dt ′


Acceleration is in the direction of ∆v ,NOT in the direction of v .


Example:

Motion with Constant Acceleration. Initial conditions: x(0) = 0,v(0) = 0. Realistic case: You stand at the top of a building. You areholding a mass m in your and release it from rest outside a window.

Let a be the constant acceleration.

v(t) = v(0) +

∫ t

0adt ′ = at (10)

The position:

x(t) = x(0) +

∫ t

0v(t ′)dt ′

=

∫ t

0(at ′)dt ′ =

12

at ′2∣∣∣∣t0

=12

at2 (11)



Dimensional Analysis

Most of the time, the final formula can be estimated unto overallcoefficients using dimensions only. Denote the dimension of anyquantity O by [O]

Dimension of x(t) is [x(t)] = mThe dimensionful parameters in the problem are the accelerationa and the time t .Assume x(t) = Aak t l where A, k and l are numbers.

[Aamt l ] = [A][a]k [t ]l = 1(m

s2

)ksl = mksl−2k (12)

x = Aak t l =⇒ k = 1 and l − 2k = 0 =⇒ x(t) = Aat2

Explicit calculation shows A = 12 .



In principle these steps can be done for the change inacceleration, change in the change in the acceleration, etc.Newton’s Laws tell us that this is not necessaryThe acceleration of an object is determined by external effects.



Compare

v(t) =dxdt⇐⇒ x(t) = x(0) +

∫ t

0v(t ′)dt ′ (13)

a(t) =dvdt⇐⇒ v(t) = v(0) +

∫ t

0a(t ′)dt ′ (14)

Integration is the inverse of differentiation


First Week Motion in 3D and Vectors

Vectors

For motion that is not confined to a line, more than a number isnecessary to describe the direction.A vector is a recipe for how to go to the point A from the origin.A vector is a number and a directionOrigin is arbitrarily chosen



y (m)

0 x (m)

~A2

3

~A = (3,2) m (15)~A = (3 m)x + (2 m)y (16)~A = (3 m)i + (2 m)j (17)

~A = (√

13 m,arctan23

) (18)

~A = (2,3) m (19)

~A = (√

13 m,arctan32

)



Vector Operations-Multiplication by a number

A vector ~A is a number (the length of the vector, |~A|) and adirection.The vector λ~A is another vector

The length of λ~A is |λ~A| = |λ||~A|The direction of λ~A is the same as the direction of ~A if λ > 0, andopposite to ~A if λ < 0



Vector Operations-Addition of Vectors

Geometrical Addition

y

O x

~A

~B

~A

~B ~C = ~A + ~B

~A− ~B

Componentwise Addition~A = Ax x + Ay y + Az z~B = Bx x + By y + Bz z~C = Cx x + Cy y + Cz zCx = Ax + Bx , Cy = Ay + ByCz = Az + Bz

Ci = Ai + Bi , i = x , y or z

Subtraction~A− ~B = ~A + ((−1)~B)





y

O x

~A

~B

~A

~B

~C = ~A + ~B

~A− ~B








y

O x

~A

~B

~A

~B ~C = ~A + ~B

~A− ~B






Vector Operations: Scalar Product

~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosα




~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosα~A · ~B = A‖B




~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosα~A · ~B = AB‖




~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosαScalar product is linear:~A · (a~B + b~C) = a(~A · ~B) + b(~A · ~C)

x · x = y · y = z · z = 1,x · y = x · z = y · z = 0




~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosαScalar product is linear:~A · (a~B + b~C) = a(~A · ~B) + b(~A · ~C)

x · x = y · y = z · z = 1,x · y = x · z = y · z = 0


~A

~B

~C~D ≡ ~B + ~C

B‖ C‖

D‖~A·~D = AD‖ = A(B‖+C‖) = ~A·~B+~A·~C



~A

~B

α

A‖

B‖

Scalar product gives a number fromtwo vectors~A · ~B ≡ |~A||~B| cosα~A = Ax x + Ay y + Az z,~B = Bx x + By y + Bz z~A · ~B = AxBx + AyBy + AzBz

Ax = ~A · x , Ay = ~A · y , and Az = ~A · z



Vector Operations: Vector Product

~A

~B

α

A⊥B⊥

Vector product gives a vector fromtwo vectors|~A× ~B| = |~A||~B| sinα

Direction of ~A× ~B is given by the righthand rule. (~A× ~B = −~B × ~A)




~A

~B

α

A⊥

B⊥


Direction of ~A× ~B is given by the righthand rule. (~A× ~B = −~B × ~A)|~A× ~B| = A⊥B




~A

~B

α

A⊥

B⊥


Direction of ~A× ~B is given by the righthand rule. (~A× ~B = −~B × ~A)|~A× ~B| = AB⊥




~A

~B

α

A⊥B⊥


Direction of ~A× ~B is given by the righthand rule. (~A× ~B = −~B × ~A)Vector product is linear:~A · (a~B + b~C) = a(~A · ~B) + b(~A · ~C)

x × x = y × y = z × z = 0,x × y = z, x × z = −y , y × z = x



Vector Operations- Vector Division

Division by a vector DOES NOT exist!



Equality of Vectors

Two vectors are equal only if all their components are equal:

~A = Ax x + Ay y + Az z (20)~B = Bx x + By y + Bz z (21)

if ~A = ~B, thenAx = Bx , Ay = By , Az = Bz (22)


Second Week Review of Integration

Review of Integration

∫ t2t1

f (t)dt is just a symbolMeaning of the symbol: What does it stand for?Value of the symbol: What does that symbol equal to?



R = 1

1 2 3

1

2

3 The area is A = πR2 ' 3.21

The number of squares lyingentirely inside the circle:The number of squares thathave a part inside the circle:Area of each square:a0A< < A < a0A>



R = 1

1 2 3

1

2

3

The area is A = πR2 ' 3.21The number of squares lyingentirely inside the circle:A< = 1The number of squares thathave a part inside the circle:A> = 9Area of each square:a0 = 1a0A< < A < a0A>=⇒ 1 < A < 9



R = 1

1 2 3

1

2

3

The area is A = πR2 ' 3.21The number of squares lyingentirely inside the circle:A< = 4The number of squares thathave a part inside the circle:A> = 16Area of each square:a0 = 1/4a0A< < A < a0A>=⇒ 1 < A < 4



R = 1

1 2 3

1

2

3

The area is A = πR2 ' 3.21The number of squares lyingentirely inside the circle:A< = 32The number of squares thathave a part inside the circle:A> = 60Area of each square:a0 = 1/16a0A< < A < a0A>=⇒ 2 < A < 3.75



R = 1

1 2 3

1

2

3 a0A< =∑

squares completelyin circle

1× a0

a0A> =∑

squares containing at leasta fraction inside the circle

1× a0

Always a0A< < A < a0A>As the grid size get smaller a0A< increases as a0A> decreases,and A is always in betweenEventually a0A< ' a0A> ' A: Mathematical expression:

lima0→0

a0A< = lima0→0

a0A> = A (23)



R = 1

1 2 3

1

2

3 a0A< =∑

squares completelyin circle

1× a0

a0A> =∑

squares containing at leasta fraction inside the circle

1× a0

In the limit a0 → 0, each sum contains infinitely many termsThe contribution of each term, i.e. 1× a0, becomes zero.In the limit a0 → 0, we are summing infinitely many zeroes: theresult is finite.rather than writing “limit as a0 → 0, sum the areas of all thesquares that lie completely inside the circle,” we write∫

disc1× da (23)



Area of a circleLet r be the inner radius of the chosen ringr + δr be the area of the outer radius of thechosen ringThe ring can be straightened out and it will fitinside a rectangle whose width is δr areheight 2π(r + δr)

The ring will contain a rectangle whose widthis δr and height 2πr .2πrδr < δA < 2π(r + δr)δr where δA is thearea of the ring



Area of a circle2πrδr < δA < 2π(r + δr)δr where δA is thearea of the ringTotal area of the disk is the sum of all suchrings from r = 0 upto r = R:

∑r

2πrδr < A =R∑

r=0

δA <∑

d

2π(r + δr)δr

(24)In the limit δr → 0

(25)

A =

∫ R

0dr2πr = πR2 (26)


Second Week POP QUIZ

POP QUIZ

~A = x + y + z (27)

~B = x + y − 2z (28)

1 Show that ~A and ~B are perpendicular.2 Calculate ~A + ~B , ~A− ~B, ~A · ~B3 Calculate |~A|, |~B|


Second Week Motion in 3D

Displacement Vector

O

~x1

~x2

∆~x = ~x12

The displacement vector is ∆~x = ~x12 = ~x2 − ~x1



Motion in 3D

The discussions on motion in 1D can be generalized to 3D by justrepresenting positions, velocities and acceleration with 3D vectors:

~xi and ~xf are initial and final positions of the particle.Displacement vector is ∆~x = ~xf − ~xi

Average velocity is ~v = ∆~x∆t . (~v is a vector times a number, hence it

is also a vector)Componentwise vx = ∆x

∆t , vy = ∆y∆t , vz = ∆z

∆t .

Instantaneous velocity ~v(t) = lim∆t→0∆~x∆t = d~x

dt

Componentwise vx (t) = dxdt , vy (t) = dy

dt , vz(t) = dzdt



Motion in 3D


~vi and ~vf are initial and final velocities of the particle.Average acceleration is ~a = ∆~v

∆t . (~a is a vector times a number,hence it is also a vector)Componentwise ax = ∆vx

∆t , ay =∆vy∆t , az = ∆vz

∆t .

Instantaneous acceleration ~a(t) = lim∆t→0∆~v∆t = d~v

dt

Componentwise ai(t) = dvidt



Motion in 3D


~v(t) = d~xdt ⇐⇒ ~x(t) = ~x(ti) +

∫ tti~v(t ′)dt ′

~a(t) = d~vdt ⇐⇒ ~v(t) = ~v(ti) +

∫ tti~a(t ′)dt ′



Motion In Earth’s Gravity

In the absence of friction, all objects have the same accelerationunder the gravitational attraction of Earth.Close to the surface of Earth, this acceleration is uniform, and isdenoted by g ' 9.8 m/s2 and points toward the center of Earth.Choose a coordinate axis: one possible choice is z pointingdownwards and x and y axis horizontal. Choose z = 0 plan to lieon the surface of earth.

z

O x , y

~a = gz




In the absence of friction, all objects have the same accelerationunder the gravitational attraction of Earth.

z

O x , y

~a = gz

~a = gz.~v(t) = ~v(ti) +

∫ tti~a(t ′)dt ′ = ~vi + gz(t − ti)




In the absence of friction, all objects have the same accelerationunder the gravitational attraction of Earth.

z

O x , y

~a = gz

~v(t) = ~vi + gz(t − ti)~x(t) = ~x(ti) +

∫ tti~v(t ′)dt ′ = ~xi + ~vi(t − ti) + 1

2gz(t − ti)2


Third Week Exam Result

0 5 10 15 20 25 300

5

10

There seems to be three different groups of students:A group around 6A group around 12A group around 16


Third Week Trajectory of a Football

Trajectory of a Football

z

x , y

~v0

α

Assume that you hit a football lying onthe ground.It’s initial speed is v0 making an angleα with the ground.Choose the origin of time such thatti = 0 and origin of coordinate axissuch that ~x(0) = 0




z

x , y

~v0

α

~x(t) = ~v0t + 12gt2z

z(t) = v0z t + 12gt2

z(t) = 0 when t = 0(initial time) andat t = −2v0z

g (when the ball hits theground)




z

x , y

~v0

α

The flight time of the ball is t = −2v0zg .

The only acceleration is along the zaxis.− v0z

g is the time it take for the zcomponent of the velocity to becomezero, i.e. the time it takes to reachmaximum heightThe time it takes to fall down is thesame (in the absence of air friction)




z

x , y

~v0

α

Assume that x and y axis are chosensuch that ~v = −v0 sinαz + v0 cosαxvy (t) = 0, y(t) = 0 for all timesx(t) = v0x t .Range is the distance the ball coversduring its flight, i.e. R = |x(tf )|

R = v0x

(−2v0z

g

)= v0 cosα

(−2(−v0 sinα)

g

)=

v20 sin 2α

g(29)




z

x , y

~v0

α

R =v2

0 sin(2α)

g

sin 2α has maximum value of 1 whenα = 45◦

Increasing v0 by a factor of 2increases the range by 4.If α1 + α2 = π

2 , their ranges are thesame



Trajectory

Trajectory is a relationship betweenthe components of the position of aparticle that does not involve time.When the particle is at the horizontaldistance x , the time that has passedis t(x) = x/v0x .The z coordinate of the particle atthat time is

z(x) = v0z t(x) +12

gt(x)2

= v0z

(x

v0x

)+

12

g(

xv0x

)2

=g

2v20 cos2 α

x(x − R) (29)

Trajectory

z

xv0x

v0y~v0

v0x

v0x

−v0y~v0


Third Week Projectile on an Inclined Plane

Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α

Q: What is the distance |OP|?



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α

Q: What is the distance |OP|?To find the point P, we will use thefact that point P is both on theparabola describing the trajectory,and also on the line that describesthe hill.



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α

Q: What is the distance |OP|?First choose a coordinate axis.



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

αQ: What is the distance |OP|?First choose a coordinate axis.The initial velocity and accelerationin these new coordinate axes are:

~v0 = v0 cos(θ − α)x + v0 sin(θ − α)y

~a = g cosα(−y) + g sinα(−x)



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α Q: What is the distance |OP|?The velocity at time t can beobtained as

~v(t) = ~v0 +

∫ t

0~a(t ′)dt ′ = ~v0 + t~a

= [v0 cos(θ − α)− gt sinα] x+ [v0 sin(θ − α)− gt cosα] y



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α

Q: What is the distance |OP|?The velocity at time t can beobtained as

~v(t) = ~v0 +

∫ t

0~a(t ′)dt ′ = ~v0 + t~a

= [v0 cos(θ − α)− gt sinα] x+ [v0 sin(θ − α)− gt cosα] y

The position at time t is

~r(t) = ~r0 +

∫ t

0~v(t ′)dt ′

(30)



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α Q: What is the distance |OP|?The position at time t is

~r(t) = ~r0 +

∫ t

0~v(t ′)dt ′

=

[v0t cos(θ − α)− 1

2gt2 sinα

]x

+

[v0t sin(θ − α)− 1

2gt2 cosα

]y



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

αQ: What is the distance |OP|?Hence, if the object reaches thepoint P at time t0,

x0 = v0t0 cos(θ − α)− 12

gt20 sinα

y0 = v0t0 sin(θ − α)− 12

gt20 cosα

(30)



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α

Q: What is the distance |OP|?At point P, y0 = 0

v0t0 sin(θ−α)−12

gt20 cosα = 0 (30)

which has solutions t0 = 0 ort0 = 2v0 sin(θ−α)

g cosα



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α Q: What is the distance |OP|?At point P, y0 = 0

v0t0 sin(θ−α)−12

gt20 cosα = 0 (30)

which has solutions t0 = 0 ort0 = 2v0 sin(θ−α)

g cosα

t0 = 0 is the beginning of motion.The second solution is the solutionwe are looking for.



Example

Oα

~v0

θ

P(x0, y0)

y

O

x

v0y

v0x

θ − αay

ax

α Q: What is the distance |OP|?The distance |OP| = x0.

Using t0 = 2v0 sin(θ−α)g cosα

x0 = v0

(2v0 sin(θ − α)

g cosα

)sin(θ − α)

− 12

g(

2v0 sin(θ − α)

g cosα

)2

cosα

(30)


Third Week Relative Motion

Relative Motion

y

O x

y ′

O′

x ′

P

~r

~r ′

~R

From the definition of vector addition~r = ~R +~r ′.The displacement of the point P in atime ∆t is

∆~r = ∆~R + ∆~r ′ (31)

The velocities in the two referenceframes are related by ~v = ~v ′ + ~V



Relative Motion

y

O x

y ′

O′

x ′

P

~r

~r ′

~R


∆~r∆t

=∆~R∆t

+∆~r ′

∆t(31)




Relative Motion

y

O x

y ′

O′

x ′

P

~r

~r ′

~R


∆~r∆t

=∆~R∆t

+∆~r ′

∆t(31)



ASSUMPTIONS

∆~r and ∆~r ′ are measured in differentreference frames. We are assumingthat they are equal. Furthermore, weare assuming the ∆t is the same inboth reference frames.


Example

Question 3.78Raindrops make an angle θ with the vertical when viewed through amoving train window. If the speed of the train is ~vT , what is the speedof the raindrops in the reference frame of the Earth in which they areassumed to fall vertically?

Solution:

Let ~vR = −vz be the speed of the raindrops in the reference frame ofEarth, ~vE be the velocity of the Earth relative to the train, i.e. ~vE = −~vT .z

x

~vR~vE

~vRT

θ



Example


Solution:


x

~vR~vE

~vRT

θ

Let ~vRT be the velocity of theraindrops relative to train



Example


Solution:


x

~vR~vE

~vRT

θ It is given that ~vRT makes θ radianswith respect to the verticalFrom the figure, it is seen that

tan θ =vE

vR=⇒ vR = vT cot θ (32)


Third Week Reference Frames

Reference Frames

event: position+timeA reference frame is a coordinate axis (to measure the position ofan event)And a clock at each point of space (to measure the time of anevent)


Third Week Dynamics-Newton’s Laws of Motion

Dynamics-Newton’s Laws of Motion

1st Law: In an inertial reference frame, in the absence of any externalinfluences, the velocity of an object is constant

This is a definition of an inertial reference frame




1st Law: In an inertial reference frame, in the absence of any externalinfluences, the velocity of an object is constantThis is a definition of an inertial reference frame



Dynamics-Newton’s Laws of MotionInertial Reference Frame

To test if a given reference frame is inertial, consider a test objectEliminate all external influecens.Check to see if the object accelerates or notIf the object is not accelerating, that reference frame is an inertialreference frame

Given one inertial reference frame, any other frame that moves atconstant velocity relative to the inertial reference frame is inertial:

~v = ~V + ~v ′ =⇒ ~a = ~A + ~a′ (33)

If a given reference frame is an inertial reference frame, all objectsobey Newton’s 1st law in that frame




2nd Law: In an inertial reference frame, the acceleration of an object isproportional to the force acting on the object. The proportionalityconstant is 1

m where m is the mass of the object

~a =~Fm

(34)

3rd Law: If an object A exerts a force ~FAB on another object B, thenobject B also exerts a force ~FBA on object A whose magnitude is equalto the magnitude of ~FAB, but opposite in direction:

~FAB = −~FBA (35)




2nd and 3rd laws define the mass of anobject

By the 3rd law, the magnitudes of theforce acting on the standard massand the unknown mass are equal:Using 2nd law:

ma = msas (36)

Accelerations can be measuredexperimentally. Hence the unknownmass can be obtained as:

m = msas

a(37)

m ms

~as~a




Once the mass is defined, 2nd Law can be considered as thedefinition of the force.Also, if the force is given (by some means), the second law can beused to obtain acceleration.Unit of Force:

[~F ] = [m~a] = [m][~a] = kgms2 ≡ N(Newton) (38)




Once the mass is defined, 2nd Law can be considered as thedefinition of the force.Also, if the force is given (by some means), the second law can beused to obtain acceleration.Unit of Force:

[~F ] = [m~a] = [m][~a] = kgms2 ≡ N(Newton) (38)


ATTENTIONThere is no force due to acceleration!The force is the cause of acceleration!


Weight

m

~w

Weight, ~w , is the force acting on anobject due to gravity.Near Earth, all object accelerate withthe same acceleration ~g.By Newton’s second law, the forceacting on an object of mass m is

~w = m~g (39)



Example 1: Mass on a scale

scale

m

~w

~N

~N: unknown force acting on the bodyby the scale~w : force of gravity acting on the body

Free body diagram: A diagram ofmasses only with the forces acting oneach body shown separately



Example 1: Mass on a scale

scale

m

~w

~N

~N: unknown force acting on the bodyby the scale~w : force of gravity acting on the bodyFree body diagram: A diagram ofmasses only with the forces acting oneach body shown separately



Example 1: Free Body Diagram

scale

m

~w

~N

Object is not accelerating:

~Fnet = ~N + ~w ≡ 0

=⇒ ~N = −~w (40)

The scale shows the magnitude of theforce acting on it: | − ~N| = |~w | = mg



Example 2: Mass on a scale inside an Elevator

z

scale

~ae

m

~w

~N

~ae is the acceleration of the elevatorThe vectors in the problem are:

~N = Nz(Unknown) (41)~w = −mgz (42)~ae = aez (43)

(In drawing the figure, it is assumedthat ae < 0)If the mass m stays on the scale,~a = ~ae = aez



Example 2: Mass on a scale inside an Elevator

z

scale

~ae

m

~w

~N Net force acting on the mass:

~FT = (N −mg)z = maez=⇒ N =m(g + ae) (41)

The force acting on the scale is −~N,Scale will show a weight m(g + ae).



14th Century Bologna University



Learners and Learning

Herb Simon Nobel laureate, Social Scientist, one of the founders of AILearning results from what the student does and thinks andonly from what the student does and thinks. The teacher canadvance learning only by influencing what the student does tolearn.

Dylan William renowned UK expert on maths education

... teachers do not create learning, and yet most teachersbehave as if they do. Learners create learning. Teacherscreate the conditions under which learning can take place.



Example 3: Mass on an Inclined Plane

The object on the inclinedsurface

Oα ~w

~N

~N

~FTα

y

x

A block sits on a frictionless incline asshown in the figureThe forces acting on the mass are itsweight and the normal force




Free Body Diagram

Oα

~w

~N

~N

~FTα

y

xA block sits on a frictionless incline asshown in the figure

Free body diagram includes only themass and the forces




Oα ~w

~N

~N

~FT

α

y

x

A block sits on a frictionless incline asshown in the figure

The net force has to be along thesurface of the inclined plane




Oα

~w

~N

~N

~FTα

y

x


In terms of their components, theforces can be written as:

~N = Ny (Unknown) (42)~w = −mg cosαy −mg sinαx (43)

The net force is:

~FT = (N −mg cosα)y −mg sinαx(44)




Oα

~w

~N

~N

~FTα

y

x


The net force is:

~FT = (N −mg cosα)y −mg sinαx(42)

The acceleration along the y directionshould be zero, henceay = 0 =⇒ FTy = 0

N −mg cosα = 0 =⇒ N = mg cosα(43)




Oα

~w

~N

~N

~FTα

y

x


The net force is:

~FT = −mg sinαx (42)

Using Newton’s second law:

~a =~FT

m= −g sinαx (43)




Oα ~w

~N

~N

~FTα

y

x


|~a|

α

g

π2

Figure : |~a| = g sinα



Tension of a String

~F ~F ′T

Tension is the magnitude of the force acting on a stringIn the above figure, if the string has negligible mass (m = 0), than

~FT = ~F + ~F ′ = m~a = 0 (42)

A massless string transfers force along its length without changingits magnitude.



Example: Two masses attached by a massless string

Tm1 m2 ~F

~F1~F2

~w1

~N1

~w2

~N2




Tm1 m2

~F~F1~F2

~w1

~N1

~w2

~N2

No motion in the vertical direction: ~w1 + ~N1 = 0 and ~w2 + ~N2 = 0The string is massless (−~F1) + (−~F2) = 0 =⇒ ~F2 = −~F1

If the elasticity of the string is neglected, both masses should havethe same acceleration: ~F1 = m1~a, ~F + ~F2 = m2~a




Tm1 m2

~F~F1~F2

~w1

~N1

~w2

~N2

Since all the forces and accelerations are in the horizontaldirection, I will only write the horizontal components of each vector

F1 = m1aF + F2 = F − F1 = m2a

}F = (m1 + m2)a =⇒ a =

Fm1 + m2

(43)



Atwood’s Machine

m1

m2

~w1

~w2

~T2

~T1

z



Atwood’s Machine

m1

m2

~w1

~w2

~T2

~T1

z

Let accelerations be ~a1 = a1z, and~a2 = a2z.~T1 = T z, ~T1 = T z where T is the(unknown) tension of the string~a1 = a1z, ~a2 = a2z (ai ’s are unknown)~w1 = −m1gz, ~w2 = −m2gzFor the masses m1 and m2:~Ti + ~wi = mi~ai → T −mig = miai

The velocities of the masses have tohave equal magnitudes but oppositedirection:~v1 = −~v2 → ~a1 = −~a2



Atwood’s Machine

m1

m2

~w1

~w2

~T2

~T1

za2 = −a1 (44)

T −m1g = m1a1 (45)T −m2g = m2a2 (46)

Subtracting the second equation fromthe third and using the first:

(m1 −m2)g = m2a2 −m1a1

= −(m2+m1)a1

=⇒ a1 = −m1 −m2

m1 + m2g (47)



Accelerometer

m

θ

~v(t)

A ball of mass m is suspended from apoint by a massless string.

If the suspension point starts to movewith constant acceleration ~a, what isthe relation between the angle θ and|~a| ≡ a?



Accelerometer

m

θ

~v(t)A ball of mass m is suspended from apoint by a massless string.If the suspension point starts to movewith constant acceleration ~a, what isthe relation between the angle θ and|~a| ≡ a?



Accelerometer

~w

~T

y

x

αOnce the oscillations settle down, theacceleration of the ball is equal to theacceleration of the suspension pointax



Accelerometer

~w

~T

y

x

α

~w = −mgy , ~T = T cosαy + T sinαx

~FT = (T cosα−mg)y + T sinαx(48)

By Newton’s second law: ~FT = max :

T cosα−mg = 0 =⇒ T =mg

cosα(49)

T sinα = ma =⇒ mg tanα = ma(50)

Hence tanα = ag


Fourth Week IMPORTANT NOTES:

IMPORTANT NOTES

Acceleration is in the direction of change in velocity (eithermagnitude and/or direction)Force is in the direction of accelerationIF there is no force, then there is no acceleration, i.e. no change invelocityALWAYS: a force is acted by something!


Fourth Week Friction

Friction

Electromagnetic in OriginVarious type of friction:

Kinetic frictionStatic frictionRolling friction

Friction always tries to oppose relative motion between surfaces incontact



Kinetic Friction

Exists if two surfaces in contact are in relative motionExperimental Observation:

|~Ff | = µk |~N| (51)

(Note: not a vectorial equation)The direction is parallel to the surface of contactµk :Coefficient of kinetic frictionIndependent of the contact area



Static Friction

The friction between two surfaces that are not in relative motion:

|~Ff | ≤ µs|~N| (52)

µs: Coefficient of static frictionUsually µs > µk


Fourth Week Examples

Example 1

The object on the inclinedsurface

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

Assume that initially the object isat rest on an inclined surface

The forces acting on the mass:

~N = Ny (Unknown) (53)~w = −mg cosαy −mg sinαx

(54)~Ffr = Ffr x (55)

(56)



Example 1

Free Body Diagram

Oα

~w

~N

~Ffr

~N

~Ffr~FT

α

y

x



(54)~Ffr = Ffr x (55)

(56)



Example 1

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x



(54)~Ffr = Ffr x (55)

The net force acting on the mass:

~FT = (N −mg cosα)y+ (Ffr −mg sinα)x (56)

(57)



Example 1

Oα

~w

~N

~Ffr

~N

~Ffr~FT

α

y

xThe net force acting on the mass:

~FT = (N −mg cosα)y+ (Ffr −mg sinα)x (53)

ay = 0 =⇒ N = mg cosα (54)

ax =Ffr −mg sinα

m(55)



Example 1

Oα

~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ay = 0 =⇒ N = mg cosα (53)

ax =Ffr −mg sinα

m(54)

If Ffr = mg sinα

No acceleration, the object stays atrest

mg sinα = Ffr ≤ µsN =µsmg cosα =⇒ tanα < µs

If α is such that tanα > µs?



Example 1

Oα

~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ay = 0 =⇒ N = mg cosα (53)

ax =Ffr −mg sinα

m(54)

The object will acceleratedownwards: ~a = ax x ,

ax =Ffr −mg sinα

m=µk N −mg sinα

m

=µk mg cosα−mg sinα

m(55)



Example 1

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ay = 0 =⇒ N = mg cosα (53)

ax =Ffr −mg sinα

m(54)


ax =Ffr −mg sinα

m=µk N −mg sinα

m

=µk 6mg cosα− 6mg sinα

6m

= −g sinα(

1− µk

tanα

)(55)



Example 1

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ay = 0 =⇒ N = mg cosα (53)

ax =Ffr −mg sinα

m(54)


ax =Ffr −mg sinα

m=µk N −mg sinα

m

=µk 6mg cosα− 6mg sinα

6m

= −g sinα(

1− µk

tanα

)(55)


NOTEIn this case, it is assumed thattanα > µs > µk . Hence

1− µk

tanα> 0


Example 1

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ax = −g sinα(

1− µk

tanα

)(53)

If µs > tanα > µk , the object willnot slide if initially at rest, but willaccelerate along the −x directionif initially moving along the −xdirection.µs > µk > tanα, will acceleratealong the +x direction, if initiallythe object is moving downwardsalong the −x direction



Example 1

Oα ~w

~N

~Ffr

~N

~Ffr~FT

α

y

x

ax = −g sinα(

1− µk

tanα

)(53)

If initially the object is movingalong the +x direction, Ffr willpoint in the opposite direction:µk → −µk

ax = −g sinα(

1 +µk

tanα

)(54)

The object will always have anacceleration along the −xdirection



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

A mass m1 is on an inclined plane. Asecond mass m2 is attached to thefirst mass through a massless stringand pulley system. The surface of theinclined plane has friction.



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

Free body diagrams for the twomasses



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′ The forces acting on the masses:

~w1 = −m1g sinαx −m1g cosαy(55)

~N = Ny (Unknown) (56)~T1 = T x(Unknown) (57)~Ffr = Ffr x(Unknown) (58)~w2 = −m2gx ′ (59)~T2 = T x ′(Unknown) (60)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′The forces acting on the masses:

~w2 = −m2gx ′ (55)~T2 = T x ′(Unknown) (56)

Newton’s Laws on the second mass:

~a2 = a2x ′ (57)~FT 2 = (T −m2g)x ′ ≡ m2a2x ′ (58)

=⇒ T −m2g = m2a2 (59)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′The forces acting on the masses:

~w1 = −m1g sinαx −m1g cosαy(55)

~N = Ny (Unknown) (56)~T1 = T x(Unknown) (57)~Ffr = Ffr x(Unknown) (58)

Newton’s Laws on the first mass:

~a1 = a1x (59)(60)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

Newton’s Laws on the first mass:

~a1 = a1x (55)~FT1 = (T −m1g sinα + Ffr )x (56)

+ (N −m1g cosα)y ≡ m1a1x(57)

=⇒N −m1g cosα = 0 (58)T −m1g sinα + Ffr = m1a1 (59)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′ Results of Newton’s Laws:

T −m2g = m2a2 (55)N −m1g cosα = 0 (56)

T −m1g sinα + Ffr = m1a1 (57)

Unknowns: T , a1, a2, N, Ffr .Total 5 unknowns, need two moreequations:

a2 = −a1 (58)|Ffr | ≤ µN (59)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

The direction of friction force shouldbe determined.If the first mass is moving, ~Ffr willpoint in the opposite direction to itsvelocityIf the first mass is at rest, ~Ffr will try toprevent it from starting to moveFirst ignore friction to determinewhich direction the system will try tomove



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

Ignoring Ffr , the equations for the twounknowns T and a1 are:

T −m2g = −m2a1 (55)T −m1g sinα = m1a1 (56)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′Ignoring Ffr , the equations for the twounknowns T and a1 are:

T −m2g = −m2a1 (55)T −m1g sinα = m1a1 (56)

Subtract the first equation from thesecond one to get:

a1 = gm2 −m1 sinα

m1 + m2(57)



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′



m1 + m2(55)

If the system is at rest andm2g ≥ m1g sinα, the first mass willtry to move in the +x direction, hencethe friction force will be in the −xdirection: Ffr ≤ 0



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′



m1 + m2(55)

If the system is at rest andm2g ≤ m1g sinα, the first mass willtry to move in the −x direction, hencethe friction force will be in the +xdirection: Ffr ≥ 0.



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′



m1 + m2(55)

To proceed, assume the system isinitially at rest and m2 > m1 sinα



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

T −m2g = −m2a1 (55)N = m1g cosα

(56)




Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

To see if the system will stay at rest ornot, set a1 = 0.T = m2gFfr = (m1 sinα−m2)g (we hadalready assumed that m2 > m1 sinα)If |Ffr | = (m2 −m1 sinα)g ≤ µsN,then the system will stay at restOtherwise if (m2 −m1 sinα)g > µsN,then the system will start moving.



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′

T −m2g = −m2a1 (55)N = m1g cosα

(56)


After the system starts moving,Ffr = −µkN = −µkm1g cosαm2g −m1g sinα− µkm1g cosα =(m1 + m2)a1



Example 2

O α

m2

~w2

m1

~w1

~N

~Ffr

~T2

~T1

α

y

x

x ′ Solving for a1:


m1 + m2(55)

− m1

m1 + m2µkg cosα (56)

Compare with the previous resultwithout the mass m2 (m2 = 0)

ax = −g sinα− gµk cosα (57)


Fourth Week Uniform Circular Motion

Uniform Circular Motion

R

~v1

~v2O

∆θ

∆`

~v1

~v2

∆~v∆θ

α

∆s

∆s = v∆θ

∆` = R∆θ

∆` = v∆t

The object moves on a circle of constant radius with constantspeed.The velocity is constantly changing =⇒ ~a 6= 0




R

~v1

~v2O

∆θ

∆`

~v1

~v2

∆~v∆θ

α

∆s

∆s = v∆θ

∆` = R∆θ

∆` = v∆t

The average acceleration is in the direction of ∆~vFor instantaneous acceleration

α = lim∆t→0

π −∆θ

2=π

2(58)

~a is perpendicular to ~v




R

~v1

~v2O

∆θ

∆`

~v1

~v2

∆~v∆θ

α

∆s

∆s = v∆θ

∆` = R∆θ

∆` = v∆t

Magnitude of ~a:

|~a| ≡ a = lim∆t→0

|∆~v |∆t

= lim∆t→0

∆s∆t

= lim∆t→0

v∆θ

∆t= v

vR

=v2

R(58)

ω = lim∆t→0∆θ∆t ≡

vr is called angular velocity



Let ~r be the position vector of the mass.R2 = ~r ·~r

0 =dR2

dt= ~r · d~r

dt+

d~rdt·~r

= 2~r · ~v (59)

Hence ~v ⊥ ~r .v2 = ~v · ~v

0 =dv2

dt= ~v · d~v

dt+

d~vdt· ~v

= 2~v · ~a (60)

Hence ~a ⊥ ~v



Let ~r be the position vector of the mass.Let θ be the angle that ~r makes with the x axis:

~r = R(cos θx + sin θy) ≡ Rr (59)

~v =d~rdt

= Rdθdt

(− sin θx + cos θy) ≡ Rdθdtθ (60)

|~v | = R dθdt ≡ v constant =⇒ ω ≡ dθ

dt = vR is constant

The acceleration:

~v = v(− sin θx + cos θy) (61)

~a ≡ d~vdt

= vdθdt

(− cos θx − sin θy) ≡ −vωr (62)



Circular Motion Dynamics

Magnitude of acceleration of a uniform circulating body is

a =v2

R(63)

Its direction points towards the center of circleA force has to be applied to the object to create this acceleration.By Newton’s second Law, the magnitude of this force:

|~F | = mv2

R(64)

This force has to be directed towards the center of circle



Conical Pendulum

θ

L

m

z

x

~Ft

~w

~wFT

θ

A mass is attached to a masslessstring. The mass makes uniformcircular motion with speed v .Calculate v and the tension on thestring



Conical Pendulum

θ

L

m

z

x

~Ft

~w

~wFT

θ


~w = −mgz (65)~Ft = T cos θz − T sin θx (66)~FT = (T cos θ −mg)z − T sin θx

(67)

≡ −mv2

Rx (68)



Conical Pendulum

θ

L

m

z

x

~Ft

~w

~wFT

θ

~FT = (T cos θ −mg)z − T sin θx

(65)

≡ −mv2

Rx = −m

v2

L sin θ(66)

=⇒{

T cos θ −mg = 0−T sin θ = −m v2

L sin θ(67)



Conical Pendulum

θ

L

m

z

x

~Ft

~w

~wFT

θ

T cos θ −mg = 0 =⇒ T =mg

cos θ(65)

T sin θ = mv2

L sin θ(66)

=⇒ gL sin θ tan θ = v2

(67)

Period of motion is T = 2πRv

Check the units and the limits v → 0and v →∞


Fourth Week Skidding on a Curve

Skidding on a Curve

car �~v

~Ffr

~N

~w

x

y

A car of mass m takes a curve whoseradius of curvature is R. What is itsmaximum velocity such that it will notskid?



Skidding on a Curve

car �~v

~Ffr

~N

~w

x

y

Forces acting on a car:

~Ffr = µsN (68)~N = Ny (69)~w = −mgy (70)~FT = (N −mg)y + µsNx (71)



Skidding on a Curve

car �~v

~Ffr

~N

~w

x

y

FT = (N −mg)y + µsNx ≡ m v2maxR x

N −mg = 0 (68)

µsN = mv2

maxR

(69)

=⇒ vmax =√µsRg (70)


Fourth Week Banked Curves

Banked Curves

car �~v

θ

~N

~wFfr

θ

x

y

What should be the value of θ, suchthat a car moving at a speed v , canturn a curve with radius R withoutskidding? Ignore friction.



Banked Curves

car �~v

θ

~N

~wFfr

θ

x

y

The forces acting on the car:

~N = N sin θx + N cos θy (71)~Ffr = 0 (72)~w = −mgy (73)~FT = N sin θx + (N cos θ −mg)y

= mv2

Rx (74)



Banked Curves

car �~v

θ

~N

~wFfr

θ

x

y

~FT = N sin θx + (N cos θ −mg)y

= mv2

Rx

=⇒{

N cos θ −mg = 0N sin θ = m v2

R

=⇒ g tan θ =v2

R



QUIZ2

m1

m2

Q: Two masses m1 = 1 kg and m2 = 2 kgare attached to each other by a masslessstring through a massless pulley as shownin the figure. Ignore air friction and takeg = 9.80 m/s2

1 Draw the free body diagrams for eachfigure

2 Write the forces acting on each of themasses in terms of their components

3 Calculate numerically theacceleration vectors of each of themasses



QUIZ2

m1

m2

Solutions:1 Draw the free body diagrams for each

figure

m1

~w1

~N

~Ffr~T1

m2

~T2

~w2



QUIZ2

m1

m2

Solutions:2 Write the forces acting on each of the

masses in terms of their components

m1

~w1

~N

~Ffr~T1

m2

~T2

~w2

x

y

~N = Ny ; ~Ffr = −Ffr x~T1 = T x ; ~T2 = T y~w1 = −m1gy~w2 = −m2gy



QUIZ2

m1

m2

Solutions:3 Calculate numerically the

acceleration vectors of each of themasses

~N = Ny ; ~Ffr = −Ffr x~T1 = T x ; ~T2 = T y~w1 = −m1gy~w2 = −m2gy

~FT 1 = T x + (N −m1g)y

≡ m1a1x

~FT 2 = (T −m2g)y

≡ m2a2ya1 = −a2



QUIZ2

m1

m2



~FT 1 = T x + (N −m1g)y ≡ m1a1x~FT 2 = (T −m2g)y ≡ m2a2y

a1 = −a2

N −m1g = 0T = m1a1

T −m2g = −m2a1



QUIZ2

m1

m2



N −m1g = 0T = m1a1

T −m2g = −m2a1

m1a1 −m2g = −m2a1

=⇒ a1 = gm2

m1 + m2=⇒ ~a1 = g

m2

m1 + m2x

~a1 =(

9.80ms2

) 23

x =(

6.53ms2

)x



COMMON MISTAKES

Ignoring vector signs on vectors: ~F = m~a is NOT the same asF = ma.Adding vector sign for objects that are not vectors: ~m for the massEquating vectors to numbers: F = m~aIgnoring units: if m = 1 kg, F = ma = aa = 5 m/s2 gets full points a = 5 gets zero points!Putting units where not necessary: a = v

r2 m/s2 gets zero points

Adding units at the end: a = 6/2 = 3 m/s2 gets zero points


Fifth Week Terminal Velocity

Terminal Velocity

In liquids and gases, friction is not constant, but velocitydependent.For small velocities ~FD = −b~vConsider a mass m left from rest at some height. (1D motion)Newton’s second law:

mg − bv = ma ≡ mdvdt

(71)

(72)

If v = mg/b, a = 0. The terminal velocity is vt = mg/b



Terminal Velocity

Newton’s second law:

mg − bv = ma ≡ mdvdt

(71)

(72)

If v = mg/b, a = 0. The terminal velocity is vt = mg/bFor any other velocity

mdv

mg − bv= dt (73)



Terminal Velocity

For any other velocity

mdv

mg − bv= dt (71)

Integration both sides from ti to t (vi to v(t))∫ t

tidt =

∫ v(t)

vi

dvmg − bv

(72)

=⇒ (t − ti) =mb

logmg − bvi

mg − bv(t)(73)



Terminal Velocity

Integration both sides from ti to t (vi to v(t))∫ t

tidt =

∫ v(t)

vi

dvmg − bv

(71)

=⇒ (t − ti) =mb

logmg − bvi

mg − bv(t)(72)

Solving for v(t):

v(t) = vt − e−bm (t−ti )(vt − vi) (73)

= vie−bm (t−ti ) + vt

(1− e−

bm (t−ti )

)(74)


Fifth Week Gravity and Planetary Motion

Kepler’s Laws

Kepler’s laws are based on observation only:1 The orbit of planets around the sun are ellipses with the Sun

positioned at one of the centers2 The vector from the sun to the planet, sweeps equal areas at

equal times3 Let si and Ti , i = 1,2 be the semi major axis and the period of

rotation respectively, of two planets. Then(T1

T2

)2

=

(s1

s2

)3

(75)

or

T 2

s3 (76)

is the same for every planet.Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 93 / 175


Kepler’s First Law

P1P2

Sun

F1

F2

b a

a (b) is the semi minor (major) axisDefinition of an ellipse:|F1P1|+ |P1F2| = |F1P2|+ |P2F2| ≡ 2b



Kepler’s Second Law

P1

P2

P3 P4A2

A1

t12 (t34) time it takes for the planet togo from P1 (P3) to P2 (P4)If t12 = t34 then A1 = A2.



Kepler’s Second Law

r

δs = |~v |δtπ − θ

The area covered in time interval δt isδA = 1

2 rδs sin(π − θ) = 12 rv sin θδt

δA is the same independent of where the planet is on its orbitAs the planet moves, rv sin θ is constant.rv sin θ = |~r × ~v |



Kepler’s Third Law

R

T 2

s3 is constantConsider a circular orbit s = RT = 2πR

v

Kepler’s Law:(2πR

v

)2( 1R3

)=

2πRv2 =

2π

R2 v2

R

=⇒ v2

RR2 = constant (77)

=⇒ |~F |R2 = constant (78)

Kepler’s second law implies that the central force decreases with thesquare of the distance



Newton’s Law of Gravitation

Kepler’s third Law =⇒ F ∝ 1r2

Law of uniform gravitational acceleration =⇒ F = mg ∝ mSymmetry of forces (action reaction pairs) −→ F ∝ mE

|~F | = GNmmE

r2 (79)

where m and mE are the masses of two gravitating objects, r isthe distance between their centres.GN = 6.67384× 10−11N(m/kg)2



Newton’s Law of Gravitation

m1

r12m2

r12

r21

~F12: Force acting on m1 due to m2

~F12 = GNm1m2

r212

r12 (80)

~F21: Force acting on m2 due to m1

~F21 = GNm2m1

r221

r21 ≡ ~F12 (81)

On the surface of the Earth, the force acting on a mass m is:

|~F | = mg = GNmmE

RE

2=⇒ g = GN

mE

R2E

(82)


Sixth Week

A MATHEMATICAL FORMULAS A-IB DERIVATIVES AND INTEGRALS A-6C MORE ON DIMENSIONAL ANALYSIS A-SD GRAVITATIONAL FORCE DUE TO A SPHERICAL MASS

DISTRIBUTION A-9E DIFFERENTIAL FORM OF MAxwELLS EQUATIONS A-12F SELECTED ISOTOPES A-14


Sixth Week

Measured value of g is position dependent:Shape of earth is not a sphereMass density is not uniformEarth is rotating, i.e. any reference frame fixed on the surface ofthe Earth is non-inertial. At the poles, g would be measured largerthan on the equator.


Sixth Week

~F12 = GNm1m2

r212

r12 is valid for point masses

If various masses mi exert gravitational attraction on a mass M,the total force acting on M is:

~F = GN∑

i

miMr2i

ri (83)

where ri is the distance of mass mi from M, and ri is the unitvector pointing from M towards mi .Superposition of forces is valid only in Newton’s Theory of gravitySuperposition of forces is not valid on General Theory of Relativity


Sixth Week

Gravitational Force of a Ring on a Mass

C

R

δθ mL

D

~F1

~F2

δm

α

δm = M δθ2π

|~F1| = |~F2| = GNδmmD2

~F1 + ~F2 will point from the mass m tothe center of the ring with amagnitude F‖ = GN

(2δm)mD2 cosα

F‖ does not depend on which δmalong the ring is consideredcosα = L

D

Summing over all δm gives

|~F | = GNMmD2 cosα = GN

MmLD3 = GN

MmL

(L2 + R2)32

(84)

~F points towards the center of the ring


Sixth Week

Gravitational Force of a Shell on a Mass

R

`

r

R

z The ring has has its center at z = kand sees and angle dθ.The area of the ring: dA = 2πRrdθR2 + (R − k)2 = `2, R2 + k2 = r2

The mass of the ring:dM = dA

4πr2 M = RM2r dθ

The force due to the ring:

d~F = −GNdMmL`3

z = −GNMm

r(R − k)R

2`3dθz (85)


Sixth Week


R

`

r

R

z The ring has has its center at z = kand sees and angle dθ.The area of the ring: dA = 2πRrdθR2 + (R − k)2 = `2, R2 + k2 = r2

The mass of the ring:dM = dA

4πr2 M = RM2r dθ

The force due to the ring:

d~F = −GNdMmL`3

z = −GNMm

r(R − k)R

2`3dθz (85)


ATTENTIONIn deciding the sphere into rings, once can either assumethat dk is constant, or rdθ is constant for each ring. Whenthe ring is opened into an approximate rectangle, rdθ is theheight of this rectangle. Hence it is better (and simpler) toassume that rdθ is constant for each ring. This implies thatdk , (i.e. the distance between the centers of the inner andthe outer circles of the ring) i not constant. If one takes dkconstant, especially closer to the point k = R, when youopen the ring, it will no longer look like a rectangle

Sixth Week


R

`

r

R

zd~F = −GN

Mmr

(R − k)R2`3

dθz (85)

`2 = r2 + R2 − 2rR cos θ (86)

k = r cos θ =R2

+r2 − `2

2R(87)

R = r sin θ (88)

`d` = rR sin θdθ = RRdθ =⇒ dθ = `RR

d`

d~F = −GNMmrR

(R − k)

2`2d` (89)

= −GNMm2rR

(R2− r2 − `2

2R

)d``2

(90)Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 104 / 175

Sixth Week


R

`

r

R

z

d~F = −GNMm2rR

(R2 −

r2−`22R

)d``2

z

If the point is outside the shell, R − r ≤ ` ≤ R + r ,~F =

∫ R+rR−r d`(· · · ) = −GN

MmR2 z

If the point is inside the shell, r − R ≤ ` ≤ R + r ,~F =

∫ R+rr−R d`(· · · ) = 0

Inside the shell, zero gravitational attraction, outside the shell,shell acts as if all its mass in at its center


Sixth Week

Planet Vulcan

Planet Vulcan was hypoth-esized to exist between theSun and Mercury


Sixth Week

Dark Matter

Consider a simple model of a galaxy as a sphere of uniform massdensity. The galaxy will spiral around its center. consider a star on itsequilateral plane at a distance r from the center. Calculate and sketchits speed as a function of r .

Rr

m

If the star is inside the galaxy, blockwill feel the the attraction for only themass inside the sphere of radius r :m(r) = 4π

3 ρr3

This is the centripetal force:

GNm(4π

3 ρr3)r2 = m

v2

r(85)

=⇒ v = r(

GN4π3ρ

) 12

(86)


Sixth Week

Dark Matter

Consider a simple model of a galaxy as a sphere of uniform massdensity. The galaxy will spiral around its center. consider a star on itsequilateral plane at a distance r from the center. Calculate and sketchits speed as a function of r .

Rr

m

If the star is outside the galaxy, then

GNm(4π

3 ρR3)r2 = m

v2

r(85)

=⇒ v =1√r

(GN

4π3

R3)

(86)


Sixth Week

Dark Matter

The galaxy appears to havea larger mass than is seen

v(r)

r

∝ r ∝ 1√r

R


Sixth Week

Gravitational Field

According to Newton’s Gravitation Law, gravity acts at a distanceTo avoid the concept of action at a distance, gravitational field ishypothesizedEvery mass, M, creates a field ~g around it given by

~g = −GNMr2 r (85)

Every other mass m placed in this field, feels a force due to thefield at its location given by

~F = m~g (86)


Sixth Week

Equivalence Principle

Mass appears in Newton’s second Law: ~F = m~a: inertial massMass appears in Newton’s law of gravity: ~g = −GN

mr2 r :

gravitational massEquivalence principle: gravitational mass and inertial mass areequal. WHY?Einstein’s theory of relativity relies on this equality


Sixth Week

Example: A mass Inside a Spherical Mass

rm

R

The shown sphere has a radius R and amass M uniformly distributed over itssurface. Another mass m is placed at adistance r < R from the center. What willbe the gravitational force that the objectwill feel?


Sixth Week

Example: A mass Inside a Spherical Mass

rm

R

Divide the sphere into an innersphere with radius r and outer shell.Outer shell will not exert any force.Inner shell will have a mass:

M(r) =M

43πR3

43πr3 = M

( rR

)3

The force exerted by the inner shell is

~F = −GNM( r

R

)3 mr2 r = −GN

MmR3

~r


Sixth Week

Example: A sphere with another sphere carved out

R2

R1

~d

A sphere of radius R2 is carved out ofanother sphere of radius R1. The positionof the center of the carved sphere isdenoted by ~d . The mass density of thesystem is ρ. If a mass m is placed insidethe cavity, what will be the force that thismass m will feel?


Sixth Week


R2

R1

~d~r2

~r1

Let ~r1 (~r2) be the position of the massm relative to the center of the large(small) sphere.


Sixth Week


R2

R1

~d~r2

~r1

Let ~r1 (~r2) be the position of the massm relative to the center of the large(small) sphere.~Ffull sphere = ~FT + ~Fcarved out mass =⇒~FT = ~Ffull sphere − ~Fcarved out mass

The cavity can be modeled as a masswith mass density −ρ.


Sixth Week


R2

R1

~d~r2

~r1

~Ffull sphere = ~FT + ~Fcarved out mass =⇒~FT = ~Ffull sphere − ~Fcarved out mass

The cavity can be modeled as a masswith mass density −ρ.Large sphere:~FL = −GN

ρ 43πr3

1r21

r1 = −4π3 GNρ~r1

Small sphere: ~Fs = −4π3 GN(−ρ)~r2

~FT = ~FL + ~Fs = −4π3 GNρ(~r1 −~r2) = 4π

3 GNρ~dGravitational attraction is uniform inside the cavity.CHALLENGE: Can you prove this without using vectors? (notrecommended)


Sixth Week

Example: Circular Orbits

R

Let ME (Ms) be the mass of Earth(satellite)What is the speed of the satellite?

m v2

R = GNMmR2 =⇒ v =

√GNM

R

The closer the satellite is to the Earth,the faster it should be.

The period of the satellite is

T =2πR

v=

2π√GNM

R32 =⇒ T 2

R3 =2π√GNM

Measuring the ratio T 2/R3, it is possible to determine the mass ofthe sun.


Sixth Week

Example: Circular Orbits

R

Let ME (Ms) be the mass of Earth(satellite)What is the speed of the satellite?

m v2

R = GNMmR2 =⇒ v =

√GNM

R

The closer the satellite is to the Earth,the faster it should be.

The period of the satellite is

T =2πR

v=

2π√GNM

R32 =⇒ T 2

R3 =2π√GNM

Geocentric orbits are those for which the relative position of thesatellite is fixed with respect to the surface of the planet.


Sixth Week

Weightlessness

z

scale

m

~w

~N

The forces acting on the mass m are:

~w = −mgz (87)~N = Nz (88)~FT = (N −mg)z (89)

where g is the gravitationalacceleration at the point of the massm.

In a satellite, whole system accelerates. If the acceleration is alsoin the z direction, ~FT = m~a ≡ mazThen N −mg = ma =⇒ N = m(g + a)

If a = −g, the object appears massless


Sixth Week

Weightlessness

z

scale

m

~w

~N The forces acting on the mass m are:

~w = −mgz (87)~N = Nz (88)~FT = (N −mg)z (89)

where g is the gravitationalacceleration at the point of the massm.

In a satellite, whole system accelerates. If the acceleration is alsoin the z direction, ~FT = m~a ≡ mazThen N −mg = ma =⇒ N = m(g + a)

If a = −g, the object appears massless


Sixth Week

QUIZ 3

Q: Consider a mass m attached at the end of a massless string. Thestring has a length L. The mass is moving uniformly around a circle ofradius R. (ignore gravity and friction)

y

xθ

m1 Draw the free body diagram at the

shown instant.2 Write the force(s) acting on the mass

m in terms of their components, i.e. inthe form ~F = Fx x + Fy y . Use thegiven coordinate axes.

3 Find a relation between T , L and v .


Seventh Week

MIDTERM

Date: November 16, 2013 (this saturday)Time: 13:30Duration: 3 HoursPlace: U1, U2 and U3Topics: Everything we have covered until the midtermFormulas will be provided in the examYou will use two points for any misplaced vector sign, units, etc.,e.g. You solve a problem of 10 points “correctly”. You misplace 5vector sign and one unit. You will get “-2” points.


Seventh Week Work and Energy

Definition

Consider a mass m and a constant force ~F acting on it. Under theinfluence of this force, the mass is displaced by ∆~r . The work done bythe force is

∆W = ~F ·∆~r

/ The unit of work is [W ] = Nm ≡ J(oule).

Scalar Product-Review

The scalar product of ~A and ~B is defined as ~A · ~B = AB cos θ whereA = |~A|, B = |~B| and θ is the angle between the vectors ~A and ~B.

~A

~B

θ



Definition

Consider a mass m and a constant force ~F acting on it. Under theinfluence of this force, the mass is displaced by ∆~r . The work done bythe force is

∆W = ~F ·∆~r

/ The unit of work is [W ] = Nm ≡ J(oule).

Scalar Product-ReviewIn terms of their components, if

~A = Ax x + Ay y + Az z (90)~B = Bx x + By y + Bz z (91)

then ~A · ~B = AxBx + AyBy + AzBz(≡ AB cos θ)



Work Done on a Block

m

~Ff

~w

~N

y

x

~F

θ

The surface has friction, and thecoefficient of kinetic friction is µk

If the block moves horizontally by ∆xthen the work done on the mass byvarious forces are:




m

~Ff

~w

~N

y

x

~F

θ



The work done by ~F :

WF = ~F ·∆~r = F∆x cos θ




m

~Ff

~w

~N

y

x

~F

θ



The work done by gravity:

Ww = ~w · ~∆r = mg∆x cosπ

2= 0




m

~Ff

~w

~N

y

x

~F

θ



The work done by the Normal force

WN = ~N · ~∆r = N∆x cosπ

2= 0




m

~Ff

~w

~N

y

x

~F

θ



The work done by the friction force is:

Wf = ~Ff ·∆~r = Ff ∆x cosπ = −Ff ∆x = −µkmg∆x




m

~Ff

~w

~N

y

x

~F

θ



The total work done on the mass is:

WT = F∆x cos θ − µkmg∆x = (F cos θ − µkmg)∆x



Some Examples

QuestionSuppose you hold your physics book, and walk on a flat surface withthe book for 5 secs. If your speed is always constant during this time,what is the work done on the book by you? (Take the mass of the bookto be 1 kg)

Answer

None. The force you apply to the book,~F , is vertical, the displacementof the book, ∆~r is horizontal. Hence W = ~F ·∆~r = F∆r cos π

2 = 0



Some Examples

QuestionSuppose you hold your physics book, and walk on a flat surface withthe book for 5 secs. If during this time, your acceleration is ~a = ax witha = 0.1 m/s2, what is the work done on the book by you?(Take themass of the book to be 1 kg)

Answer

Since the book is accelerating, You should be applying an additionalhorizontal force with magnitude F = (1 kg)(0.1 m/sec2) = 0.1 N. Thehorizontal distance covered in the x direction during 5 sec is∆x = 1

2at2 = 12(0.1 m/s2)(5 s)2 = 2.5 m. The force and the

displacement are in the same direction, hence the work done on thebook by you is W = (0.1 N)(2.5 m) = 0.25 J



Some Examples

QuestionSuppose you hold your physics book, and walk on a flat surface withthe book for 5 secs. If during this time, your acceleration is ~a = ax witha = 0.1 m/s2, what is the work done on the book by you?(Take themass of the book to be 1 kg)

AnswerSince the book is accelerating, You should be applying an additionalhorizontal force with magnitude F = (1 kg)(0.1 m/sec2) = 0.1 N. Thehorizontal distance covered in the x direction during 5 sec is∆x = 1

2at2 = 12(0.1 m/s2)(5 s)2 = 2.5 m. The force and the

displacement are in the same direction, hence the work done on thebook by you is W = (0.1 N)(2.5 m) = 0.25 J



Some Examples

QuestionSuppose you take the elevator with the book from the ground floor tothe first floor which is 3 m above. What is the work done on the bookby you? by its weight? Assume that the the elevator is always movingvery slowly so that neglect any acceleration.

Answer

If the upward direction is taken as the positive z direction, then theforces acting on the book are: The force of the hand: ~Fh = mgz; theweight of the book: ~w = −mgz; the displacement of the book:∆~r = ∆xz where ∆x = 3 m. The work done by these forces are:

Wh = ~Fh ·∆~r = mh∆x (92)Ww = ~w ·∆~r = −mg∆x (93)



Some Examples

QuestionSuppose you take the elevator with the book from the ground floor tothe first floor which is 3 m above. What is the work done on the bookby you? by its weight? Assume that the the elevator is always movingvery slowly so that neglect any acceleration.

AnswerIf the upward direction is taken as the positive z direction, then theforces acting on the book are: The force of the hand: ~Fh = mgz; theweight of the book: ~w = −mgz; the displacement of the book:∆~r = ∆xz where ∆x = 3 m. The work done by these forces are:

Wh = ~Fh ·∆~r = mh∆x (92)Ww = ~w ·∆~r = −mg∆x (93)



Some Examples

QuestionIf you are just holding your book, but not doing anything else, you arenot doing any work. Why do you get tired if you are not doing anywork?

Answer


Seventh Week Work Done By a Variable Force

Work Done By a Variable Force

P1

P2

∆~

~F

Divide the path of the object into infinitesimal segments ∆~

Within each segment, the force is (almost) constant

∆W = ~F (~r) ·∆~

The total work is the sum of all ∆W in the limit |∆~| → 0

W =

∫ Pf

Pi

~F · d ~



Work Done By a Spring

x

If the mass is displaced by x from its equilibrium position, thestring exerts a force of magnitude, F = kx . Its direction is towardsthe equilibrium position (Hooke’ Law) :

~F = −kxx .

k is called the spring constant.




x

If the mass is displaced by x from its equilibrium position, thestring exerts a force of magnitude, F = kx . Its direction is towardsthe equilibrium position (Hooke’ Law) :

~F = −kxx .

k is called the spring constant.If the mass is initially at the position x and is displaced by dx , then∆~r = dxx , hence,

∆W = ~F ·∆~r = (−kxx) · (dxx) = −kxdxAltug Özpineci ( METU ) Phys109-MECHANICS PHYS109 119 / 175



x

If the mass is initially at the position x and is displaced by dx , then∆~r = dxx , hence,

∆W = ~F ·∆~r = (−kxx) · (dxx) = −kxdx

Summing over x from x = 0 upto x = L, the total work done by thestring is

Ws =

∫ L

0(−kxdx) = −k

x2

2

∣∣∣∣x=L

x=0= −1

2kL2




x

Summing over x from x = 0 upto x = L, the total work done by thestring is

Ws =

∫ L

0(−kxdx) = −k

x2

2

∣∣∣∣x=L

x=0= −1

2kL2

If a force ~Fext is applied to displace the mass with zeroacceleration, then ~Fext = −~F and hence

Wext = −Ws =12

kL2



Kinetic Energy and Work-Energy Principle

In the previous exercise, assume that at time t , the mass is atposition x , with velocity ~vi = vi x . Its acceleration is −kx/m.In time dt , it will be displaced by δx = vi +vf

2 dt . Its new speed willbe vf = vi − kx/mdtThe work done on the mass is

dW = −kxdx = −kxvi + vf

2dt = kx

vi + vf

2mkx

(vf − vi) (92)

=m2

v2f −

m2

v2i = d

(12

mv2)

=⇒W = ∆

(12

mv2)

(93)

12mv2 is defined as the kinetic energy of the particleThe work done on a mass is equal to the change in its kineticenergy.



General Derivation of Work-Energy Principle

Definition of Work, and Newton’s second Law (note that ~F is thetotal force):

W =

∫~F · d ~=

∫m~a · d ~ (94)

d ~ is the displacement of object (in a time dt), hence d ~= ~vdt :

=

∫m

d~vdt· d ~

dtdt =

∫m

d~vdt· ~vdt (95)

Simplifying:

=

∫ddt

(12

m~v2)

dt =

∫d(

12

m~v2)

= ∆

(12

m~v2)

(96)


Seventh Week Example: Roller Coaster

Roller Coaster

Track of a Roller Coaster

~v0

R

~v

θ

~N ~w

θ x

y

12mv2 + mgh = const , ~Ffr = ~0

Q: Assume that the cart is going with a very slow speed v . What is themaximum height that it can reach on the circular track?

At the maximum height, its speed is zero v = 0:12mv2

0 + mg0 = 12m02 + mghmax =⇒ hmax =

v20

2g

Same as if it was thrown vertically upwards!



Roller Coaster

Track of a Roller Coaster

~v0

R

~v

θ

~N ~w

θ x

y


Q: Assume that the cart is going with a very slow speed v . What is themaximum height that it can reach on the circular track?

At the maximum height, its speed is zero v = 0:12mv2

0 + mg0 = 12m02 + mghmax =⇒ hmax =

v20

2g

Same as if it was thrown vertically upwards!Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 122 / 175


Roller CoasterTrack of a Roller Coaster

~v0

R

~v

θ

~N ~w

θ x

y


Q: What is the minimum speed with which the cart should have at thebottom so that it can go around the top of the loop?

hmax = 2R =⇒ v20 = 2g(2R) = 4gR

WRONGIf v2

0 = 4gR, the cart reaches the top with zero velocity. It can notgo round!




~v0

R

~v

θ

~N ~w

θ x

y



hmax = 2R =⇒ v20 = 2g(2R) = 4gR

WRONG

If v20 = 4gR, the cart reaches the top with zero velocity. It can not

go round!




~v0

R

~v

θ

~N ~w

θ x

y



hmax = 2R =⇒ v20 = 2g(2R) = 4gR

WRONGIf v2

0 = 4gR, the cart reaches the top with zero velocity. It can notgo round!




~v0

R

~v

θ

~N ~w

θ x

y



At the top ~N = −Ny , ~w = −mgy =⇒ ~FT = −(N + mg)yFor circular motion at the top~F ≡ −m v2

R y =⇒ v2 = (N + mg)R/m ≥ gRAt the threshold of falling off the track, N = 0 =⇒ v2

min = gR.At the bottom of the roller coaster 1

2mv20min = 1

2mv2min + mg(2R)

=⇒ v20min = 5gR




~v0

R

~v

θ

~N ~w

θ x

y


Q: If the cart is moving with a speed less then the minimum speed, atwhat point will it leave the track?

Assume it falls at angle θ > 0.At that point, the central force is only due to ~w , wr = mg sin θ

For circular motion: m v2

R = wr = mg sin θ =⇒ v2 = gR sin θ.12mv2

0 = 12mv2 + mgR(1 + sin θ) = 1

2mgR(3 sin θ + 2)

=⇒ sin θ =v2

0−2gR3gR



MOON

It is known that moon always shows its same face to the Earth. This isbecause the period of rotation of the moon around its axes is equal toits period of rotation around Earth. Can you come up with anexplanation of this equality?Keywords to consider: tides, friction, work


Eighth Week

Ask questions: let me know how you perceive nature so that ifthere is a misperception, we can correct itAsk simple questionsif you have any doubt, repeat what you have understood. It neednot be in the form of a questionDon’t try to guess the type of questions that I can ask, try tounderstand natureBad habits that you have learned in years takes more than a fewmonths to correct!


Eighth Week

The only forces that we will study in this year are gravity, and EM force.EM force represents itself through

FrictionAny pull or a push (e.g. Normal Force, forced due to the tensionon a string, force acting by a spring)


Eighth Week

TRACKER Software (http://www.cabrillo.edu/ dbrown/tracker/)


http://www.cabrillo.edu/~dbrown/tracker/

Eighth Week

Review of Work Done By Gravity

Wtot = ∆T where T = 12mv2

WG = −mg∆h depends only on the height difference, and onnothing elseIf an object moves under the influence of gravity only thenthroughout the motion

12

mv2 + mgh = const (97)


Eighth Week Work Done By Gravity

Work Done By Gravity

The force acting on an object of mass m is ~Fw = −mgz (z pointsupwards)If the object is displaced by d ~, then dW = ~Fw · d ~= −mgdz, i.e.the work done by its weight is proportional to the change in its zcoordinate (its height)As the object goes from P1 to P2, the to calculate the total work,just sum the changes in its height. Hence total work isWtot = −mg∆h


Eighth Week Work Done by Friction

Work Done By Friction

P1

a

O b P2

c

Assume a constant friction force ofmagnitude Ff .a, b and c are the correspondinglength.WP1→O = −Ff aWO→P2 = −Ff bWP1→O→P2 = −Ff (a + b)

WP1→P2 = −Ff c 6= −Ff (a + b)

Hence the work done by friction depends on how one goes fromthe initial point to the final point



Example: ~F = axx + byy

y

O x

y0 A

x0

y1B

x1

C

~F = axx + byy

Work done by the force as one goes fromA to B through C:WA→C→B = WA→C + WC→B

Along the path A→ C, x = x0 andhence ~F = ax0x + byy , d ~= dyydW = ~F · d ~= bydy∫ WA→C

0dW =

∫ y1

y0

bydy (98)

WA→C =12

by21 −

12

by20 (99)

Work done as one goes from A to B is the same in both paths. Andcan be written as WA→B = U(x0, y0)− U(x1, y1)




y

O x

y0 A

x0

y1B

x1

C

~F = axx + byyWA→C = 1

2by21 −

12by2

0

Work done by the force as one goes fromA to B through C:WA→C→B = WA→C + WC→B

Along the path A→ C, y = y1 andhence ~F = axx + by1y , d ~= dxxdW = ~F · d ~= axdx∫ WC→B

0dW =

∫ x1

x0

axdx (98)

WC→B =12

ax21 −

12

ax20 (99)





y

O x

y0 A

x0

y1B

x1

C


2by21 −

12by2

0WC→B = 1

2ax21 −

12ax2

0

Work done by the force as one goes fromA to B through C:WA→C→B = WA→C + WC→BWA→C→B =(1

2ax21 + 1

2by21)−(1

2ax20 + 1

2by20)





y

O x

y0 A

x0

y1B

x1

C

~rAB


2by21 −

12by2

0WC→B = 1

2ax21 −

12ax2

0

Work done by the force as one goesfrom A to B through a straight line:

Let ~rA = x0x + y0y , ~rB = x1x + y1y ,~rAB = ~rB −~rA

Any point on the trajectory can bewritten as ~r = ~rA + λ~rAB; 0 ≤ λ ≤ 1d ~= (dλ)~rAB,dW = ~F · d ~= dλ~F ·~rAB =dλ(a(x1λ+ x0(1− λ))(x1 − x0) +b(y1λ+ y0(1− λ)(y1 − y0))

WA→B =∫WA→B

0 dW =∫ 1

0 dλ(· · · ) =(12ax2

1 + 12by2

1)−(1

2ax20 + 1

2by20)



Eighth Week Conservative Forces

Conservative Forces

Definition: A force is conservative if the work done by that forceas an object moves from a point P1 to a point P2 is independent ofthe path that the object takes.Friction is an example of a non-conservative force.Gravity (any constant force in general) is a conservative force.~F = axx + byy is a conservative force



If ~F is conservative W =∫ P2

P1~F · d ~ is independent of how one

goes from P1 to P2.To evaluate W , one can choose any pathDefine a function U(P) such that U(P) = U(P0)−

∫ P1P0~F · d ~.

W = ∆TSuppose the object moves from P1 to P2 under the influence ofthe conservative force ~F .The work done by ~F is:

W =

∫ P2

P1

~F · d ~=

∫ P0

P1

~F · d ~+

∫ P2

P0

~F · d ~ (98)

= (U(P1)− U(P0)) + (U(P0)− U(P2)) = U(P1)− U(P2) (99)

W = T2 − T1:

T2 − T1 = U(P1)− U(P2) (100)T1 + U(P1) = T2 + U(P2) (101)



Conservation of Energy

U is called the potential energyT + U is called the mechanical energyFor an object moving under the influence of a conservative forceonly T + U is always conserved.An object that is raised by h, has potential to do work, it has alarger potential energyPotential energy is NOT a property of a single object, but aproperty of the system as a whole.



Newton’s three laws are vector relationsConservation of energy is a scalar relationIf the expression for potential energy is known, speed at any pointcan be determined without solving any differential equation orintegrals.



Force from Potential

Consider two nearby points separated by the vector d~r .The difference in their potential energies is:

U(~r + d~r)− U(~r) = −~F (~r) · d~r (102)

Above is valid for any d~r . Denote U(~r) ≡ U(x , y , z) if~r = xx + yy + zzIf d~r = dxx , then

U(x + dx , y , z)−U(x , y , z) = −Fx (x , y , z)dx (103)

=⇒ Fx (x , y , z) =− U(x + dx , y , z)− U(x , y , z)

(x + dx)− x≡ −∂U

∂x(104)

Similarly Fy = −∂U∂y and Fz = −∂U

∂z

Hence ~F = −(x ∂U∂x + y ∂U

∂y + z ∂U∂z ) ≡ −~∇U

where ~∇ is the nabla operator.



Interpreting Potential Graphs

U

x

E

−x0 x0

P1 P2

Consider one dimensional exampleFx = −dU

dx , i.e. Fx points in thedirection that U(x) is decreasingConsider a particle with total energyE . Then K = E − U > 0, i.e. Theparticle can only be at points forwhich U(x) < E .points x such that U(x) = E arecalled turning points




U

x

E

−x0 x0

P1 P2

At points x = ±x0, and x = 0, theforce acting on an object is zero: theyare called equilibrium points.If an object is slightly displaced fromx = ±x0, they try to move towards±x0: they are called stableequilibrium pointsIf an object is slightly displaced fromx = 0, they try to move away fromx = 0: x = 0 is an unstableequilibrium point.




U

x

E

−x0 x0

P1 P2

If an object is slightly displaced fromx = ±x0, they try to move towards±x0: they are called stableequilibrium pointsIf an object is slightly displaced fromx = 0, they try to move away fromx = 0: x = 0 is an unstableequilibrium point.If an object is displaced slightly froman equilibrium point, it neither goestowards nor away from theequilibrium point, it is called neutralequilibrium point.



Dissipative Forces

Dissipative forces in fact convert mechanical energy to internalenergyThe total energy of the universe is constant∆(T + U) = Wdissipative forces



Gravitational Potential Energy and Escape Velocity

Gravitational force acting on a mass m, due to a mass M is:

~F = −GNmMr2 r (105)

The work done when m is displaced by d ~ is

dW = ~F · d~r = −GNmMr2 r · d ~= −GN

mMr2 dr (106)

where dr is the change in the radial distance, i.e. radialcomponent if d ~.Potential energy difference is

U(P)− U(∞) = −∫ P

∞

(−GN

mMr2 dr

)(107)

= −GNmM

r

∣∣∣∣r=rP

r=∞= −GN

mMrP

(108)

In general U(∞) is chosen to be zero U(∞) = 0Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 138 / 175


Gravitational Potential Energy and Escape Velocity

Q: What should be the initial speed of an object on the surface of aplanet of mass M and radius R, if it is to go until infinity?

Let v0 and v∞ be the initial and final speeds.Initial mechanical energy is E = 1

2mv20 + GN

mMR .

Final mechanical energy is E = 12mv2

∞Conservation of mechanical energy:

12

mv20 −GN

mMR

=12

mv2∞ (109)

=⇒ v20 =

2m

(12

mv2∞ + GN

mMR

)(110)

The minimum possible speed is called the escape velocity:

vesc =

√GN

2MR

; vesc,Earth = 11.2 km/s = 40320 km/h (111)



Potential Energy of a Spring

The work done by a spring on an object as it moves from x = 0 tox = L was calculated as W = −1

2kL2.The potential energy of a spring that is stretched by L isU(L) = 1

2kL2



Power

Power is the rate of doing work.

P =dWdt

= ~F · d ~

dt= ~F · ~v (112)

NOTE: The above is in general NOT the derivative of W !The unit of power is Watt: [P] = J/s ≡WThe efficiency of an engine is

e =Pout

Pin(113)



QUESTION TO THINK OVER

The cars are categorized in terms of the volume of their engine. Thesmaller the engine, the less fuel it uses. When you are driving a carwith a small engine, it will be difficult to go up a hill, whereas for a carwith a larger engine, it is much easier. This difficulty of the small carcan be over come if you use it at a much higher rpm (rotations perminute). Why is it that the car with smaller engine finds it difficult to gouphill but using it with a large rpm resolves this issue? (Assume thatthe efficiencies of both cars are more or less equal)



Discussion ProblemsSend your answer as an SMS to: 4660

x

U(x)

x0

A particle is released from rest at theposition x = x0 in the potential describedbelow.

U(x) =

{−ax x < 0bx2 x > 0

(114)

Determine whether the followingstatements are true (A) or false (B). (sendA or B)

The subsequent motion is periodic.




x

U(x)

x0


U(x) =

{−ax x < 0bx2 x > 0

(114)


The velocity is a continuous functionof time.




x

U(x)

x0


U(x) =

{−ax x < 0bx2 x > 0

(114)


The acceleration is a continuousfunction of time.




x

U(x)

x0


U(x) =

{−ax x < 0bx2 x > 0

(114)


The force is conservative.



Discussion ProblemsSend your answer as an SMS to: 4660Question 5

Consider the following sketch of potential energy for a particle as a function of position. There are no dissipative forces or internal sources of energy.

If a particle travels through the entire region of space shown in the diagram, at which point is the particle's velocity a maximum?

1. a 2. b 3. c

4. d

5. e

Consider the above sketch of potential energy for a particle as afunction of position. There are no dissipative forces or internal sourcesof energy. If a particle travels through the entire region of space shownin the diagram, at which point is the particle’s velocity a maximum?(A) a (B) b (C) c (D) d (E) e




Question 6 Consider the following sketch of potential energy for a particle as a function of position. There are no dissipative forces or internal sources of energy.

What is the minimum total mechanical energy that the particle can have if you know that it has traveled over the entire region of X shown?

1. -8 2. 6 3. 10

4. It depends on direction of travel 5. Can’t say - Potential Energy uncertain by a constant

Send SMS to 4660

Consider the above sketch of potential energy for a particle as afunction of position. There are no dissipative forces or internal sourcesof energy. What is the minimum total mechanical energy that theparticle can have if you know that it has traveled over the entire regionof X shown?(A) -8 (B) 6 (C) 10 (D) It depends on direction of travel(E) Can’t say - Potential Energy uncertain by a constant




You lift a ball at constant velocity from a height hi to a greater height hf .Considering the ball and the earth together as the system, which of thefollowing statements is true?

A The potential energy of the system increases.B The kinetic energy of the system decreases.C The earth does negative work on the system.D You do negative work on the system.E The source energy of the ball increases.F Two of the above.G None of the above.



QUIZ5

An object is dropped to the earth from a height of 10m. Which of thefollowing sketches best represent the kinetic energy of the object as itapproaches the earth (neglect friction).

3. An object is dropped to the earth from a height of 10m. Which of the following sketches best represent the kinetic energy of the object as it approaches the earth (neglect friction).

1.a 2.b 3.c 4.d 5.e


Ninth Week

Momentum

Rewrite Newton’s second law as:

~F = md~vdt≡ d(m~v)

dt

Define the momentum of the particle as ~p = m~v :

~F =d~pdt

i.e. force is the rate of change of momentumCan also be written as

d~p = ~Fdt

i.e., if there is force, the change in momentum in dt time is ~Fdt(Note: Compare the form of this equation with d~v = ~adt)


Ninth Week

Momentum Conservation

d~p = ~FdtConsider two masses m1 and m2 exerting forces ~F12 and ~F21 oneach other.d~p1 = ~F12dt and d~p2 = ~F21dtd(~p1 + ~p2) = (~F12 + ~F21)dt

Newton’s third law: ~F12 = −~F21, i.e. ~F12 + ~F21 = 0d(~p1 + ~p2) = 0, i.e. ~p1 + ~p2 is constant.Newton’s third law is valid for any system under any condition.Hence momentum conservation is valid under any conditionEnergy conservation for a system is valid only under the absenceof dissipative forces.


Ninth Week

Consider two masses m1 and m2 moving along a line withvelocities v1 and v2 (since they are moving along a line, these arethe components along the line)Initial total momentum of the system: Pi = m1v1 + m2v2

Two masses collide and after collision they move with velocities v ′1and v ′2.Final momentum of the system is Pf = m1v ′1 + m2v ′2Momentum conservation:

m1v1 + m2v2 = m1v ′1 + m2v ′2


Ninth Week

Consider two masses m1 and m2 moving along a line withvelocities v1 and v2 (since they are moving along a line, these arethe components along the line)Initial total momentum of the system: Pi = m1v1 + m2v2

Two masses collide and after collision they move with velocities v ′1and v ′2.Final momentum of the system is Pf = m1v ′1 + m2v ′2Momentum conservation:

m1v1 + m2v2 = m1v ′1 + m2v ′2


NOTEIf a quantity is conserved, it is conserved independent ofhow complicated the interactions are in the intermediatestages!

Ninth Week

m1v1 + m2v2 = m1v ′1 + m2v ′2Unknowns: v ′1, v ′2: two unknowns but a single equationWe need one more condition to determine v ′1 and v ′2Elastic collision: energy is conserved:12m1v2

1 + 12m2v2

2 = 12m1v ′21 + 1

2m2v ′22

Completely inelastic collision: The two masses stick together:v ′1 = v ′2Partially inelastic collision: Coefficient of restitution:

CR =relative speed after collision

relative speed before collision=|v ′1 − v ′2||v1 − v2|

(114)


Ninth Week

1D Collision of Two Masses

Consider an elastic collision: 12m1v2

1 + 12m2v2

2 = 12m1v ′21 + 1

2m2v ′22

Energy conservation can be re written as

m1(v1 − v ′1)(v1 + v ′1) = m2(v ′2 − v2)(v ′2 + v2)

Momentum conservation: m1v1 −m1v ′1 = m2v ′2 −m2v2

Combined v1 + v ′1 = v2 + v ′2, or v1 − v2 = −(v ′1 − v ′2)

Consider the special case: v2 = 0 =⇒ v ′2 = v1 + v ′1m1v1 −m1v ′1 = m2(v1 + v ′1)

v ′1 = (m1−m2)(m1+m2)v1

v ′2 = 2m1m1+m2

v1


Ninth Week

1D Collision of Two Masses

v ′1 = (m1−m2)(m1+m2)v1

v ′2 = 2m1m1+m2

v1

Special case: m1 = m2: v ′1 = 0 and v ′2 = v1

Special case: m2 � m1: v ′1 = −v1 and v2 = 0Special case: m1 � m2: v ′1 = v1, v2 = 2v1 (discuss also in thereference frame in which v1 = 0)


Ninth Week

General Momentum Conservation

System of point particles of mass mi .Let ~F int

ij be the force acting on mass mi due to the mass mj (canbe gravitational attraction, EM attraction, push, pull, etc)Let ~F ext

i be the force acting on mass mi due to objects that are notpart of my system.Newtons’s second law: d~pi

dt=∑

j 6=i~F int

ij + ~F exti

Sum over all i ∑i

d~pi

dt=∑i,j 6=i

F intij +

∑i

~F exti (115)

ddt

∑i

pi = ~0 + ~F extT (116)

d~PT

dt= ~F ext

T (117)


Ninth Week

General Momentum Conservation

Sum over all i ∑i

d~pi

dt=∑i,j 6=i

F intij +

∑i

~F exti (115)

ddt

∑i

pi = ~0 + ~F extT (116)

d~PT

dt= ~F ext

T (117)

Note that for any ~Fij , there is a ~Fji = −~Fij because of Newton’sthird law. Hence the first sum is zero.~PT =

∑i ~pi is that total momentum and F ext

T =∑

i~F ext

i is the totalforce acting on the system


Ninth Week

~F extT =

d~PT

dtNewton’s second law is also valid for extended objects if themomentum is taken as the total momentum of the system, and theforce is taken as the total force acting on the system.Define ~vCM as ~PT = M~vCM where M =

∑i mi and

~vCM =1M

∑i

~pi =1M

∑i

mi~vi (118)

=1M

∑i

mid~ri

dt=

ddt

(1M

∑i

mi~ri

)≡ d~rCM

dt(119)

~rCM = 1M∑

i mi~ri is the position of the center of mass of thesystem.A system of particles behaves like a point mass of mass M andposition ~rCM with regards to translational motion.


Ninth Week

Collisions in 3D

~v

~vB

~vA

y

xθφ

θ > 0 and φ < 0

Initial state: mB is at rest, mA moveswith velocity ~vFinal state: the velocities of mA andmB are ~vA and ~vB.Initial momentum:

~Pi = mA~v = mAvx (120)

Final momentum:

~Pf = mA~vA + mB~vB

= (mAvA cosφ+ mBvB cos θ) x+ (mAvA sinφ+ mBvB sin θ) y

Unknowns vA, vB, φ, θ are related through:

mav = mavA cosφ+ mBvB cos θ (121)0 = mAvA sinφ+ mBvB sin θ (122)

We need two more equations


Ninth Week

Collisions in 3D

~v

~vB

~vA

y

xθφ

θ > 0 and φ < 0

Initial momentum:

~Pi = mA~v = mAvx (120)

Final momentum:

~Pf = mA~vA + mB~vB

= (mAvA cosφ+ mBvB cos θ) x+ (mAvA sinφ+ mBvB sin θ) y





Ninth Week

Collisions in 3D

~v

~vB

~vA

y

xθφ

θ > 0 and φ < 0

In elastic collisions:12mAv2 = 1

2mAv2A + 1

2mBv2B

One more equation involving vA andvB.Still we need to measure one of theunknowns to determine the otherthreeIn completely inelastic collisions~vA = ~vB, hence vA = vB and θ = φ





Ninth Week

System of Two Masses

In terms of ~vCM and ~v = ~v1 − ~v2, ~v1 = ~vCM + m2m1+m2

~v and~v2 = ~vCM − m1

m1+m2~v

Total kinetic energy:

K =12

m1v21 +

12

m2v22

=12

m1

(~vCM +

m2

m1 + m2~v)2

+12

m2

(~vCM −

m1

m1 + m2~v)2

=12

(m1 + m2)v2CM +

12

m1m2

m1 + m2v2

≡ 12

Mv2CM +

12µv2 (122)

The first term is the kinetic energy of the whole systemThe second term is the internal energy of the whole system.In elastic collisions, only the direction of ~v can change


Ninth Week

Impulse

d~p = ~Fdt =⇒ ~pf − ~pi =∫ tf

ti~Fdt ≡ ~J

~J is called the impulseThe average force acting on a system is defined as

~Fav =~J

∆T(123)

where ∆T is the length of time during which the impulse is givento the system.


Ninth Week

CM of a Continuous Mass Distribution

Q: What is the CM of a thin rod of length L and mass M uniformlydistributed over its length?

y

0 x

dx

x L dm = ML dx

~rCM = 1M∑

i mi~ri = 1M∑

i mixi x

Each segment has mass dm = ML dx .

~rCM = 1M

∫ L0

(ML dx

)xx = L

2 x


Ninth Week


Q: Where is the CM of a semi disk with radius R and total mass M?

y

L/2

y

x−R

R

L = 2√

R2 − y2

ρ = 2MπR2

The mass of the red strip: dm = ρLdyThe CM of red strip: ~ri = yyCM of the semi disk:

~rCM =1M

∑i

dm~ri =1M

∑i

(ρLdy)(yy)

=1M

∫ R

0ρ2√

R2 − y2ydyy

(124)


Ninth Week



y

L/2

y

x−R

R

L = 2√

R2 − y2

ρ = 2MπR2

CM of the semi disk:

~rCM =1M

∑i

dm~ri =1M

∑i

(ρLdy)(yy)

=1M

∫ R

0ρ2√

R2 − y2ydyy

(124)

Make a change of variables:y = R sin θ, dy = R cos θdθ:

~rCM =2R3ρ

M

∫ π2

0cos2 θ sin θdθy

(125)


Ninth Week



y

L/2

y

x−R

R

L = 2√

R2 − y2

ρ = 2MπR2

Make a change of variables:y = R sin θ, dy = R cos θdθ:

~rCM =2R3ρ

M

∫ π2

0cos2 θ sin θdθy

=2R3

M2MπR2

13

y =4

3πRy (124)


Ninth Week

Systems of Variable MassRocket Propulsion

Q: A rocket can expel exhaust at a speed u relative to the rocket.Suppose the it starts with a mass M and expels exhaust at a rate of r(in units of kg/s). What will be its speed when its mass becomes Mf ?

At the moment that the rocket has a mass m, suppose it expels amass dm of exhaust.Assume that initially it has a speed v , and after it expels theexhaust, it has a velocity v + dv .Initial momentum: mvFinal momentum(m − dm)(v + dv) + (−dm)(v − u) = mv + mdv − dmdv − dmu(dm < 0)


Ninth Week

Systems of Variable MassRocket Propulsion

Q: A rocket can expel exhaust at a speed u relative to the rocket.Suppose the it starts with a mass M and expels exhaust at a rate of r(in units of kg/s). What will be its speed when its mass becomes Mf ?

Conservation of momentum:mv + mdv + dmdv + dmu = mv =⇒ mdv + dmu = 0dv/u = dm/m. Integrating both sides from ti (when the speed is viand mass M) until tf (when speed is vf and mass Mf )

vf − vi

u= − ln

Mf

M

vf = vi + u ln MMf


Ninth Week

Discussion QuestionsSend SMS to 4660

Drop a stone from the top of a high cliff. Consider the earth and thestone as a system. As the stone falls, the momentum of the system

A increases in the downward direction.B decreases in the downward direction.C stays the same.D not enough information to decide.


Ninth Week


Consider yourself and the Earth as one system. Now jump up. Doesthe momentum of the system

A increase in the downward direction as you rise?B increase in the downward direction as you fall?C stay the same?D dissipate because of friction?E Not enough information is given to decide.


Ninth Week


Suppose you are on a cart, initially at rest on a track with very littlefriction. You throw balls at a partition that is rigidly mounted on thecart. If the balls bounce straight back as shown in the figure, is the cart

put in motion?

Question 3. Suppose you are on a cart, initially at rest on a track with very little friction. You throw balls at a partition that is rigidly mounted on the cart. If the balls bounce straight back as shown in the figure, is the cart put in motion?

1. Yes, it moves to the right.

2. Yes, it moves to the left.

3. No, it remains in place.

4. Not enough information is given to decide.

A Yes, it moves to the right.B Yes, it moves to the left.C No, it remains in place.D Not enough information is given to decide.


Ninth Week


The greatest acceleration of the center of mass of a baseball bat willbe produced by pushing with a force F at

Question 5: Pushing a Baseball Bat

The greatest acceleration of the center of mass of a baseball bat will be produced by pushing with a force F at

1. Position 1

2. Position 2

3. Position 3

4. All the same

5. Not enough information is given to decide.

A Position 1B Position 2C Position 3D All the sameE Not enough information is given to decide.


Ninth Week


A compact car and a large truck collide head on and stick together.Which undergoes the larger momentum change?

A carB truckC The momentum change is the same for both vehicles.D Can’t tell without knowing the final velocity of combined mass.


Ninth Week


Two balls that are dropped from a height hi above the ground, one ontop of the other. Ball 1 is on top and has mass m1 , and ball 2 isunderneath and has mass m2 with m2 � m1 . Ball 2 first collides withthe ground and rebounds with speed v0 Then, as ball 2‘starts to moveupward, it collides elastically with the ball 1 which is still movingdownwards also with speed v0 . The final relative speeds after ball 1and ball 2 collide is

A ZeroB v0

C 2v0

D 3v0

E None of the above


Ninth Week


An explosion splits an object initially at rest into two pieces of unequalmass. Which piece has the greater kinetic energy?

A The more massive piece.B The less massive piece.C They both have the same kinetic energy.D There is not enough information to tell.


Ninth Week


A spacecraft with speed v1i approaches Saturn which is moving in theopposite direction with a speed vs. After interacting gravitationally withSaturn, the spacecraft swings around Saturn and heads off in theopposite direction it approached. What is the final speed of thespacecraft v1f after it is far enough away from Saturn to be nearly freeof Saturn’s gravitational pull?


Ninth Week


Suppose rain falls vertically into an open cart rolling along a straighthorizontal track with negligible friction. As a result of the accumulatingwater, the speed of the cart

A increases.B does not change.C decreases.D not sure.E not enough information is given to decide.


Ninth Week

LEARNING

Measurable and relatively permanent change in behavior throughexperience, instruction, or study. · · · Learning itself cannot bemeasured, but its results can be. In the words of Harvard BusinessSchool psychologist Chris Argyris, learning is "detection and correctionof error" where an error means "any mismatch between our intentionsand what actually happens."Read more: http://www.businessdictionary.com/definition/learning.html


Ninth Week

LEARNING

There are three components to the definition of Learning 1:“Learning is a process, not a product.”Exam scores and term papers are measures of learning, but theyare not the process of learning itself.

1http://www.cidde.pitt.edu/ta-handbook/teaching-and-learning-principles/definition-learning


Ninth Week

LEARNING

There are three components to the definition of Learning 1:“Learning is a change in knowledge, beliefs, behaviors orattitudes.”This change requires time, particularly when one is dealing withchanges to core beliefs, behaviors, and attitudes. Don’t interpret alack of sea change in your students’ beliefs or attitudesimmediately following a lesson as a lack of learning on their part,but instead, consider that such a change will take time – perhapsa few weeks, perhaps until the end of the term, or even longer.



Ninth Week

LEARNING

There are three components to the definition of Learning 1:“Learning is not something done to students, but somethingthat students themselves do.”If you have ever carefully planned a lesson, only to find that yourstudents just didn’t “get it,” consider that your lesson should bedesigned not just to impart knowledge but also to lead studentsthrough the process of their own learning (Ambrose 2010:3).



Tenth Week

Outline of Topics CoveredWill be Redone for Rotational Motion

Kinematics-how to describe the state (position and velocity) of thesystemDynamics–why the state of the system changes (acceleration)Work done by forceConserved quantities

Energy ConservationMomentum Conservation


Rotational Motion

Rotational Variables

Rigid body: The distances between parts of the object are fixed.General motion of a rigid body: translation+ rotationPure rotation around a fixed axis: all the points on the objectrotate around a given axis-axis of rotationThe orientation of an object can be completely specified byspecifying the position of a determined point.


Rotational Motion

Rotation Angle(Kinematics)

O

P

y

x

Rθ

`

~F

γ

In radians θ = `R , or ` = θR (θ =

theta)∆θ: change in θ, angulardisplacementAverage angular velocity: ω = ∆θ

∆t (ω= omega)Average angular acceleration:α = ∆ω

∆t(α = alpha)Instantaneous angular velocity:ω = dθ

dt

Instantaneous angular acceleration:α = dω

dt


Rotational Motion


O

P

y

x

Rθ

`

~F

γ

The speed of point P isv = d`

dt = d(Rθ)dt = R dθ

dt = RωThe further the point is from the axisof rotation, the faster it movesatan = dv

dt = Rα (note that these arenot vectors)frequency: How many full rotationsthe object completes in one second:f = ω

2π =⇒ w = 2πfPeriod: How long one full rotationtakes: T = 2π

ω = 1f


Rotational Motion


O

P

y

x

Rθ

`

~F

γ

Angular velocity has a magnitude anda direction (the axis of rotation) henceit is a vector.The direction of angular velocityvector ~ω can be found by the righthand rule.Angular acceleration vector ~α = d~ω

dt~v = ~ω ×~r (see vector products )~atan = ~α×~r


Rotational Motion


O

P

y

x

Rθ

`

~F

γ

For variable angular acceleration

ω(t) = ω0 +

∫ t

0α(t ′)dt ′ (125)

⇐⇒ ~v(t) = ~v0 +

∫ t

0~a(t ′)dt ′ (126)

θ(t) = θ0 +

∫ t

0ω(t ′)dt ′ (127)

⇐⇒ ~r(t) = ~r0 +

∫ t

0~v(t ′)dt ′ (128)

For constant angular acceleration:

ω(t) = ω0 + αt (129)

θ(t) = θ0 + ω0t +12αt2 (130)


Rotational Motion

Torque(Dynamics)

O

P

y

x

Rθ

`~F

γ

If a force ~F is applied at the point P,such that it makes an angle γ with theline connecting P to O, the torque isdefined as: τ = FR sin γ = |~F × ~R|


Rotational Motion Torque

Torque on Point Mass

O

~R

w

~w

~Fγ

Ftan = Fsinθmatan = Ftan =⇒ mRα = Ftan

τ = FtanR = mR2α ≡ IαI = mR2 is called the moment ofinertia.

Note that the net force acting on the mass m also contain theforce due to tension. But this force does not have a tangentialcomponent, hence does not contribute to the tangentialacceleration, and hence to angular acceleration.~τ = ~r × ~F has the same magnitude as I~α, and is in the samedirection.Hence ~τ = I~α



Torque on Continuous Mass

For a system formed by mi , the torque acting on mi is

~τi = ~Ri ×

∑j 6=i

~Fij + ~F exti

= m2i Ri~α

(Note that for a right body α is the same for all parts of the system)Total torque acting on the system

~τ ≡∑

i

~τi =∑

i

miR2i ~α =

(∑i

miR2i

)~α ≡ I~α

I =∑

i miR2i



Rotational Kinetic Energy

Assume that a rigid body is rotating around a fixed axis withangular velocity ω.mi located at a distance Ri from the axis will have a speed ωRi .The kinetic energy of mi is K = 1

2mi(Riω)2 = 12miR2

i ω2.

Summing the kinetic energy of all the masses, the kinetic energyof the rigid body is:

K =∑

i

12

(miR2i )ω2 ≡ 1

2Iω2



Kinetic Energy of A Rotating Object That also HasTranslational Motion

Let ~rCM be the position of the CM.Let ~ri and ~Ri be the position of mass mi in the rigid body relative toa fixed coordinate axis and relative to the CM respectively:~ri = ~rCM + ~Ri~vi = ~vCM + ~Vi , where ~Vi is the velocity relative to the CM.Total Kinetic Energy of the rigid body is:

K =∑

i

12

mi~v2i =

∑i

12

mi(~v2CM + 2~vCM · ~Vi + ~V 2

i )

=12

(∑

i

mi)V 2CM +

12

∑i

mi~V 2

i + ~vCM ·∑

i

mi~Vi

The first terms is the translational kinetic energyThe second term is the rotational kinetic energy around the CM.



Kinetic Energy of A Rotating Object That also HasTranslational Motion

Total Kinetic Energy of the rigid body is:

K =∑

i

12

mi~v2i =

∑i

12

mi(~v2CM + 2~vCM · ~Vi + ~V 2

i )

=12

(∑

i

mi)V 2CM +

12

∑i

mi~V 2

i + ~vCM ·∑

i

mi~Vi

The first terms is the translational kinetic energyThe second term is the rotational kinetic energy around the CM.The third term is zero since

∑i mi

~Vi is the total momentumrelative the the CM which is zero.K = 1

2Mv2CM + 1

2 Iω2

Note that this simple form is valid only if one considers a rotationaxis passing through the CM.



Work Done On a Rotating Object

W =∫ Pf

Pi~F · d ~

d` = Rdθ~F · d ~= FRdθ cos γ, where γ is the angle between ~F and d ~

F cos γ is the component of ~F along d ~

d` is perpendicular to ~R.Hence F cos γ = F⊥

W =

∫ θf

θi

F⊥Rdθ =

∫ θf

θi

τdθ

The power is:

P =dWdt

= τdθdt

= τω



Work Energy Principle For Rotations

W =

∫ θf

θi

τdθ =

∫ tf

tiIdωdt

dθdt

dt

=

∫ tf

ti

dωdt

Iωdt =

∫ tf

ti

ddt

(12

Iω2)

dt

=12

Iω2f −

12

Iω2i


Rotational Motion Examples

Example-Calculating Moments of InertiaMoment of Inertia Of a Rigid Rod Rotating Around CM

y

0 x

dx

x

CM

Ldm = M

L dx

I =∑

i

miR2i =

∑i

ML

dx(

L2− x

)2

=ML

∫ L

0

(L2− x

)2

=M3L

(L2− x

)3∣∣∣∣∣x=L

x=0

=M3L

(L2

)3

− M3L

(−L

2

)3

=M12

L2



Parallel Axis TheoremProof

Let ICM be the moment of inertia around an axes going throughthe CM.Choose z axis to be along the this axes.Choose a second axes that goes through the point (x0, y0).The distance of point with coordinates (xi , yi , zi) from the secondaxis is R2 = (xi − x0)2 + (yi − y0)2

The moment of inertial with respect to the second axis is

I =∑

i

mi

[(xI − x0)2 + (yi − y0)2

](125)

=∑

i

mi(x2i + y2

i ) +∑

i

mi(x20 + y2

0 )− 2∑

i

mi(xix0 + yiy0) (126)

= ICM + Md2 (127)

where d2 = x20 + y2

0 and∑

i mixi =∑

i miyi = 0



Parallel Axis TheoremExample

y

0 x

dx

x

CM

Ldm = M

L dx , ICM = 12ML2

Moment of inertia of a thin rod around one end:

I = ICM + M(

L2

)2

=1

12ML2 +

14

ML2 =13

ML2



Example: Thin Rod Rotating Around One End

xdx

L

The net torque on the rod around the fixed axes:

~τ =∑

i

~ri × (mi~g) =

(∑i

mi~ri

)× ~g = (M~rCM)× ~g = ~rCM × ~w

(125)

Hence the center of gravity for an object in uniform gravitationalfield is its CM.τ = MgL

2

α = τI =

MgL2

13 ML2 = 3

2gL

If the rod is initially at rest: aCM = aCMtan = αL

2 = 34g < g

aCM = F ext

M . Which other force is acting on the rod?Altug Özpineci ( METU ) Phys109-MECHANICS PHYS109 170 / 175


Perpendicular Axis TheoremProof

Consider a very thin object in the xy plane.For any point in the object zi ' 0Ix =

∑i mi(y2

i + z2i ) '

∑i miy2

i

Similarly Iy =∑

i mi(x2i + z2

i ) '∑

i mix2i

Then Iz =∑

i mi(x2i + y2

i ) =∑

i mix2i +

∑i miy2

i = Ix + Iy



Perpendicular Axis TheoremExample

The moment of inertia of a loop of mass M and radius R forrotations around an axis that is perpendicular to its plane andgoing through its CM: Iz =

∑i miR2 = MR2

Its moment of inertia around any axis that goes through its CMand is in its plane:

If x and y axes are the two axes in the plane of the loop, due tosymmetry Ix = IyBy perpendicular axis theorem: Iz = Ix + Iy = 2IxHence Ix = 1

2 MR2.

Moment of inertia for rotation around an axis that goes through theedge and is in the plane of the loop:I′ = ICM + Md2 = 1

2MR2 + MR2 = 32MR2



Heavy Pulley

m1

m2

Since the pulley has a mass, thetensions at each end of the pulley willbe different.

~w1

m1

~T1

~w2

m2

~T2

−~T1 −~T2

z

Let x axis be out of the screen



Heavy Pulley

m1

m2

~w1

m1

~T1

~w2

m2

~T2

−~T1 −~T2

z

The net forces acting on mass m1and m2 are:

~F1T = (T1 −m1g)z (126)~F2T = (T2 −m2g)z (127)



Heavy Pulley

m1

m2

~w1

m1

~T1

~w2

m2

~T2

−~T1 −~T2

z

The torque acting on the pulley is~τ = R(T1 − T2)x



Heavy Pulley

m1

m2

Let ~a1 = ai z and ~α = α~x . Then

T1 −m1g = m1a1 (126)T2 −m2g = m2a2 (127)

R(T1 − T2) = Iα (128)

Unkowns: T1, T2, a1, a2, and α: 5unknownsThe remaining two eqns are:

a1 = −a2 (129)αR = −a1 (130)



Heavy Pulley

m1

m2

The solutions of these equations are:

a1 = −a2 =(m2 −m1)R2

I + (m1 + m2)R2 g

(126)

α =(m1 −m2)R

I + (m1 + m2)R2 g (127)

T1 =m1g(I + 2m2R2)

I + (m1 + m2)R2 (128)

T2 =m2g(I + 2m1R2)

I + (m1 + m2)R2 (129)



Rolling Without Sliding

θ

If the object rolls without sliding, thepoint contact is at restSuppose the object starts at rest at aheight h: ME = MghWhen it reaches the bottom of theincline, it has a velocity ~v .Then, its angular speed is ω = v

R

Its final mechanical energy is:

ME =12

Mv2+12

Iω2 =12

(M +

IR2

)ω2




θ

Its final mechanical energy is:

ME =12

Mv2+12

Iω2 =12

(M +

IR2

)ω2

Conservation of Energy:

Mgh =12

(M +

IR2

)v2 (130)

=⇒ v =

√2gh

MM + I

R2

(131)




θ

Using concepts of torque:The weight of the rolling objectcreates a torque τ = MgR sin θThe moment of inertia of the objectaround the point of contactI′ = I + MR2

The angular acceleration:α = MgR sin θ

I+MR2

The linear acceleration along theincline is a = MR2

I+MR2 g sin θ(< g sin θ)




θ

Using concepts of torque:The angular acceleration:α = MgR sin θ

I+MR2

The linear acceleration along theincline is a = MR2

I+MR2 g sin θ(< g sin θ)

Along the incline, its position as afunction of time is x(t) = 1

2at2.The time it takes to reach the bottomfrom a height h is

hsin θ = 1

2at2 =⇒ t =√

2ha sin θ




θ

Using concepts of torque:Along the incline, its position as afunction of time is x(t) = 1

2at2.The time it takes to reach the bottomfrom a height h is

hsin θ = 1

2at2 =⇒ t =√

2ha sin θ

Its translational speed at the bottomis:

v = at =

√2hasin θ

=

√2gh

MM + I

R2




θ

Using concepts of torque:Its translational speed at the bottomis:

v = at =

√2hasin θ

=

√2gh

MM + I

R2




θ

Q: We have shown thata = MR2

I+MR2 g sin θ(< g sin θ) for linearacceleration of the CM. We also know thatthe acceleration of the CM for any systemof particles is determined completely bythe total force acting on the system,independent of whether the object isrotating or not. If the object was notrotating, its acceleration would have beengiven by a = g sin θ. Which force causesits acceleration to further reduce?



Concept Questions

Object A sits at the outer edge (rim) of a merry-go-round, and object Bsits halfway between the rim and the axis of rotation. Themerry-go-round makes a complete revolution once every thirtyseconds. The magnitude of the angular velocity of Object B is

A half the angular speed of Object A .B the same as the angular speed of Object A .C twice the angular speed of Object A .D impossible to determine



Concept Questions

Which has the smallest I about its center?A Ring (mass m , radius R )B Disc (mass m , radius R )C Sphere (mass m , radius R )D All have the same I.



Concept Questions

In this problem ignore any friction/drag. Suppose that you release(from rest) an object from a very high building. Where does it fall?

A straight downB a bit to the northC a bit to the southD a bit to the eastE a bit to the west



QUIZ 6

A mass m1 = 2 kg that has a velocity ~v1 = (3 m/s)x + (4 m/s)ycollides with a mass m2 = 3 kg that moves with a velocity~v2 = (2 m/s)y . The two masses stick together.

1 What is their common velocity after the collision?2 How much kinetic energy is converted into internal energy in this

collision?


phys109-mechanics - middle east technical...

Documents