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PHY 113 C Fall 2013 -- Lecture 17 110/24/2013

PHY 113 C General Physics I11 AM – 12:15 PM TR Olin 101

Plan for Lecture 17:Review of Chapters 9-13, 15-16

1. Comment on exam and advice for preparation

2. Review3. Example problems


Webassign questions – Assignment #15

Consider the sinusoidal wave of the figure below with the wave function y = 0.150 cos(15.7x − 50.3t)where x and y are in meters and t is in seconds. At a certain instant, let point A be at the origin and point B be the closest point to A along the x axis where the wave is 43.0° out of phase with A. What is the coordinate of B?

xoo 7.15

18043



A transverse wave on a string is described by the following wave function. y = 0.115 sin ((π/9)x + 5πt)where x and y are in meters and t is in seconds.

(a) Determine the transverse speed at t = 0.150 s for an element of the string located at x = 1.50 m.

(b) Determine the transverse acceleration at t = 0.150 s for an element of the string located at x = 1.50 m.

tx

ttxy 5

9cos

9115.0),(

tx

ttxy 5

9sin

9115.0),( 2

2

2



A sinusoidal wave in a rope is described by the wave function y = 0.20 sin (0.69πx + 20πt)where x and y are in meters and t is in seconds. The rope has a linear mass density of 0.230 kg/m. The tension in the rope is provided by an arrangement like the one illustrated in the figure below. What is the mass of the suspended object?

mgTk

c 69.0

20

tkxy sin0

T

mg


Comment about exam on Tuesday 10/29/2013


iclicker questionWhat is the purpose of exams?

A. Pure pain and suffering for all involved.B. To measure what has been learned.C. To help students learn the material.D. Other.


Advice on how to prepare for the exam

Review lecture notes and text chapters 9-13,15-16 Prepare equation sheetWork practice problems

Topics covered

Linear momentumRotational motion and angular momentumGravitational force and circular orbitsStatic equilibriumSimple harmonic motionWave motion


What to bring to exam:

Clear headCalculatorEquation sheetPencil or pen


iclicker question:Have you looked at last year’s exams? A. Yes B. No


Linear momentum What is it? When is it “conserved”? Conservation of momentum in analysis of collisions Notion of center of mass

dtd

dtmd

dtdmm

m

pvvaF

pv

sNm/skg :momentumlinear of Units :momentum"linear " Define

if

f

i

f

i

ddt

ddt

pppFI

pF

:Impulse


Linear momentum -- continued

Physics of composite systems

ii

i

ii

i

ii

iii

ii dt

ddtmd

dtdmm pvvaF

:law second sNewton'

ii

ii

ii

ii

ii

dtd

final initial

(constant)

0

: then,0 if that Note

pp

p

p

F


Example – completely inelastic collision; balls moving in one dimension on a frictionless surface

i

iiv

iviv

vvv

vvv

pp

ˆ 0.125

/5.03.0

ˆ15.0ˆ23.0

ˆ/1 ,ˆ/2

5.0 ,3.0For

21

21

21

2211

212211

final initial

m/s

sm

smsm

kgmkgmmmmm

mmmm

f

ii

iif

fii

ii

ii


Examples of two-dimensional collision; balls moving on a frictionless surface

-- equations more 2 Need

,,, :Unknowns,, :Knowns

sinsin0

coscos

21

121

2211

221111

final initial

ff

i

ff

ffi

ii

ii

vvvmm

vmvm

vmvmvm

pp


The notion of the center of mass and the physics of composite systems

2

2

2

2

2

2

:Define

:law second sNewton'

dtdM

mMM

mdtmd

dtdmm

CMtotal

ii

ii

iii

CM

i

ii

i

ii

iii

ii

rFF

rr

rraF


Finding the center of mass

ii

iii

CM mMM

m

rr

ji

jiir

jiir

ˆ00.1ˆ750 4

ˆ22ˆ21ˆ)1)(1(

ˆˆˆ2 ;1 :example In this

321

332211

321

mm.

mmm

mmmymxmxm

kgmkgmm

CM

CM


Rotational motion and angular momentum Angular variables Newton’s law for angular motion Rotational energy Moment of inertia Angular momentum

dtddtd


Review of rotational energy associated with a rigid body

iii

iii

iii

iiirot

rmI

Irm

rmvmK

2

222

22

here w21

21

21

21

:energy Rotational


Moment of inertia: i

iirmI 2

22MaI 22 22 mbMaI


22

21

21

:object rolling ofenergy kinetic Total

CM

CMrollingtotal

MvI

KKK

CMvRdtdR

dtds

dtd

: thatNote

22

222

21

21

21

CM

CM

CMrollingtotal

vMRI

MvRRI

KKK

22

21

21

:object rolling ofenergy kinetic Total

CM

CMrollingtotal

MvI

KKK

CMvRdtdR

dtds

dtd

: thatNote

22

222

21

21

21

CM

CM

CMrollingtotal

vMRI

MvRRI

KKK

CMCM


iclicker exercise:Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?

AB C

2MRI A 22 5.0

21 MRMRIB

22 4.052 MRMRIC

2

22

/12

01210

MRIghv

vMR

IMMgh

UKUK

CM

CM

ffii


How can you make objects rotate?

Define torque:

t = r x F

t = rF sin

r

F

αarτFraF

Imm

sinr

F sin

ατ I:motion rotationalfor law sNewton'


Example form Webassign #11

X

t1

t3

t2

iclicker exerciseWhen the pivot point is O, which torque is zero?

A. t1?B. t2?C. t3?


Vector cross product; right hand rule

sinBACBAC

ˆ ˆ ˆ ˆ ˆ ˆ 0ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ

i i j j k ki j j i kj k k j ik i i k j


From Newton’s second law – continued – conservation of angular momentum:

(constant)

0 0 If

:Define

L

Lτ

prL

prτFr

dtd

dtd


Example of conservation of angular momentum

wheelbf

wheelwheelbf

wheelibiwheelfbf

LL

LLL

LLLL

2

0


Summary – conservation laws we have studied so far

Conserved quantity Necessary conditionLinear momentum p Fnet = 0Angular momentum L tnet = 0

Mechanical energy E No dissipative forces


Fundamental gravitational force law and planetary motion Newton’s gravitational force law Gravity at Earth’s surface Circular orbits of gravitational bodies Energy associated with gravitation and orbital motion


Universal law of gravitation Newton (with help from Galileo, Kepler, etc.) 1687

212

122112

ˆrmGm rF

2

211

kgmN 10674.6

G


Gravitational force of the Earth

RE m

2226

2411

2

2

m/s8.9m/s)1037.6(

1098.51067.6

E

E

E

E

RGMg

RmGMF

Note: Earth’s gravity acts as a point mass located at the Earth’s center.


days 27.4 s 32367353.951098.51067.6

)1084.3(π2

π2

π2ω

:Earth todueMoon for law sNewton'

2411

38

3

2

2

E

EM

EMEM

EM

E

EM

MMM

GMRT

RT

Rv

RGM

Rva

M aF

Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth)

REM

F

a

v


m1

R2

R1

m2

v1

v2

Circular orbital motion about center of mass

CM

21

321

21

2

111

1

2

1

11

1

21

1

2211

2

22

2221

21

1

21

1

2

212

mmGRRTT

TRm

RTRm

Rvm

RmRmRvm

RRmGm

Rvm

2

31

21

321

21

1212

22

then if that Note

GmR

mmGRRTT

RRmm


(const)

0

L

Lτdtd

m1

R2

R1

m2

v1

v2

L1=m1v1R1

L2=m2v2R2

L = L1 + L2

2

2

1

1

RL

RL

Note: More generally, stable orbits can be elliptical.


Gravitational potential energy

rmGm dr

rmGmrU

rmGmdrU

r

gravity

r

rgravity

ref

212

21

221

''

)(

ˆ )(

rFrF

hRmGMhRrU

E

SEEgravity

)(

Example:


Analysis of static equilibrium

Meanwhile – back on the surface of the Earth:

Conditions for stable equilibrium

0 : torqueof Balance

0 :force of Balance

ii

ii

τ

F


0)()2( :Torques 1 CMg RmgmF


T

Mgmg

**X

2/

x

t

sin2//

0sin2

0

MgmgxT

TMgmgx

NTNMgNmgmxmo

313200 600

2 8 53For


Some practice problems


From webassign:

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.800 rev/s in 2.00 s? (State the magnitude of the force.)

FR

view from top:

2

21 MRI

I

αFrτ


From webassign:A 10.3-kg monkey climbs a uniform ladder with weight w = 1.24 102 N and length L = 3.35 m as shown in the figure below. The ladder rests against the wall and makes an angle of θ = 60.0° with the ground. The upper and lower ends of the ladder rest on frictionless surfaces. The lower end is connected to the wall by a horizontal rope that is frayed and can support a maximum tension of only 80.0 N.

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