Physics 121 Mechanics Lecture notes are posted on Instructor Karine Chesnel April 14, 2009 Final...

Physics 121

Mechanics

Lecture notes are posted on

www.physics.byu.edu/faculty/chesnel/physics121.aspx

InstructorKarine Chesnel

April 14, 2009

Final Review

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Do not forget to evaluate this classonline until April 15th

Thank you!

Quiz # 40

Final review 04/14/09

Physics 121 Winter 2009

* Today * last lecture

Last assignment:

• Online homework 24: (70pts)

until midnight today!

http://gardner.byu.edu/121w2/homework.html

For any question after this class:

Karine Chesnel N319 ESC 801-422-5687 [email protected]

I will be available until Friday April 24th

Physics 121 Winter 2009 - section 2

Class Average

First test 83 /100Second test 68/100Third test 74/100

Test average 75/100 (30%)Final (20%)Homework 85/100 (25%)Labs 96/100 (15%)quizzes 113/100 (10%)

Prepare well for the FINAL test!Try to increase your average test score

Class statistics

Final exam

• Friday April 17 through Wednesday April 22

• At the testing center : 8 am – 9 pm (Mo- Fri) 10 am – 4 pm (Sat)

• Closed Book

• Only bring: - Phys 121 (Chesnel)Memorization sheets (5pages)- Math reference sheet- Pen / pencil- Calculator- your CID

• No time limit

• Multiple Choice Questions: 30 questions

Final Review

ch 1 – ch 15

Part I: Kinematics Ch. 1- 4

Part II: Laws of Motion (material points) Ch. 5- 8

Part III: Laws of Motion (Solids rotation) Ch. 9 - 13

Part IV: Oscillatory motion Ch. 15

Part I

Kinematics

Ch. 2 Motion in one dimension• Position, velocity, acceleration• Case of constant velocity• Case of constant acceleration• Free falling motion

Ch. 4 Motion in two dimensions• Position, velocity, acceleration• Case of constant acceleration• Projectile motion• Circular motion (uniform & non-uniform)• Tangential and radial acceleration

Ch. 3 Vectors• Coordinate systems• Algebra

Ch. 1 Physics & measurments• Standard units• Dimensional analysis

Kinematics

The displacement is the difference between two positions

x1

x2

x= x2 – x1

Average accelerationt

Vaavg

(Instantaneous) acceleration dt

dV

t

Va

ttx

0

, lim

t

xVavg

Average velocity

(Instantaneous) velocity dt

dx

t

xV

ttx

0

, lim

The speed is the amplitude of the velocity


Motion under constant velocity

V = constant

x(t)

tt

x

0

x0

= Slope x/t

Position at t=0

x(t) = x0 + v.t


Motion under constant acceleration

a = constant

a(t)

t0

v(t) = v0 + a.t

v(t)

tt

v

0v0

V(t) is linear

x(t)

t0

x0

x(t) = x0 +v0.t+ ½ a.t2

x(t) is parabolic

Application: Free Fall a = g


Vectors components

Cartesian to polar conversion in 2D

x

y

yx

tan

22

)sin(

)cos(

y

x

x

y

(0,0)

A=(x,y)

x

y


Projectile Motion

r(t) = r0 + V0t + ½ g t2

r0

V0

y

x(0,0)

a = g = (0, -g)

V2

V1

V3

x(t) = x0 + V0,x t

y(t) = y0 + V0,y t - ½ gt2

v(t) = v0 + gt

Uniform velocity along x

Free falling along y

vertical

horizontal


Projectile

V0

y

x(0,0)

a = g

Vmax

Performances

H

Hits the ground

R

R = V02

sin (2 /g

The particle is projected with a speed V0 at angle

The maximal height is

The horizontal range is

H = (V0 sin )2 / 2g

ReviewCh.4 Motion in two dimensions 1/27/09Final review 04/14/09

Uniform Circular Motion

Angular speed is constant:

= .t

Velocity:

Vx = - R0 sin (tVy = R0 cos (t

Position

x = R0 cos (ty = R0 sin (t

Acceleration

ax = - R0 cos (tay = - R0 sin (t

a = - r

|V| = R0

The acceleration is centripetal. Its magnitude is

|a| = R0

(0,0) x

y

R0

V

a


Tangential and radialacceleration

General case

V1

V2

V3

a

a a

V is tangential to the trajectory

The components of the acceleration in the Frenet frame are:

• TangentialThe sign tells if the particlespeeds up or slows downat= dV/dt

• centripetalAlways directed toward the center of curvatureR radius of curvature

ac= V2/R


Generalization

positionr (t)

velocity

First derivative

dt

rdV

acceleration

Second derivative

2

2

dt

rd

dt

Vda

First integration

t

dtaV0

.

Second integration

t tt

dttadttVtr0 0

2

0

).().()(

a(t)


Quiz # 41Two racquet balls are thrown in the air at the same time

from the same height H. One ball (yellow) is thrown at some angle, with a vertical velocity V0,y, and horizontal velocity V0,x.

The other ball (blue) is thrown vertically with the same vertical velocity V0,y Which ball will hit the floor first?

A The blue ballB The yellow ballC Both of them

V0,yV0

x(t) = x0 + V0,x t

y(t) = y0 + V0,y t - ½ gt2

For the yellow ball the motion will be given by

For the blue ball the motion will be given by

y(t) = y0 + V0,y t - ½ gt2

Both balls touch the ground at the same time!

y

H


Part II

Laws of motion

Ch. 6 Newton’s laws applications• Circular Motion• Drag forces and viscosity• Friction• Fictitious forces

Ch. 8 Conservation of Energy• Mechanical energy• Conservation of energy

Ch. 7 Work and energy• Work• Kinetic energy• Potential energy• Work- kinetic energy theorem

Ch. 5 The Laws of Motion• Newton’s first law• Newton’s second law• Newton’s third law

Summary ofthe Laws of Motions

First Law: Principle of InertiaIn a inertial frame,

an isolated system remains at constant velocity or at rest

Second Law: Forces and motionIn an inertial frame

the acceleration of a systemis equal to the sum of

all external forcesdivided by the system mass

Fam

m

Fa

Third Law: Action and reaction If two objects interact,

the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction

to the force exerted by object 2 on object 1.

F1 F2


Forces of Friction

F < fs,max= s N

s is called the coefficient of static friction

• Static regime

fk= k N

k is called the coefficient of kinetic friction

• Kinetic regime

F

f

Static regime Kinetic regime

F

f

maxf

kf

Two regimes


amF

Work and kinetic energy

B

A

B

ABA rdamrdFW

..

Defining the kinetic energy

2

2

1mVK

Using Newton’s second law

Work- Kinetic energy theorem

ABBA KKKW


Mechanical energy

nccons WWK

ncmech WE

ncWUK

ncWUK

ncWUK )(

UKEmech

We define the mechanical energy Emech

as the sum of kinetic and potential energies

Sometimes, the work of non conservatives forces (friction, collision)is transformed into internal energy

intEWnc 0int EEmechthus


Closed System with conservative forces only

0 ncmech WE

Fcons

cstEmech

cstUK

There are no non-conservative forces working

The mechanical energy is constant

iiff UKUK

The mechanical energy is conserved between initial and final points


Examples ofPotential energy

2

2

1kxU k x

L0

Spring

Gravity

g

mghU g h

r

mMGU g

Gravitational fieldm

r

M


Quiz # 42

`F

mg

N

F

This man is pushing this box of 85kg on the carpet at a constant speed.

How is the magnitude of the force he needs to apply?

A Larger than the weight of the boxB Same than the weight of the boxC Same than the friction forceD Larger than the friction forceE None of the answers

amF

According to Newton’s law

Here the velocity is constant, so 0

F

fF

Thus and on the other hand

mgmgNf kk

This is true both vertically

and horizontally

gmN

fF

f


Part III

Laws of motion for Solids

Ch. 10 Rotation of solid• Rotational kinematics• Rotational and translational quantities• Rolling motion• Torque

Ch. 12 Static equilibrium and elasticity• Rigid object in equilibrium• Elastic properties of solid

Ch. 11 Angular momentum• Angular momentum• Newton’s law for rotation• Conservation of angular momentum• Precession motion

Ch. 9 Linear Momentum & collision• Linear momentum• Impulse• Collisions 1D and 2D

Ch. 13 Universal gravitation• Newton’s law of Universal gravitation• Gravitational Field & potential energy • Kepler’s laws and motion of Planets

Linear Momentum & Impulse

• The Newton’s second law can now be written as F

dt

pd

Vmp

• The linear momentum of a particle is the product of its mass by its velocity

Units: kg.m/s or N.s

cstp

• For an isolated system 0

dt

pd

• The impulse is the integral of the net force, during an abrupt interaction in a short time

f

idtFI

Ip

• According to Newton’s 2nd law:


Collisions

iiff pppp ,2,1,2,1

1. Conservation of linear momentum

2,22

2,11

2,22

2,11 2

1

2

1

2

1

2

1iiff VmVmVmVm

2. Conservation of kinetic energy

(2)

Elastic collision

iiff VmVmVmVm ,22,11,22,11

(1)

V1,i

V2,i

V1,

f

V2,f

Inelastic collision: change in kinetic energyPerfectly inelastic collision: the particles stick together


Collisions 1D

if Vmm

mmV ,1

21

21,1

if V

mm

mV ,1

21

1,2

2

• If one of the objects is initially at rest:

iif Vmm

mV

mm

mmV ,2

21

2,1

21

21,1

2

• Combining (1) and (2), we get expression for final speeds:

iif Vmm

mV

mm

mmV ,1

21

1,2

21

12,2

2

V1,i

V1,f V2,f

V2,i

Collisions 2D

• 3 equations

• 4 unknow parametersV1,i

x

y V1,f

V2,f


Solid characteristics

M OC dmr

• The center of mass is defined as:

C

O

dm

rdVdm

Ctot VMp

FaM C

• The moment of inertia of the solid about one axis:

dmrI 2

I

2' MDII

I’

D


Rotational kinematics

dt

d

dt

d

• Solid’s rotation

Angular position

Angular speed

Angular acceleration

RVt

• Linear/angular relationship

Velocity

Acceleration

• Tangential

• Centripetal

Rat 2RaC

For any point in the solid

• Rotational kinetic energy2.

2

1 IK


Motion of rolling solid

P

C

R

Non- sliding situation

• The kinetic energy of the solid is given by the sum of the translational and rotational components:

Ksolid = Kc + Krot

22

2

1

2

1 IVMK Csolid

222

2

1

2

1 IMRK solid

22 )(2

1 IMRK solid

cstKU sol If all the forces are conservative:


Torque & angular momentum

Fr

We define the torque

F

The angular momentum is defined as

dt

Ld

Deriving Newton’s second law in rotation

angular momentum Linear momentum

prL

sinrF

When a force is inducing the rotation of a solid about a specific axis:

For an object in pure rotation

Inet IL


Solving a problem

Static equilibrium

• Define the system

• Locate the center of mass (where gravity is applied)

• Identify and list all the forces

0

F• Apply the equality

• Choose a convenient point to calculate the torque (you may choose the point at which most

of the forces are applied, so their torque is zero)

• List all the torques applied on the same point.

0

• Apply the equality


Gravitational laws

rMg ur

mMGrgmF

2

)(

Any object placed in that field experiences a gravitational force

Any material object is producing a gravitational field

rM ur

MGrg

2

)( M

r

ur

m

Fg

The gravitational field created by a spherical object is centripetal

(field line is directed toward the center)

The gravitational potential energy is

r

mMGU g

Ug

0 r


Kepler’s Laws

“The orbit of each planet in the solar system is an ellipse with the Sun as one focus ”

First Law

0LcstL

“The line joining a planet to the sun sweeps out equal areas during equal time intervals as the planet travels along its orbit.”

Second Law

cstmL

dtdA

20

“The square or the orbital period of any planet is proportional to the cube of the semimajor axis of the orbit”

Third Law

32

2 4R

GMT

S


Satellite Motion

Tsat =TEarth

= 1 day

Geostationary orbit

T

RV GSGS

2

• Satellite speed

(1)GS

EGS R

MGV 2

From Newton’s law

(2)

Escape speed

R

GMVesc

2

GS

E

R

mMGE

2The mechanical energy of

The satellite on orbit is


Quiz # 43

A planet has a mass twice the mass of the Earthand a diameter 0.7 times the Earth diameter.What would be the weight, in Newtons, of a 82kg personstanding at the surface of this planet?

A 334 NB 803 NC 3280ND 483 NE 4830 N

2p

ppp

R

mMGmgF

The weight of this person at the surface of this planet is

Compared to the weight on earth2

p

E

E

pEp R

R

M

MmgF

9.8 m/s2

Gravitational field on Earth

2

7.0

128.982

N3280

(Equivalent “mass’ on earth: 334kg!)


Part IV

Oscillatory motion (ch15)

• Harmonic equation and solutions

• Energy of harmonic oscillator

• Spring motion • Pendulum motions

• Damped oscillation

• Forced oscillation

x

Harmonic motionSpring

Equation ofthe motion is 0

2

2

xm

k

dt

xd Harmonicequation

A general solution to this harmonic equation is

)cos()( tAtx

Amplitudeangular

frequency

Phase constant(phase at t=0)

m

k

with

Unit = rad/s

frequency

m

kf

21

Unit = Hz

period

k

mT 2

Unit = s

F


Position, velocity and acceleration

)cos()( tAtx

)sin()( tAdt

dxtV

)cos()( 2 tAdt

dVta

Position, velocity and acceleration are all sinusoidal functions

Tx(t)

t

Position

Velocity

Acceleration

t

V(t)

Phasequadrature

t

a(t)

Opposite phase


Energy

cstkAUKEmech 2

2

1

0 t

U

K

0 x

U

K


The pendulum(punctual)

L

m

0The equation for motion is

02

2

L

g

dt

d

The solution for the angle position is

)cos()( max tt

L

gwith

The oscillation frequency does not depend on the mass m

The period of the oscillation is

g

LT

22


Torsional pendulum

A torsional pendulum uses the torqueinduced by torsion to oscillate

0 (rest)

For an angular displacement , the torque is k

The equation for the motion is then 2

2

dt

dI

dt

dLk

02

2

I

k

dt

d

A general solution is:

)cos()( max ttI

kwith

L

dt

Ld

According to Newton’s second law


Damped oscillations

If the oscillator is moving in a resistive medium:friction, viscosity…. The oscillation will be damped.

L0

x0

FA

Compressed

The expression of the spring force is xk ukxF

Applying the Newton’s second law

dg FFam

dt

dxbkx

dt

xdm

2

2Equation ofthe motion

The expression of resistive force is VbFd

Damping

coefficient


Damped oscillations

)cos(2

exp)(

tt

m

bAtx

with 2

20 2

m

b

The general solution for the motion is

T

x(t)

t

Underdamped

02

m

b

x(t)

t

overdamped

02

m

b

TCritical

02

m

b

x(t)

t

0


Forced oscillations

A general solution to this equation is

)cos()( tAtx

Amplitude Frequencyforced

Phase constant(phase at t=0)

kxdt

dxbtF

dt

xdm sin02

2

An external force is applied to the system, forcing the oscillation to a frequency

If any, resistive force is VbFd

xk ukxF

The spring force is

The external force is tFFosc sin0

dgosc FFFma


Forced oscillations

A general solution to a forced oscillation motion

)cos()( tAtx

with

2

220

2

0

mb

mF

A

m

k0and

Resonance

In absence of resistive forces, the amplitude of the oscillation is amplified to infinity when the force frequency

approaches the proper frequency 0

Low damping b

No damping b=0

Large damping b


GOOD LUCK With the FINALS !

To contact me:

Karine CHESNEL

[email protected]

Office: N319 ESC

801 – 422 – 5687


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