lecture non inertial system

8/2/2019 Lecture Non Inertial System

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Non Inertial System

Non Inertial System

Physics of Magnetism and PhotonicsInstitut Teknologi Bandung FI 2101 Mekanika 2

Non inertial SystemSo far we have only considered coordinate system (referenceframe) that are stationary or moving with constant velocity,i.e. Inertial System. And hence, we can apply Newtons Lawof dynamics.

When a coordinate system (reference frame) is accelerated,

i.e. Non Inertial System, we may not apply the Newtons Lawto determine the dynamics of a system.

A Non Inertial Observer that applies the Newtons Law, willcreate fictitious forces to explain the dynamics of thesystem that he observes.


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Non Inertial System


Accelerated System

The non inertial(accelerated) observersees a stationary ball

a

Ffict

T

mgmg

Ta

0sin

0cos

fictFTmgT

The inertial observer seesthe ball swings because it

is being accelerated

maTmgT

sin0cos

amFmaFfictfict

Non Inertial System


Rotating System

)(tr

Consider 2 reference frame, S(black) and S (red) which isrotating around the Z axis withangular velocity

Using the right hand rule, wecan define an angular velocity

vector

Consider a particle that initiallyat )(tr

)( ttr

r

)(tr

t

After time t, coordinate S has rotated, and the particle hasmoved to )( ttr

Rotating coordinate system is also anon inertial frame centripetal acc.


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Non Inertial System


Rotating System )( ttr r)(tr

t

)(tr

r )( ttr

According to S, the displacementof the particle is

)()( trttrr

S also measures the displacement,

but according to S, the initialposition of the particle is at )(tr

Hence, S measures the

displacement as)()( trttrr

Note that )()( ttrttr

t

Non Inertial System

Physics of Magnetism and PhotonicsInstitut Teknologi Bandung

)(tr

)(tr

t

FI 2101 Mekanika 6

Rotating SystemWe want to find the relation ofand

r

r

)()()()(

trtrr

trttrr

sinr )()( trtr

)()( trtrrr

or

Note that

sin)()( trtrtr

tsinr

sintr

)()( trtr

Taking the direction into account,we can write

trtrtr

)()(


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Non Inertial System


Rotating SystemThus we have trr

trtrrr

)()(

or

Divide through with t

rt

r

t

r

r

dt

rd

dt

rd

rotin

This is a GENERAL relation that relates the rate of changeas observed by an inertial observer (S) and by a rotating

observer (S)

In particular

dt

d

dt

d

dt

d

rotin

Non Inertial System


Dynamics in Rotating SystemAfter knowing how to evaluate a rate of change in a rotatingsystem, we can now calculate the acceleration as observedby both observer

rdt

drdt

rd

dt

rd

rdt

rd

dt

d

dt

rd

rotrot

inrotin

22

2

2

2

rdt

dmrm

dt

rdm

dt

rdm

dt

rdm

rotinrot

2

2

2

2

2

Thus the rotating observer S measures

fictitiousFFF

in

dt

rdmF

2

2

rotdt

rdmF

2

2


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Non Inertial System


Rotating SystemWhere the fictitious forces that are observed by the rotatingobserver are

rotdt

rdm

2 : Coriolis force that is only observed if

the particle is moving with respect to therotating observer

rm

rdt

dm

: Centrifugal force that have an equalmagnitude as the Centripetal force

: a fictitious force that is only observed ifthe angular velocity is not constant

Non Inertial System


A cylindrical bucket half filled with water is rotated withconstant angular velocity along its axis. Determine theshape of the water surface.

Example

z


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Non Inertial System


According to an observer that is rotating with the bucket, thewater is stationary, hence the sum of all forces acting on awater element must vanish. Obviously the water elementhas a mass, and because it is in contact with other water

element, there is a contact (normal) force . Thus, theobserver concluded (created) that there should be a(fictitious) force acting radially on the water element.

Example

centrifF

contactF

0sin0cos

centrifcontact

contact

FFmgF

contactF

gm

centrifF

z

Non Inertial System


With then

Recall that the shape of water surface has a gradient of

Example 2mFcentrif

2sin

cos

mFF

mgF

centrifcontact

contact

g

2

tan

contactF

gm

centrifF

zgd

dz

2

tan

Integrating yields

In other words, thesurface of the water is aparabolic.

22

2

gz


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Non Inertial System


A ball can move along a rod with out friction. The rod isrotating with constant angular velocity (assume that thereis no gravity). Because of the rotation, the ball move awayfrom the axis of rotation. Determine all forces (and draw the

forces) that acts on the ball according to an observer in thereference frame of the ball. Determine the position of theball as a function of time .

Example

r

)(tr

Non Inertial System


rrmrmF lcentrifuga )(2

222 vmrmdt

rdmF

rot

Coriolis

CoriolisFN

According to a rotating observer, the forces on the ball arecentrifugal force, Coriolis force (since the ball is movingwith respect to this observer) and the normal force from therod to the ball.

Example

Fcentrifugal

FCoriolis

N

r


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Non Inertial System


Thus, the ball is moving along the rod because of thecentrifugal force, hence

or,

Suppose the initially of the ball is at rest a distance A from

the axis, thus

FI 2101 Mekanika 15

rrmFrm lcentrifuga 2

Example

Fcentrifugal

FCoriolis

N

r

tt beaetrrr )(02

)cosh()( tAtr

Non Inertial System


X

YZ

According to the player (the XYZcoord. rotating with angular

velocity ), the position of

the ball, its velocity and itsacceleration are

Soccer on a rotating tableA football player kicks a ball along

the surface of a rotating table withinitial velocity v0. The radius of the

table is Rand it is rotating with

angular velocity .

Determine v0 if the ball leaves the

table at a point that is exactly 90from the direction that it was kicked.

0v

k

jyixrjyixrjyixr ,,

X

Y

r

rCor

F

cfF


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Non Inertial System


Soccer on a rotating table

)(2

)]()[(2

0

00

22

jxiym

xjiym

yx

kji

mrmFCor

][)]([

0

00

)]()([)(

22 yjximyjxim

xy

kji

m

xjyimrmFsf

The Coriolis force,

The centrifugal force,

X

Y

r

rCor

F

cfF

Non Inertial System



sfCor FFrm

yxy

xyx

2

2

2

2

ttvy

ttvxtxtyty

ttvy

ttvxtxtytx

x

y

x

y

)sin()(

)cos()()sin()cos()(

)cos()(

)sin()()cos()sin()(

00

0000

00

0000

The equation of motion of the ball is

or in components

With a general solution given by

X

Y

r

rCorF

cfF


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Non Inertial System


Soccer on a rotating table0,,0

0000 yx vvvyx

)sin()(

)cos()(

tvtty

tvttx

2)tan()(

)(

2tan tttx

ty

RvvtytxR

2

4)2()2(

2

22222

With initial condition :

we obtained

Thus the ball leaves the rotatingtable 90 from initial direction

Hence,

X

Y

r

rCor

F

cfF

Non Inertial System




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Non Inertial System


Effects of the Earth Rotation

7 27 10 5, rad / s

Angular velocity of Earth rotation about its axis is rathersmall :

Then the centrifugal force can be neglected

2)( rmF lcentrifuga

Non Inertial System


Projectile MotionConsider the Earth with rotatingcoordinate system as depictedin the figure. The Z axis isperpendicular to the Earthsurface. The X axis is directedto the East and the Y is

directed to the North. Neglectthe centrifugal force observedby the Non Inertial observer onthe Earth surface, since themagnitude of this force isproportional to 2 which is verysmall.

N

S

Y Z

X

m


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Non Inertial System


Projectile MotionPosition of a particle with massmaccording to an observer thatrotates with the Earth is

thus,

N

S

Y Z

X

m

r x e y e z e x y z ' ' '

d rdt

x e y e z erot

x y z

2

2

' ' '

dr

dtx e y e z e

rot

x y z

' ' '

YZ

X

cos sin ' 'e ey z

Non Inertial System


Projectile MotionAccording to the observer, the equation of motion of theparticle is

At the latitude angle of , components of the Earths angularvelocity vector is

Hence, the equation of motion in the X, Y and Z axis are:

md r

dtmg F mg e m

dr

dtrot

coriolis z

rot

2

22

'

cos sin ' 'e ey z

( cos sin ) x z y2

sin y x2

cos z g x2


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Non Inertial System


Projectile MotionIntegrate the y and z components of equation of motion withrespect to twith initial velocity condition

to obtain

Substitute this result to the x component of the equation ofmotion and integrate (neglecting terms that is proportional

to 2) we obtain

Integrate once again with initial position

v v e v e v e x x y y z z0 0 0 0' ' ' ' ' '

sin ' y x vy2 0

cos ' z g t x vz2 0

cos ( cos sin )' ' ' x g t t v v v z y x 2

0 0 02

x x v t g t t v v x z y0 01

3

3 2

0 0' ' 'cos ( cos sin )

r x e y e z e x y z0 0 0 0 ' ' '

Non Inertial System


Projectile MotionUse the last result to integrate the y and z components ofthe equation of motion (neglecting terms that is proportionalto 2) yield

Which together with the previous result

Specify the position of the mass m as observed by therotating observer.

x x v t g t t v v x z y0 01

3

3 2

0 0' ' 'cos ( cos sin )

y y v t v t x t y x0 0 02

02' ' sin sin

z z v t g t v t x t z x0 01

2

2

0

2

02' ' cos cos


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Non Inertial System


Projectile Motion ExampleA marble is dropped from a high h on a location that has alatitude angle . Determine the position of the marble whenit reaches the Earth surface.

The initial condition is

Using this values in the general solution we get

0,0,0

,0,0

000

000

zyx vvv

hzyx

cos)sincos(cos3

3

1'0'0

23

3

1'00 tgvvttgtvxx yzx

0sin2sin 02

'0'00 txtvtvyy xy

2

2

10

2

'0

2

2

1'00

cos2cos tghtxtvtgtvzz xz

Non Inertial System


Projectile Motion ExampleSuppose h = 100 m and = 45 LU, then when the marblereached the Earth surface means z = 0, thus

and

mtgx 015.0cos3

3

1

5210

2002221

g

httghz


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Non Inertial System


Anti Cyclone andCyclone system

rotv

rotv

rotv

CoriolisF

HH

Northern Hemisphere Southern Hemisphere

Cyclones develop due to the

Coriolis force

rotCoriolisvmF

2

L L

Non Inertial System


Cyclone System

Tropical Cyclone in theNorthern Hemisphere

A large low pressure area swirls off thesoutheastern coast of Iceland

L


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Non Inertial System


The Great Red Spot of JupiterThe Great Red Spot is a large high pressure storm in thesouthern hemisphere of Jupiter that has lasted over 600years

Mass 1.90 x 1027 kgDiameter 142,800 kmMean density 1314 kg/m3

Escape velocity 59500 m/sAverage distance from Sun 5.203 AURotation period 9.8 hrRevolution period 11.86 yr

Non Inertial System


The Great Red Spot of Jupiter


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Non Inertial System


The Great Red Spot of Jupiter

Hhigh pressure storm inthe southern hemisphere

Non Inertial System


Foucault PendulumThe effect of Earth rotation inthe form of Coriolis was firstdemonstrated by Foucault in1851. In his demonstration,Foucault used a heavy mass

to minimize the effect of air

flow.

Even though the magnitude ofthe Coriolis force is small andgravity is dominant force inthe motion of the pendulum

Z

X

Y

gm

T

vT

hT

Foucault showed that the plane of oscillation rotates with anangular velocity that depends on the latitude.


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Non Inertial System


Foucault PendulumAssume that the angle of oscillation issmall and the length of the string is l,

then the equation of motion of the ball is

with Earth angular velocity given as

The eq. of motion in components are

Z

X

Y

gm

T

vT

hT

rotrotdt

rdmTgm

dt

rdm

22

2

cos sin ' 'e ey z

sin2

sincos2

2

2

2

2

rotrot

rotrotrot

dt

xdmT

l

y

dt

ydm

dtyd

dtzdmT

lx

dtxdm

Non Inertial System


Foucault PendulumWe neglect the z-component of thevelocity,

And since the oscillation angle issmall, the string tension is similar to

weight of the ball, hence

rotrot

rotrot

dt

xdy

l

g

dt

yd

dt

ydx

l

g

dt

xd

2

2

2

2

2

2

rotdt

zd

Where is the z-component of the Earth localangular velocity.

sin' z

Z

X

Y

gm

T

vT

hT


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Non Inertial System


Foucault PendulumThese last two equations arecoupled differential equation of thecoordinates x and y. In order to be

able to solve this coupled equation,consider another coordinate(X,Y,Z) that is rotating with respectto coordinate (X,Y,Z) with angular

velocity . Then the relationbetween these coordinates are

Z

X

Y

gm

T

vT

hT

X

Z

Y

t

)cos()sin()sin()cos(tytxy

tytxx

Non Inertial System


Foucault PendulumTransform the two coupledequations of motion with theabove transformation anddropping the small terms

proportional to 2 we have

0)sin()cos(2

2

2

2

ty

l

g

dt

ydtx

l

g

dt

xd

rotrot

Z

X

Y

gm

T

vT

hT

X

Z

Y

t


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Non Inertial System


Foucault PendulumThis equation will be satisfied ifthe following is satisfied

Z

X

Y

gm

T

vT

hT

X

Z

Y

t0

0

2

2

2

2

yl

g

dt

yd

xl

g

dt

xd

rot

rot

Non Inertial System


Foucault PendulumThus in the coordinate frame(X,Y,Z) the ball is oscillating

harmonically, while thiscoordinate system is rotating withrespect to the coordinate system(X,Y,Z) with angular velocity

And the period of this precession

is

sin' z

hrT

sin

24

sin

22

Z

X

Y

gm

T

vT

hT

X

Z

Y

t


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Non Inertial System


Foucault Pendulum (1)

Non Inertial System


Foucault Pendulum (2)


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Non Inertial System


Rotation of Rigid Body From previous lecture, we have seen the problem thatarises when we want to study the dynamics of a rigidbody that is in motion in space.

xy

z

xy

z

IL

dt

Ld

labinertial

,

,

The moment ofinertial tensorkeeps changingas the rigidbody is inmotion

Non Inertial System


Rotation of Rigid Body We can calculate a constant moment of inertial tensor

using a coordinate system that is fixed to the rigid body.

However, if now we apply the Newtons Law of motion,

we must restate it in a rotating coordinate system, calledthe Eulers equation

Idt

dI

dt

Id

IIdt

d

LdtLd

dtLd

rot

axisprincipal

axisprincipal

axisprincipal

rot

axisprincipal

axisprincipalrotlabinertial ,,


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Non Inertial System


ExampleChoose the coordinate system as depicted in the previousfigure, thus they are the principal axis of the rod. Then,

Note also that in this problem, the total torque is zero ( ),since external force (gravity) acts at the center of mass whichis passed by the axis of rotation.Thus, Euler equation yields

cos,sin, zyx

0

zzzyyyxxx

zyx

zyx

z

y

x

zzz

yyy

xxx

axisprincipalaxisprincipal

III

eee

dtd

dtd

dtd

Ie

Ie

Ie

I

dt

dI

00

00

00

Non Inertial System


ExampleWe are interested to find (t), thus we just look at the x-

component of the torque

Note that because the torque is zero ( ) then we have

Or,

With small angle approximation ( ), we have

0x

yzyyzz

xxxx II

dt

dI

sincos02

yyzzxx III

02sin22

xx

yyzz

I

II

22sin

02

xx

yyzz

I

II


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Non Inertial System


ExampleWhich is a harmonic oscillator equation with solution,

0cos)( t

I

IIAt

xx

yyzz

Non Inertial System


lecture non inertial system

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