Mechanics in Noninertial Frames

Why Frames of Reference?

Newton’s Equations :

2

2

dt

rdmamF

==

are meaningless without frame of reference, w.r.t which, the position vector, and consequently the acceleration vector, are measured.

A reference frame is a rigid body with three reference directions attached to it

x

y

z

r

v

a

Natural Question to ask :

Are Newton’s Equations (Consequently whole of mechanics) valid in all kinds of reference frames?

Answer :

Newton’s equations are valid in a special class of frames known as the inertial frames.

Inertial Frame (Uncritical Definition) :

A frame that moves with constant velocity without rotation

With respect to what ?

Another Inertial Frame!!!

A frame in which a body, not acted upon by any force, is either at rest or moves uniformly

A more fundamental definition is provided by Newton’s First Law :

Q. Does there exist such a frame?

A. If there exists one such frame, then there exist a whole class of infinitely many such frames, as provided by Galelian Principle of Invariance

Galelian Principle of Invariance

If Newton’s equations of motion are valid in one frame, then they are valid in any other frame, moving with uniform velocity without rotation, w.r.t. it

r r ′

R

S

S’

2

2

2

2

2

2

2

2

⇒dt

rd

dt

Rd

dt

rd

dt

rdRrr

S : Inertial Frame in which Newton’s Laws are valid

S’ : Another Frame, moving uniformly w.r.t S

FFdt

rdm

dt

rdm

2

2

2

2

Proof :

(force does not change)

A frame, that is very close to being an inertial frame, is the Celestial Sphere

Newton’s Grand Belief : There is one such frame, the absolute frame. The space, with respect to which this frame is at rest, is the absolute space

Red Frame : Frame of the Celestial Sphere

Yellow Frame : Frame fixed to the Earth

Consequence : Earth is a little less close to being an inertial frame, as it slowly rotates with respect to the celestial sphere

Newton’s Equation in an accelerated frame

r r ′

R

S

S’

S : Inertial Frame

S’ : Accelerated Frame

2

2

2

2

2

2

dt

Rd

dt

rd

dt

rd⇒Rrr

+

′=+′=

0aaa,or

+′=

0am-amam∴

=′

Fam∴ ′=′

Where,fictitioustrue FFF

+=′

and, 0fictitious am-F

=

However,

bodytheonactingforceactualthe,Fam true

=

Thus, Newton’s equations of motion are still valid in an accelerated frame, provided, a fictitious force is added to the real force

Mpua0 Mpia0 Mpiaa0

0a

Fictitious forces (also called inertial force) on objects in an accelerated frame are proportional to their masses!

Mpug Mpig Mpiag

Earth

Uniform Gravity!

The two forces, inertial and uniform gravity are totally indistinguishable

One can banish uniform gravity by accelerating his frame in an appropriate manner

Earth

g

One can take two viewpoints

1. Inertial forces are as real as gravity

2. Gravity is as fictitious as inertial forces

Einstein took the second viewpoint and created general theory of relativity

One can make uniform gravity completely disappear by dropping his frame in this gravitational field

What about nonuniform gravity?

Can it be also made to vanish?

Ans : The major part of it can be made to vanish. However, a residual part will still remain, and this residual gravity is the well known Tidal Force

Tidal Force

Earth

Earth

g0

MoonEarth

Tidal Force of the Moon on Earth

RrP1 P2

322TidP R

GMr2

R

1

)rR(

1GMg

1

322TidP R

GMr2

R

1

)rR(

1GMg

2

M

3R

GMr

Distribution of Tidal Gravity on the surface of Earth

Moon

Earth

Moon

Low tide

Low tide

High tide High tide

Earth

The Moon Wins Over the Sun in its Tidal Effects on Earth !!

Sun’s Gravity on Earth is Much Stronger than that of Moon :

mM

sMSR

mR

32m

2s

s

m

s

m 106R

R

M

M

g

g

8

s

m 107.3M

M

2

m

s 104R

R

4.2R

R

g

g

g

g

m

s

s

m.Tid

s

.Tidm However,

Roche Limit

Roche Limit

3R

GMd2

2d

Gm

Equating the two :3/1

m

M

3/12

am

M2dR

R

a

Mm

d

If the tidal gravity of the primary over a secondary becomes stronger than the self gravity of the secondary, the secondaty is torn apart.

Inside the Roche Limit, no object can be held together by its gravitational attraction alone

If the densities of the primary and its satellite are the same, then

a26.1R

A more realistic calculation, taking into account the deformation of the secondary before break up gives

a45.2R

Roche Limit for Saturn

For a comet, whose density is low, the Roche Limit is much larger

The comet Scoemaker Levy, Breaking up after entering the Roche limit of Jupiter

Prob. 8.2

A truck at rest has one door fully open.

The truck accelerates forward at constant rate A, and the door begins to swing shut. The door is uniform and solid, has total mass M, height h, and width w.

a. Find the instantaneous ang. Velocity of the door about its hinges, when it has swung through 90 degrees.

b. Find the horizontal force on the door when it has swung through 90 deg.

w

dt

dI

MAw cos2

Writing ,

d

d

dt

d

d

d

dt

d

3

MwI

2

d

dI

MAwcos

2

Integrating,

0

2/

0

cos2

dIdMAw

Or, I

MAw

a)mA

A

Truck’s Frame

I2

AwMMAMaMAF

22

cm

Truck’s Frame

FMAb)

cmMaMAF

2cm 2

wa

Ground Frame

FMA

The net acceleration (w.r.t. the ground) of the CM of the door in the forward direction and in the position shown, is :

A2

wa 2

cm

I2

AwMMAMaF

22

cm

It is a pendulum, which is such that, it always points towards the centre of the earth, no matter what the motion of its point of suspension is, so long as the point of suspension moves in the local horizontal direction

Prob. 3 The Schuler Pendulum

Ordinary Pendulum (The Plumb line)

The plumb line defines the local vertical, so long as the point of suspension is at rest or is moving uniformly

a

g

atan 1

mg

ma

The plumb line will no more follow the local vertical

Under the bouncy movement of Professor Calculus, the pendulum has a tough time finding the local vertical

Captain Haddock can be a more serious non-inertial frame!

Thundering typhoons! Why cannot your pendulum behave, Professor Calculus?

Prob. 3 Find the time period of the Schuler pendulum

P

CGFFI

FFII as

S : Distance between point of suspension and CG

2)(; LRMFFMaFF III

FFI & FFII : fictitious forces on the pendulum

MasP

If the pendulum has to always point towards the centre of the earth, it must rotate about P with an angular acceleration that must be the same as the ang. acc. of P about the centre of the earth.

LR

askMIMas

)( 22

LRsk

s

122

Now, the period of a compound pendulum is :

g

R

g

LR

gs

skT 222

22

Plugging in the values :

26 /10&104.6 smgmR

we get :

.min84T

However, solving for s, we get

R

k

LR

ks

22

If msthenmk 15.0~,1~

Thus, Schuler pendulum is just a concept, however, a very useful concept in Inertial Navigation System

Rotating Coordinate System

x

y

z

x’

y’

z’(x,y,z) : Inertial frame

(x’,y’,z’) : Frame rotating w.r.t the inertial frame

Goal : Find the equation of motion of a particle in frame (x’,y’,z’)

Change in a vector that undergoes infinitesimal rotation about a fixed axis

A

A

sinAA

)ˆ( AnA

n

n : Unit vector along axis of rotation

If the vector is rotating about the axis with angular velocity , then

A

t

Result I

Adt

Ad

t)An(A

t)A(

Conversely, if the coordinate frame is rotating with angular velocity , a fixed vector will appear to rotate in the reverse sense. So,

Adt

Ad

Relationship between rates of change of a vector in two frames, rotating w.r.t. each other

x

y

z

x’

y’

z’

(x,y,z) : Inertial frame

(x’,y’,z’) : Frame rotating w.r.t the inertial frame

Green arrow : A vector changing with time)(tA

A

Result II

Let :

:in

dt

Ad

Rate of change of in the inertial frame

:rot

dt

Ad

Rate of change of in the rotating frame

A

A

Q : How are & related to each other?in

dt

Ad

rotdt

Ad

Claim : Adt

Ad

dt

Ad

inrot

Proof :

inA)(

rotA)(

Let & be changes in the vector as observed in the two frames, in time . t

The two would be the same, if there were no relative rotation between the two

However, due to the rotation of the primed frame, there will be an additional change in this frame

This additional change is : tAA app )()(

tAAA inrot )()()(

Adt

Ad

dt

AdOr

inrot

,

inrot dt

d

dt

dOr,

Velocity and Acceleration Vectors in the Two Frames

rot

rotrot dt

vda

Substituting from the first equation,

inin

in

in

rot

dt

)r(d

dt

vd

dt

vd

)r(vra rotin

rotrot dt

rdv

r

dt

rd

in

rvin

rotin

rot vdt

vd

inin vra

rv2)r(aa rotinrot

Equation of Motion in the Rotating Frame

Multiplying both sides of the above eq. by the mass of the particle :

)r(m)v(m2))r((mamam rotinrot

However, Newton’s eq. being valid in an inertial frame,

Fam in

Therefore, Eq. of motion in the rotating frame is :

)r(m)v(m2))r((mFam

.fict3

.fict2

.fict1realeff FFFFF

Thus, eq. of motion of a particle in a rotating frame is in the form of Newton’s equation, provided, following fictitious forces are added :

))r((m.1

The Centrifugal Force

)v(m2.2

The Coriolis Force

)r(m.3

Unnamed, as mostly absent

Centrifugal Force

)n)rn(r(m))r((mF 2cf

)axisfromaway(rm 2

n

r cfF

n

r

Coriolis Force

cF

v

Marble on a Roulette Wheel

On a merry-go-round in the night

Coriolis was shaken with fright

Despite how he walked

‘Twas like he was stalked

By some fiend always pushing him right

David Morin, Eric Zaslow ……

Example I : Car on a Revolving Platform (Chapter 3)

Equation of motion of the car in the frame of the platform :

cf.cor.fricrot FFFam

Car driven along a fixed radial line with uniform velocity

Since in the platform’s frame the car is moving with uniform velocity,

cfcorrfric FFF

))r((m)v(m2

ˆvm2rtvm 002

Rotating Vessel of WaterExample II :

)r(zgg 0eff

z

)yyxx(zg 20

The surface of a fluid must follow a gravitational equipotential surface

)z,y,x(

)0,0,0(

effeff dg)z,y,x(V (x,y,z)

222

0 yx2

zg

Constant potential surface is given by :

22

0

2

0 yxg2

zz

0z

If R is the radius of the cylindrical vessel, then the depth of the surface is :

0

22

g2

Rh

hR

Prob. 8.12 : A pendulum is fixed on a revolving platform as shown. It can swing only in a plane perpendicular to the horizontal axle.

M : Mass of pendulumL : Length of massless rod : Const. Ang. Vel. of platformSolution

The forces on the pendulum are shown. The frame of reference is the rotating platform.

mg

T

cfF

L

corF

cossinMLsinMgL 22rest

2cf )sinL(MF

)Lg(ML 2

L

Lg 2

g2

L1

g

L1

2

0

2

0

mg

T

cfF

L

This component alone, is mainly responsible for Coriolis effects

x

y

z

sin

cos

The Coriolis force on a body moving on the surface of the earth is as if the earth is rotating with angular velocity about the vertical

sin

Coriolis Effect on Earth’s Surface

Coriolis Effect on Cruise Missiles

vsinv2a c (Taking )

030

For a cruise missile fired at a target distant L away, the deviation S is :

v

L

2

1ta

2

1s

22

c

,s/m700v,km10LFor

m5s

8.9. A 400 tons train runs south at a speed of 60 mi/h at a latitude of 60. What is the force on the tracks? What is the direction of the force?

0cor 60sinvmF

The force, being to the right of direction of motion, is to the west (red arrow)

x

y

z

060

v

Foucault Pendulum

The only force in the direction is the centrifugal force :

ˆrm2Fcor

rm2)r2r(m

Eq. of motion in the direction :

r

corF

v

x

y

Foucault pendulum on a rotating platform

A solution for is :

t

Thus, the plane of oscillations of the pendulum, rotates about the vertical with an angular velocity, which is the same as that of the rotating platform, but in the opposite sense.

x

y

z

sin

On the surface of the earth, at latitude , the rate of rotation is : , where is earth’s angular velocity. The rotation is clockwise.

sin

Foucault Pendulum on Rotating Earth

At the north pole, the plane of the pendulum, rotates once a day.

Coriolis force and hurricane formation

Northern hemisphere Southern hemisphere

Low

Northern Hemisphere

Southern Hemisphere

High Pressure Hurricanes Do not Form