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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
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