Rot Mech_last Part (1)
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Transcript of Rot Mech_last Part (1)
![Page 1: Rot Mech_last Part (1)](https://reader034.fdocuments.net/reader034/viewer/2022051402/5695d0d71a28ab9b0294193e/html5/thumbnails/1.jpg)
• Precession of top
• Euler Equations
• Euler Angles
• HEAVY Symmetrical Top
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The Precession of Top
3
2
1
00
00
00
I
Top spinning about z and gravity
is switched off
zL;z 3
As no torque acting, the top continues
to spin indefinitely about the same axis
with angular velocity
…..(1)
R
z
x
y
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Gravity ON
Lets turn on the gravity causing a torque (make it weak..how?)
gMRN
KEY Points:
Magnitude is RMgsin and direction normal to z and
axis of the top
Existence of N indicates that L STARTS to change
Change of L implies starts to change. Components
x and y cease to be zero, though small as torque is
weak and eq (1) remains a good approx.
L changes in direction but not in magnitude zL 3
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Precession
Changing the arguments to mathematical expressions:
zzMgR
zgMzR
gMRz
NL
3
zzMgR
dt
zdˆˆ
ˆ
3
?????
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zzMgR
dt
zd
3
The top-axis rotates about the
vertical with angular velocity
given by the bracketed term.
This is precession. Note
is in the denominator ! !
Simple Physics
If z is a unit vector fixed in the body,
then its rate of change, as seen from
a fixed frame is:
zdt
zd
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Consequence
The earth spins on its axis, much like a spinning top & the
axis of spin is inclined at an angle 230 from the normal to
the earth’s orbit around the sun. Because of earth’s bulge
at the equator, sun & moon exert small torques on earth &
the torques cause earth’s axis to precess slowly (one full
turn in 26000 yrs), tracing out a cone of ½ angle 230 around
the normal to the orbital plane… precession of equinoxes
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Euler Equations
The Euler equations are the equations of motion of a rotating
body like motion of a top
KEY:
Before we launch Euler’s equation, there is a complication we
must now face up. To take advantage of principal axes, we like
to use those as our coordinate axes. But they are fixed in the
body and they rotate with the body and therefore noninertial !!
The way out is the following relation:
rotfix dt
d
dt
d
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Euler Equations
rotrotfixrotfix
Ldt
Ld
dt
Ld
dt
d
dt
d
In body (rot) frame The Euler equation zyx ,,Principal Axes
333222111 ILIL;IL
332211
321
III
zyx
L z
y
x
3
2
1
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Euler Equations
Combining:
2112333
1331222
3223111
IIIN
IIIN
IIIN
These are Euler equations for a rigid body
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Torque-free motion
For torque-free motion
211233
133122
322311
III
III
III
Consider:
a)
b) Body happens to rotate about one principal axis
say z-prime at t = 0 at t = 0
321 III
021
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The Consequence
211233
133122
322311
III
III
III
Under the above-mentioned conditions, rhs of Euler eq 0.
This means, all components of remain constant. In other
words if the body starts rotating about one principal axis, it
will continue to do so with constant angular velocity
Q. If you give a tiny kick to the rigid body, what happens ????
… use 3rd Euler eq. and consider small values of 1, 2
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The Consequence
131322
233211
III
III
1
2
3
21
13231
II
IIII
Motion is stable against small disturbances as long as the
bracketed term is positive i.e. for the cases when 213 I,II
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Transformation
The transformation from one coordinate system to another can
be represented by a matrix equation:
The rotation matrix completely describes the relative orientation of the two
systems. For a rigid body, contains three independent angles, (,,).
We will see sequence of three rotations about three different axes allows to
take us from fixed axis to body axis
XX Body system Fixed system
(all rotation anti clockwise)
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Euler Angles
X
N
Z Z
X
O
ON: Intersection of XOY and XOY’ planes
and hence perpendicular to Z’OZ
This is called line of nodes
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The Steps
Step 1: Rotate system about Z axis by
This takes OX to ON:
100
0cossin
0sincos
X
N
Z Z
X
O
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The Steps
Step 2: Rotate the body thro’ about the new X, i.e. ON. OZ OZ
cossin0
sincos0
001X
N
Z Z
X
O
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The Steps
Step 3: Rotate about new Z axis i.e. about OZ’ by . ON OX’ and we
have come to the body system
100
0cossin
0sincos
X
XX
X
N
Z Z
X
O
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Angular Vel in terms of Euler Angles
We express the angular velocities (about body
axes) in terms of . Cosines of the angles between the
components is given below:
zyx ,, ,,
X
N
Z Z
X
O
100
coscossinsinsin
0sincos
zyx
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Components
The components of these angular velocities along the body
set axes are:
cos
sincossin
cossinsin
z
y
x
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Heavy Symmetrical Top with one point fixed
1. Symmetry axis is z-prime and is one
of the principal axes.
2. Since one point is fixed, configuration
of top is completely specified by the
Euler angles
3. The distance of the CG on the axis of
symmetry from the fixed point is l
X
Y
Z
Axes drawn in green color correspond to
body axes
Z
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Rotational Angles
Rotn. about z prime “Spin”
‘Precession’ of z-prime about z
keeping constant
Rise/fall of symmetry axis wrt
the change in “nutation”
X
Y
Z Z
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Solution
We take the Lagrangian route:
cosmglV
1...........
cosI2
1sinI
2
1
IIII2
1I
2
1T
2
3
222
1
321
2
z3
2
y
2
x1
2.......
cosmglcosI2
1sinI
2
1L
2
3
222
1
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The Lagrangian
cosmglcosI2
1sinI
2
1L
2
3
222
1
Clearly, L is cyclic in and and does not depend on t explicitly
EmglIIH
aIL
p
bIIL
p
coscos2
1sin
2
1
cos
coscossin
2
3
222
1
3
3
2
1
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Top Motion
A general rotational motion of the top is a combination of those
motions….in terms of a and b, we have
3..............sinI
cosab2
1
This means that if is known as a function of time, and hence
will be determined
We now eliminate and from the above equations, write
u=cos and make little rearrangements,
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Final Equations
2
131
2
1
2
1
2 u1I
mglu2
II
a
I
E2
I
aubu
To make it look neater, we substitute:
1
31
2
111I
mgl2;IIaIE2;Ia;Ib
222 u1uuu
22 uu1uuf;dt
uf
du
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Comments
22 uu1uuf;dt
uf
du
Unfortunately this integration cannot solved analytically.
f(u) is cubic in u and the integral is known as an elliptic
function which can be obtained by computation.
We are not going to do any computation. Instead we try to have
general idea of the shape of the things:
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Looking into it
22 uu1uuf u=cos
As u=cos , it lies between +1 and -1 but for the moment we
forget that & consider for all values of u
a) As u , f(u) as the dominating term is +u3
b) When u = 1, only 2nd term in f(u) survives so that f(u) < 0,
unless
……consistent with a) & b), one gets two possible plots of
f(u) as a function of u
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Plots
Fig 2 corresponds to realizable motion but fig 1 is NOT
Hint f(u) must be non-negative in the domain where u
corresponds to real
f(u) has two zeros at u1 and u2 and between these values
of u (within 1), is real uf
So the motion will be limited between the corresponding
value of , i.e. nutation 2
1
21
1
1 ucos&ucos
22 uu1uuf;dt
uf
du
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Apparent conclusion
Lets rewrite some useful equations;
;dtuf
du;
sinI
cosab2
1
acosI3
In principle, the problem is solved. 1st eq is integrated numerically,
with given values of the constants , , and . Thus one has
known as a function of t. Using this, 2nd eq yields again by
computation . Knowledge of and suffices to determine
from the 3rd eq, so that , & are known for different times.
…..but we are interested to extract physics out of it….
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Some Interesting Physics
1) Top starts with only spin, no nutation or precession
0and0
2
0
2
0
2
222
u1uuuu
u1uuu
0002
1
uuaubsinI
cosab
(Unless u0 = 0)
;; 11 IaIb
Initial value of u
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More Physics
PE << KE (or equivalently total E) FAST TOP
3
22
1
3
22
3
1
I2
I
I2
aI
2
1T
2
ImglcosmglV
1II 31
Now, as PE < KE,
2
3
12
3
22
11
I
I
I2
I
2
I
…but not a big number
0
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Fast Top
2
0
2
0
22 u1uuuuu
Obviously, vanishes for . Let it vanish for also. u0uu 1u
2
110
2
10
2 u1uuuu0
2
2
110
u1uu
01
2 uu
nutational motion takes place within
a very small region of
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Fast Top continues….
Without much error, one can write,
0
22
0
2 sinu1u1
2
422
2
0
42
2
0
2
0
2
0
2
0
2
0
2
2
0
2
0
22
4
sinuu
4
sin
2
sinuu
sinuuuu
u1uuuuu
2
0
2
02
sinuu
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Fast Top continues
2
422
2
2
4222
4
sinuuuu
dt
d
4
sinuuu
This is just the equation for SHM telling us u varies harmonically
about the mean value u^bar with period 2/. The time period of
the nutation motion is given by:
acosI3
1Iaz3
1
I
I2T
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Conclusion
z3
1
I
I2T
1. For the ‘fast’ top, nutation is a SHM with small spread of value
of u.
2. T varies inversely as spin velocity. If latter is large, T will be small
correspondingly. Thus the nutational range is small and takes
place rapidly so one would hardly be able to observe it. One then
would think that there is no nutation at all and the motion is termed
it as pseudo-regular precession.
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The ‘Sleeping’ Top
Top starts with its axis vertical u0 = 1 and continues
SLEEPING TOP
u1u1
u1uuu
22
222
u = 1 is a double root of f(u) = 0 and then the question is how
f(u) vs u curve would look like…..????
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f(u) vs u curve
One can argue that curve II not curve I is the sleeping top
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Argument
With curve I, 3rd root of f(u) corresponds to real , say 1, so the top may
have a nutational motion between 1 and 0 but for curve II 3rd root give an
imaginary and f(u)>0 only for u=1 so the motion is restricted to u=1,
there is no nutation, nor even at precession but simply the top goes on
spinning with its axis vertical. Such a motion is called a sleeping motion
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Mathematical Condition
Condition: 3rd root of f(u)=0 must occur at u1>1
2
1
2
z
1
2
3
2
1
222
I
mgl4
I
I
2u1
u1u1u
Condition to sleep: 2
3
1z
I
mglI4
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“Wake up”
Q. Does a top “wake up” some time?
A. The top does “wake up”. This is because the frictional
force cause a dissipation of the KE so that spin velocity
goes on decreasing and ultimately crosses the critical
value . The sleeping motion can then no
longer go on, nutation and precession set in
2
31 ImglI4
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Problem
Find the torque needed to rotate a rectangular plate of length
b and breadth a about a diagonal with constant
x
j
yba
bx
ba
a
yyxx
zyx yx
ˆˆ
ˆˆ.ˆˆ.
ˆ0ˆˆ
2222
zba
ababMN ˆ
120
0
22
22
3
21
a
b
y