• Precession of top

• Euler Equations

• Euler Angles

• HEAVY Symmetrical Top

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

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

Precession

Changing the arguments to mathematical expressions:

zzMgR

zgMzR

gMRz

NL

3

zzMgR

dt

zdˆˆ

ˆ

3

?????

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

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

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

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

Euler Equations

Combining:

2112333

1331222

3223111

IIIN

IIIN

IIIN

These are Euler equations for a rigid body

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

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

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

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)

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

The Steps

Step 1: Rotate system about Z axis by

This takes OX to ON:

100

0cossin

0sincos

X

N

Z Z

X

O

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

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

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

Components

The components of these angular velocities along the body

set axes are:

cos

sincossin

cossinsin

z

y

x

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

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

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

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

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,

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

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:

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

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

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….

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

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

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

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

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

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.

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…..????

f(u) vs u curve

One can argue that curve II not curve I is the sleeping top

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

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

“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

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