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Torque-free motion: Symmetric Top based on FW-28 torque-free ω 3 = const. symmetric initial conditions: solution: The perpendicular projection of ω is constant! The magnitude of ω and the angle between ω and the symmetry axis is fixed. (the axis of rotation precesses about the symmetry axis with angular velocity Ω) 148 Torque-free motion: Asymmetric Top based on FW-28 In general we have to solve all three Euler’s equations: we can also use conservation of kinetic energy: and conservation of angular momentum squared: L is constant in the inertial frame, L is constant in any frame! 2 For rotation axis near one of the principal axis the problem simplifies: simple harmonic motion with frequency: An asymmetric body can rotate stably about two of its principal axis, those with the largest and smallest moments of inertia! 149 Euler Angles based on FW-29 We need three angles to specify the orientation of a rigid body (or relate the inertial frame to a body fixed frame ): 2. Rotate about the line of nodes through an angle β till you bring to its final position. 3. Rotate about the new through an angle γ till you bring to its final position. 1. Starting initially with the frame coinciding with the inertial frame, rotate about through an angle α till you bring to the line of nodes (perpendicular to both and ) 150 In order to construct the lagrangian, we want to write the kinetic energy: in terms of Euler angles: angular velocity along the instantaneous body-fixed principal axes although the vectors corresponding to Euler angles are not orthogonal, still, the total angular velocity can be written as a vector sum: thus, to get the kinetic energy we have to express the unit vectors corresponding to Euler angles in terms of 151

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### Transcript of Torque-free motion: Symmetric Top Euler Anglesdermisek/CM_14/CM-11-4p.pdf · Torque-free motion:...

Torque-free motion: Symmetric Topbased on FW-28

torque-free

�3 = const.

symmetric

initial conditions:

solution: The perpendicular projection of ω is constant!The magnitude of ω and the angle between ω

and the symmetry axis is fixed. (the axis of rotation precesses about the symmetry axis with angular velocity Ω)

148

Torque-free motion: Asymmetric Topbased on FW-28

In general we have to solve all three Euler’s equations:

we can also use conservation of kinetic energy:

and conservation of angular momentum squared:

L is constant in the inertial frame, L is constant in any frame! 2

For rotation axis near one of the principal axis the problem simplifies:

simple harmonic motion with frequency:

An asymmetric body can rotate stably about two of its principal axis, those with the largest and smallest moments of inertia!

149

Euler Anglesbased on FW-29

We need three angles to specify the orientation of a rigid body (or relate the inertial frame to a body fixed frame ):

2. Rotate about the line of nodes through an angle β till you bring to its final position.

3. Rotate about the new through an angle γ till you bring to its final position.

1. Starting initially with the frame coinciding with the inertial frame, rotate about through an angle α till you bring to the line of nodes (perpendicular to both and )

150

In order to construct the lagrangian, we want to write the kinetic energy:

in terms of Euler angles:

angular velocity along the instantaneous body-fixed principal axes

although the vectors corresponding to Euler angles are not orthogonal,

still, the total angular velocity can be written as a vector sum:

thus, to get the kinetic energy we have to express the unit vectors corresponding to Euler angles in terms of

151

Translation of unit vectors:

we can plug these to the formula for kinetic energy:

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Symmetric Top: torque-free motionbased on FW-30

the motion of an object in free space or gravitational field can be understood as the motion of the center of mass and a torque-free motion about the center of mass gravity exerts no torque about the CM!

Lagrangian describing an internal motion of a symmetric top:

Equations of motion:

, β, γ Euler angles are generalized coordinatesα and γ are cyclic!

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Equations of motion: α and γ are cyclic!

projection of ω or L along the symmetry axis remains constant

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By straightforward but lengthy calculation using

it can be shown that generalized momenta are projections of the L:

line of nodes

are constant!

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Description of motion in inertial frame:we choose space-fixed (inertial frame) such that L points in 3rd direction, and we get:

the only momentum which was not constant is now identically zero! β is constant

from Eqs. of motion we find:

ω, L and are coplanar!156

ω, L and are coplanar!

157

Focusing on case, e.g. the Earth:

Warning! This animation is for the orbital precession. But the same applies for almost

daily precession of the symmetry axis about the space-fixed L, with much smaller angle!

constant precession of the symmetry axis about L at fixed polar angle β

constant rotation of the object about the symmetry axisit is negative - inertial observer sees a backward motion about (body fixed observer sees a positive precession of ω about the symmetry axis)

For the Earth:

perpendicular projection of ω:

L is between ω and

158

Focusing on case, e.g. the football:

constant precession of the symmetry axis about L at fixed polar angle β

constant rotation of the object about the symmetry axis

it is positive - inertial observer sees a forward motion about

perpendicular projection of ω:

ω is between L and

For (e.g. spinning football):I3 ⌧ I1

159

For a symmetric object:

0 < < 2

+1 > > �1

property of the trace

< 20 <

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