Torque-free motion: Symmetric Top Euler Anglesdermisek/CM_14/CM-11-4p.pdf · Torque-free motion:...
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)
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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!
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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 )
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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
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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!
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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
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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
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For a symmetric object:
0 < < 2
+1 > > �1
property of the trace
< 20 <
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