sect 5.6: torque-free motion of a rigid body euler’s eqtns of motion for a rigid body with 1 pt....

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Page 1: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)
Page 2: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

Sect 5.6: Torque-Free Motion of a Rigid Body

• Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed:

I1(dω1/dt) - ω2ω3(I2 -I3) = N1 (1)

I2(dω2/dt) - ω3ω1(I3 -I1) = N2 (2)

I3(dω3/dt) - ω1ω2(I1 -I2) = N3 (3)

• Interesting & useful special case: Motion of a rigid body not subject to any net forces or torques:– Object is at rest or moving uniformly (Newton’s 1st Law!)

– Rotational motion started by some means, then force or torque is removed.

– Can discuss in reference from where CM is at rest.

Angular momentum comes from rotation about CM. Euler’s eqtns, (1), (2), (3), apply with right sides = 0!

Page 3: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• Force & torque free conditions Euler’s Eqtns:

I1(dω1/dt) = ω2ω3(I2 -I3) (1)

I2(dω2/dt) = ω3ω1(I3 -I1) (2)

I3(dω3/dt) = ω1ω2(I1 -I2) (3)– Note, assumed that (1), (2), (3) are written in principal axes system. So

are most subsequent eqtns!

– Also describe motion of rigid body with one pt. fixed & no forces, torques.

From earlier conservation theorems can immediately know that 2 integrals or constants of the motion are:

Total KE: T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2 = const.

Total ang. momentum: L = I1ω1 + I2ω2 + I3ω3 = const.– Using these, can completely integrate (solve!) (1), (2), (3) in terms of elliptic integrals.

“Not very illuminating!”

Page 4: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• Euler’s Eqtns under force, torque free conditions:

I1(dω1/dt) = ω2ω3(I2 -I3) (1)

I2(dω2/dt) = ω3ω1(I3 -I1) (2)

I3(dω3/dt) = ω1ω2(I1 -I2) (3)

T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω 3)2 (½)Iω2 = const.

L = I1ω1 + I2ω2 + I3ω3 Iω = const.

• Alternative to a “brute force” solution to (1),(2),(3): A geometric (& perhaps

more physical?) description of motion

Poinsot’s Construction. • Coord system with principle axes. n unit vector along rotation axis. n ω/|

ω|. Consider the vector ρ n/(I)½ from before. I nIn = moment of inertia. T (½)Iω2 = const.

ρ = ρ(t) = ω/|ω(I)½| = ω/(2T)½ (time dependent!)

Define: F(ρ) ρIρ = (ρ1)2I1+(ρ2)2I2+(ρ3)2I3

Page 5: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

Poinsot’s Construction • F(ρ) ρIρ = (ρ1)2I1+(ρ2)2I2+(ρ3)2I3 (Note: ρ = ρ(t)!)

ρ = ω/(2T)½ . Surfaces of const F(ρ) ellipsoids in “ρ” space. F(ρ) = 1 Inertia Ellipsoid from before.

• Body rotates. Directions of rotation axis n & of ρ change with time. The tip of ρ always defines a point on the inertia ellipsoid. The gradient of F(ρ) evaluated at that point defines the direction of the normal to the inertia ellipsoid at that point.

• Take the gradient of F(ρ) & use the definitions:

ρF(ρ) = 2Iρ = 2(Iω)/(2T)½

Note: L Iω = const.

ρF(ρ) = (2/T)½ L

Defines the direction to the inertia ellipsoid

Page 6: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• F(ρ) ρIρ = (ρ1)2I1+(ρ2)2I2+(ρ3)2I3

ρF(ρ) = (2/T)½ L (parallel to the L vector!) (A)

By the definition of , (A) defines a direction to the inertia ellipsoid. Also, since T & L are constant, so is ρF(ρ)!

PHYSICS: The angular

velocity vector ω always

moves such that the

corresponding to the

inertia ellipsoid is in the

direction of the angular

momentum L. Here, we have torque free motion

L =const. (in direction & magnitude). The inertia ellipsoid,

fixed in the body, must move in space to preserve (A). See figure.

ρ = ρ(t) !

Page 7: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• Also: The distance between the ellipsoid origin & the plane tangent to it at the point ρ is constant in time. See figure:

Geometry That

distance = The projection of

ρ on L or: h = (ρL)/L

But: ρ = ω(2T)-½ (= ρ(t)!)

So: h = [(ωL)(2T)½]/(L)Note: (ωL) = ωIω = 2T

h = (2T)½/(L) = const.

(since T = const. & L = const.)

Page 8: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• The distance between ellipsoid origin & the plane tangent to it at ρ:

h = (2T)½/L = const. • PHYSICS: The tangent plane is always

a fixed distance h from origin of the inertia

ellipsoid. Normal to that plane (|| L) has a

fixed direction. The tangent plane the invariable plane

Visualize the force & torque free motion of a rigid body in

“ρ” space as the inertia ellipsoid rolling without slipping on the

invariable plane. Ellipsoid center a const height h above plane. Rolls

without slipping because the point of contact is defined by the position

of the tip of ρ. Curve traced out on ellipsoid by the point of contact of

the ellipsoid with plane polhode. Curve traced out on the plane by the

point of contact of the ellipsoid with the plane herpolhode. (“The polhode

rolls without slipping on the herpolhode lying in the invariable plane!”).

Page 9: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• Poinsot’s construction completely describes the force & torque free motion of a rigid body.

• The direction of the invariable plane & the height h of origin of the inertia ellipsoid are completely determined by the (constant) values of T & L. – T & L are determined by the initial conditions of problem!

• Then, it’s just geometry to trace out the polhode & the herpolhode. – The direction of the (time dependent) angular velocity ω is the same

as the direction of ρ (= ρ(t) ) – The instantaneous orientation of the body is the same as the

orientation of the inertia ellipsoid, fixed in the body!– Is it necessary to state that the shape of the inertia ellipsoid has nothing whatsoever

to do with the shape of the body?

Page 10: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

Torque-Free Motion of a Symmetrical Body (Say, symmetric about one or more axes)

• Special Case: Torque Free motion of a symmetrical body.

Inertia Ellipsoid = Ellipsoid of Revolution The polhode on the ellipsoid is a circle about symmetry axis.

The herpolhode is a circle on the invariable plane.• An observer, fixed in the body axis system, sees the angular velocity

vector ω move on the surface of a cone ( the body cone). Intersection of this cone with inertia ellipsoid polhode

• An observer, fixed in the space axis system, sees ω move on the surface of a cone ( the space cone). Intersection of this cone with invariable plane herpolhode

The motion can also be described as the rolling of the body cone on the space cone.

Page 11: Sect 5.6: Torque-Free Motion of a Rigid Body Euler’s Eqtns of Motion for a Rigid Body with 1 pt. fixed: I 1 (dω 1 /dt) - ω 2 ω 3 (I 2 -I 3 ) = N 1 (1)

• Let rotation the axis be axis 3 in the principal axis system.

– If I3 < I1, I2 the inertia ellipsoid

is prolate (~ football shaped)

as in the figure: The body

cone is outside the space cone.

– If I3 > I1, I2 the inertia ellipsoid is oblate (shaped like football on its side). In this case, the body cone rolls around inside the space cone.

– In both cases, the direction of the angular velocity vector ω precesses in time about the symmetry axis of the body.